Selected Solutions to Artin’s Algebra, Second Ed. Takumi Murayama July 22, 2014 These solutions are the result of taking MAT323 Algebra in the Spring of 2012, and also TA-ing for MAT346 Algebra II in the Spring of 2014, both at Princeton University. This is not a complete set of solutions; see the List of Solved Exercises at the end. Please e-mail [email protected]with any corrections. Contents 2 Groups 4 2.1 Laws of Composition .......................... 4 2.2 Groups and Subgroups ......................... 4 2.4 Cyclic Groups .............................. 5 2.6 Isomorphisms ............................... 7 2.8 Cosets ................................... 8 2.10 The Correspondence Theorem ..................... 9 2.11 Product Groups ............................. 9 2.12 Quotient Groups ............................. 10 6 Symmetry 13 6.3 Isometries of the Plane ......................... 13 6.4 Finite Groups of Orthogonal Operators on the Plane ......... 14 6.5 Discrete Groups of Isometries ...................... 16 6.6 Plane Crystallographic Groups ..................... 17 6.7 Abstract Symmetry: Group Operations ................ 17 6.8 The Operation on Cosets ........................ 18 6.9 The Counting Formula ......................... 19 6.10 Operations on Subsets .......................... 19 6.11 Permutation Representations ...................... 20 1
Transcript
Selected Solutions to Artin’s Algebra, Second Ed.
Takumi Murayama
July 22, 2014
These solutions are the result of taking MAT323 Algebra in the Spring of 2012,and also TA-ing for MAT346 Algebra II in the Spring of 2014, both at PrincetonUniversity. This is not a complete set of solutions; see the List of Solved Exercisesat the end. Please e-mail [email protected] with any corrections.
Exercise 2.1.2. Prove the properties of inverses that are listed near the end of thesection.
Remark. The properties are listed on p. 40 as the following:(a) If an element a has both a left inverse l and a right inverse r, i.e., if la = 1
and ar = 1, then l = r, a is invertible, r is its inverse.(b) If a is invertible, its inverse is unique.(c) Inverses multiply in the opposite order: If a and b are invertible, so is the
product ab, and (ab)−1 = b−1a−1.(d) An element a may have a left inverse or a right inverse, though it is not
invertible.
Proof of (a). We see l = lar = r.
Proof of (b). Let b, b′ be inverses of a. Then, b = bab′ = b′, by (a).
Proof of (c). Consider ab. We see that b−1a−1 is the inverse of ab since (b−1a−1)(ab) =b−1a−1ab = b−1b = 1 by associativity. Uniqueness follows by (b).
Proof of (d). Consider Exercise 2.1.3 below. s is not invertible since it does not havea two-sided inverse, but it does have a left inverse.
Exercise 2.1.3. Let N denote the set {1, 2, 3, . . .} of natural numbers, and let s : N →N be the shift map, defined by s(n) = n + 1. Prove that s has no right inverse, butthat it has infinitely many left inverses.
Proof. s does not have a right inverse since s does not map any element of N backto 1; however, we can define a left inverse rk(n) = n− 1 for n > 1, and rk(1) = k forsome k ∈ N; we see that this is a left inverse of s, i.e., that rk ◦ s = idN. Since k isarbitrary this implies that there is an infinite number of rk’s.
2.2 Groups and Subgroups
Exercise 2.2.3. Let x, y, z, and w be elements of a group G.(a) Solve for y, given that xyz−1w = 1.(b) Suppose that xyz = 1. Does it follow that yzx = 1? Does it follow that
yxz = 1?
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Solution for (a). We claim that y = x−1w−1z. This follows since
x(x−1w−1z)z−1w = xx−1w−1zz−1w = 1.
Solution for (b). Suppose xyz = 1. This implies x−1 = yz, and by Exercise 2.1.2(a),this left inverse is a right inverse, and so 1 = xyz = x(yz) = (yz)x = yzx.
Now consider yxz; the example
x =
[1 10 1
], y =
[0 11 0
], z =
[0 11 −1
], xyz =
[1 00 1
], yxz =
[1 −11 0
]in GL2(R) shows that xyz = 1 does not imply yxz = 1.
Exercise 2.2.4. In which of the following cases is H a subgroup of G?(a) G = GLn(C) and H = GLn(R).(b) G = R× and H = {1,−1}.(c) G = Z+ and H is the set of positive integers.(d) G = R× and H is the set of positive reals.
(e) G = GL2(R) and H is the set of matrices
[a 00 0
], with a 6= 0.
Solution for (a). H is a subset since R ⊂ C implies H ⊂ GLn(R). H is a subgroupsince GLn(R) is a group, hence contains an identity and is closed under multiplicationand inversion.
Solution for (b). H is a subgroup since it is clearly a subset, contains the identity 1,and −1×−1 = 1 implies H is closed under multiplication and inversion.
Solution for (c). H is not a subgroup since −1 /∈ H, though it is the inverse of 1.
Solution for (d). H is a subgroup since it is clearly a subset, contains 1, is closedunder multiplication since the product of two positive real numbers is a positive realnumber, and since x ∈ H has inverse 1/x ∈ H, which is still positive and real.
Solution for (e). H is not a subgroup since it is not even a subset of G.
2.4 Cyclic Groups
Exercise 2.4.1. Let a and b be elements of a group G. Assume that a has order 7and that a3b = ba3. Prove that ab = ba.
Exercise 2.4.3. Let a and b be elements of a group G. Prove that ab and ba havethe same order.
Proof. Suppose (ab)n = 1. We note that b = b(ab)n = (ba)nb, but this implies(ba)n = 1, and so both have order n.
Exercise 2.4.6.(a) Let G be a cyclic group of order 6. How many of its elements generate G?
Answer the same question for cyclic groups of order 5 and 8.(b) Describe the number of elements that generate a cyclic group of arbitrary orders
n.
Solution for (b). By Prop. 2.4.3, if x generates G a cyclic group of order n, anotherelement xi ∈ G generates G if and only if gcd(i, n) = 1 for 1 ≤ i ≤ n, since then|xi| = n. Thus, the number of elements that generate G is equal to the number ofnumbers less than n that are coprime to n.
Solution for (a). By (b), it suffices to count the number of numbers less than n thatare coprime to n. For 6, {1, 5} are coprime to 6, hence two elements generate thecyclic group of order 6. For 5, {1, 2, 3, 4} are coprime to 5, hence four elementsgenerate the cyclic group of order 5. For 8, {1, 3, 5, 7} are coprime to 8, hence fourelements generate the cyclic group of order 8.
Exercise 2.4.9. How many elements of order 2 does the symmetric group S4 con-tain?
Solution. The order 2 elements of S4 consist of(42
)= 6 two-cycles, and
(42
)× 1
2= 3
products of disjoint two-cycles, and so there are 9 elements of order 2.
Exercise 2.4.10. Show by example that the product of elements of finite order in agroup need not have finite order. What if the group is abelian?
Solution. Consider GL2(R), and the following matrices in GL2(R):
A =
[−1 10 1
], B =
[−1 00 1
].
We see that A2 = B2 = 1, and so they are of order 2, whereas
AB =
[1 10 1
]=⇒ (AB)n =
[1 n0 1
],
and so AB has infinite order.Now suppose the group is abelian. Suppose a, b are our elements of finite order,
of order n,m respectively. Then, (ab)nm = anmbnm = (an)m(bm)n = 1, and so ab isnecessarily of finite order.
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2.6 Isomorphisms
Exercise 2.6.2. Describe all homomorphisms ϕ : Z+ → Z+. Determine which areinjective, which are surjective, and which are isomorphisms.
Solution. By the definition of homomorphism, for all positive n ∈ Z+, we have
ϕ(n) = ϕ(1) + · · ·+ ϕ(1)︸ ︷︷ ︸n times
, ϕ(−n) = −ϕ(n), ϕ(0) = ϕ(n) + ϕ(−n) = 0.
Thus, ϕ is fully determined by what 1 maps to. By the above, we then have that ϕn :z nz for n ∈ Z+ are all the homomorphisms of Z+. The injective homomorphismsconsist of those ϕn for n 6= 0. The surjective homomorphisms consist of those ϕn forn = ±1; these are also the isomorphisms of Z+ since they are injective.
Exercise 2.6.3. Show that the functions f = 1/x, g = (x−1)/x generate a group offunctions, the law of composition being composition of functions, that is isomorphicto the symmetric group S3.
Solution. We define
f1 = x, f2 =1
x, f3 = 1− x, f4 =
1
1− x, f5 =
x
x− 1, f6 =
x− 1
x.
Then, we can construct the multiplication table:
f1 f2 f3 f4 f5 f6f1 f1 f2 f3 f4 f5 f6f2 f2 f1 f4 f3 f6 f5f3 f3 f6 f1 f5 f4 f2f4 f4 f5 f2 f6 f3 f1f5 f5 f4 f6 f2 f1 f3f6 f6 f3 f5 f1 f2 f4
This proves closure since every combination of factors is accounted for, identity sinceevery row/column contains e = f1, and associativity since associativity holds forcomposition of rational functions. We claim that this is isomorphic to S3. Thisfollows since if we let f1 e, f2 (12), f6 (123), we get the following table:
e (12) (13) (132) (23) (123)e e (12) (13) (132) (23) (123)
(12) (12) e (132) (13) (123) (23)(13) (13) (123) e (23) (132) (12)(132) (132) (23) (12) (123) (13) e(23) (23) (132) (123) (12) e (13)(123) (123) (13) (23) e (12) (132)
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This proves it is a homomorphism since all of the multiplications are accurate, andis an isomorphism since every element in S3 is mapped to, with inverse defined bymatching entries. This shows f2, f6 generate the group of functions since (12), (123)generate S3 as on p. 42.
Exercise 2.6.6. Are the matrices
[1 1
1
],
[11 1
]conjugate elements of the group
GL2(R)? Are they conjugate elements of SL2(R)?
Solution. We explicitly calculate the conjugation for the conjugation matrix A:[a11 a12a21 a22
] [1 10 1
]=
[1 01 1
] [a11 a12a21 a22
][a11 a11 + a12a21 a21 + a22
]=
[a11 a12
a11 + a21 a12 + a22
]This equality requires a11 = 0, a12 = a21; however, we see that then detA < 0 in thiscase, so the matrices are not conjugate in SL2(R).
We see that they are, however, conjugate elements of the group GL2(R), since[0 11 0
] [1 10 1
]=
[1 01 1
] [0 11 0
].
2.8 Cosets
Exercise 2.8.4. Does a group of order 35 contain an element of order 5? of order7?
Solution. Any element in G has order in {1, 5, 7, 35} by Cor. 2.8.10. Suppose Ghad no elements of order 5; then, all non-identity elements must have order 7, for if|x| = 35, then |x7| = 5. Let h have order 7, and H = 〈h〉; since |H| = 7, pick g /∈ H.Then, g 6= e and has order 7. The left cosets H, gH, g2H, . . . , g6H must be disjoint,for, if gahi = gbhj, then ga−b = hi−j, and so picking r such that r(a− b) ≡ 1 mod 7,we have that g = gr(a−b) = hr(i−j) ∈ H, a contradiction. But this contradicts thecounting formula (2.8.8), since |G| = 35 6= 49 = 7 ·7 = |H|[G : H], and so G containsan element of order 5.
Now suppose G had no elements of order 7; then all non-identity elements haveorder 5 as before. Letting h have order 5 and H, g as before, the same argument givesthat H, gH, g2H, . . . , g4H are disjoint left cosets in G. This contradicts the countingformula (2.8.8) again, since |G| = 35 6= 25 = 5 · 5 = |H|[G : H], hence G contains anelement of order 7.
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Exercise 2.8.8. Let G be a group of order 25. Prove that G has at least one subgroupof order 5, and that if it contains only one subgroup of order 5, then it is a cyclicgroup.
Proof. Any element in G has order in {1, 5, 25} by Cor. 2.8.10. If G had no elementsof order 5, it must have an element x of order 25, but then |x5| = 5, hence 〈x5〉 is asubgroup of order 5.
Now suppose H is the only subgroup of order 5 in G, and pick x /∈ H. x hasorder 5 or 25 by Cor. 2.8.10 again, and since e ∈ H. But |x| = 5 implies H = 〈x〉,hence x ∈ H, and so we know |x| = 25, i.e., G = 〈x〉 is cyclic.
Exercise 2.8.10. Prove that every subgroup of index 2 is a normal subgroup, andshow by example that a subgroup of index 3 need not be normal.
Proof. If H 6 G with [G : H] = 2, we see that, picking a /∈ H, G = H q aH =H qHa, since this is the only we can form cosets of a in G. Thus, aH = Ha, henceH CG by Prop. 2.8.17.
Now we consider S3; the 2-cycle y = (12) generates a subgroup H of order 2, andtherefore of index 3 by the counting formula (2.8.8). However, by the multiplicationtable constructed last week, we see that, from (2.8.4) and (2.8.16),
xH = {x, xy} 6= {x, x2y} = Hx.
2.10 The Correspondence Theorem
Exercise 2.10.3. Let G and G′ be cyclic groups of orders 12 and 6, generated byelements x and y, respectively, and let ϕ : G→ G′ be the map defined by ϕ(xi) = yi.Exhibit the correspondence referred to in the Correspondence Theorem explicitly.
Solution. We note that K = kerϕ = {e, x6}. The subgroups of G that contain theseare 〈x〉, 〈x2〉, 〈x3〉, 〈x6〉, and each correspond to 〈y〉, 〈y2〉, 〈y3〉, e respectively, whichare all the subgroups of G′.
2.11 Product Groups
Exercise 2.11.1. Let x be an element of order r of a group G, and let y be anelement of G′ of order s. What is the order of (x, y) in the product group G×G′?
Solution. The order is lcm(r, s), since (x, y)n = (xn, yn) = (e, e) implies r, s | n.
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Exercise 2.11.3. Prove that the product of two infinite cyclic groups is not infinitecyclic.
Proof. Recall that a cyclic group must be generated by a single element; since allinfinite cyclic groups are isomorphic to Z, we can consider Z×Z. Suppose that (a, b)is this single element. But then, we see that (2a, b) cannot be obtained from adding(a, b) to itself, which implies that Z× Z is not infinite cyclic.
2.12 Quotient Groups
Exercise 2.12.2. In the general linear group GL3(R), consider the subsets
H =
1 ∗ ∗0 1 ∗0 0 1
, and K =
1 0 ∗0 1 00 0 1
,where ∗ represents an arbitrary real number. Show that H is a subgroup of GL3, thatK is a normal subgroup of H, and identify the quotient group H/K. Determine thecenter of H.
Proof. H 6 GL3 since clearly I ∈ H,1 a11 a210 1 a220 0 1
1 b11 b210 1 b220 0 1
=
1 a11 + b11 a21 + b21 + a11b220 1 a22 + b220 0 1
∈ H,
and since 1 a11 a210 1 a220 0 1
1 −a11 −a21 + a11a220 1 −a220 0 1
= I.
We now show that K is a normal subgroup of H. Define ka as the matrix in Kthat has the parameter a; we see k0 = I ∈ K, kakb = ka+b, kak−a = I, and so itonly remains to show normality by showing hk = kh for h ∈ H, k ∈ K, which isequivalent to hkh−1 = k:1 0 a
0 1 00 0 1
1 b11 b210 1 b220 0 1
=
1 b11 a+ b210 1 b220 0 1
=
1 b11 b210 1 b220 0 1
1 0 a0 1 00 0 1
.By the additive nature of the ka relations, we see that K ≈ R+.
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The quotient group H/K is then represented by matrices of the form1 b 00 1 c0 0 1
,i.e., the cosets are these times K, since multiplying by ka only alters the top-rightentry, and so multiplying by ka will keep b, c constant, and therefore remain in thesame coset. Moreover, this generates the entire space since we can multiply arbitraryka to a coset with arbitrary parameters b, c.
The center of H contains K, since K commutes with elements of H as shownabove; we claim the center is solely K. If A,B ∈ H, then by the above AB = BA is1 a11 + b11 a21 + b21 + a11b22
0 1 a22 + b220 0 1
=
1 a11 + b11 a21 + b21 + a22b110 1 a22 + b220 0 1
.This implies a11b22 = a22b11. But if A is fixed and B is an arbitrary matrix in H, thena11b22 = a22b11 must hold for all matrices B, and so a11 = a22 = 0, hence A ∈ K.Thus, the center of H is K ≈ R+.
Exercise 2.12.4. Let H = {±1,±i} be the subgroup of G = C× of fourth roots ofunity. Describe the cosets of H in G explicitly. Is G/H isomorphic to G?
Solution. By (2.8.5), cosets aH, bH are equal if and only if b = ah for some h ∈ H,and so if a = re2πiθ, b = se2πiη for r, s ∈ R>0, θ, η ∈ [0, 1), then aH = bH if andonly if r = s and θ− η ∈ {0, 1/4, 1/2, 3/4}. Hence the cosets of H are {re2πθH | r ∈R>0, θ ∈ [0, 1/4)}.
Now consider the map ϕ : G → G, x x4. Then, this is trivially a homomor-phism of G; it is moreover surjective since any nonzero complex number has a fourthroot. We see that kerϕ = H, and so G/H ≈ G by the isomorphism theorem.
Exercise 2.12.5. Let G be the group of upper triangular real matrices
[a b0 d
], with
a and d different from zero. For each of the following subsets, determine whether ornot S is a subgroup, and whether or not S is a normal subgroup. If S is a normalsubgroup, identify the quotient group G/S.(i) S is the subset defined by b = 0.(ii) S is the subset defined by d = 1.(iii) S is the subset defined by a = d.
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Solution for (i). S 6 G since the diagonal entries would multiply with each otherand not affect the other entries of the matrices, I ∈ S, and the inverse can be foundby letting the inverse have entries a−1, d−1. S is not a normal subgroup since[
1 10 1
] [a 00 d
]=
[a d0 d
]6=[a a0 d
]=
[a 00 d
] [1 10 1
].
Solution for (ii). This forms a subgroup since[a b0 1
]−1
=
[1/a −b/a0 1
],
implies it is closed under inversion, and[a b0 1
] [a′ b′
0 1
]=
[aa′ ab′ + b0 1
],
implies it is closed under multiplication; the identity is trivially in S.We now see that it is a normal subgroup:[
a b0 d
] [a′ b′
0 1
] [1/a −b/ad0 1/d
]=
[a′ (b+ ab′ − a′b)/d0 1
]∈ S.
We see that the quotient group G/S would be represented by matrices of the form[1 00 d
],
for d ∈ R× since this would cover all of G:[1 00 d
] [a b0 1
]=
[a b0 d
],
and since each matrix of the form above gives a different coset because multiplyingby elements in S keep d constant.
Solution for (iii). This is a subgroup since it satisfies closure:[a b0 a
] [c d0 c
]=
[ac bc+ ad0 ac
]∈ S,
inverse: [a b0 a
]−1
=
[1/a −b/a20 1/a
]∈ S,
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and clearly the identity is in S.We now check this subgroup is normal:[
a b0 d
] [a′ b′
0 a′
] [a b0 d
]−1
=
[a′ ab′/d0 a′
]∈ S.
The quotient group G/S would be represented by matrices of the form[1 00 d
],
for d ∈ R× since this would cover all of G:[1 00 d
] [a b0 a
]=
[a b0 ad
],
and since each matrix of the form above gives a different coset since multiplying byelements in S cannot give another matrix of the same form unless d is the same.
6 Symmetry
6.3 Isometries of the Plane
Exercise 6.3.2. Let m be an orientation-reversing isometry. Prove algebraicallythat m2 is a translation.
Proof. By Thm. 6.3.2 and (6.3.3), m2 = tvρθrtvρθr = tvρθrtvrρ−θ = tvρθtv′ρ−θ =tv+v′′ , where v
′′ = ρθ(r(v)).
Exercise 6.3.6.(a) Let s be the rotation of the plane with angle π/2 about about the point (1, 1)t.
Write the formula for s as a product taρθ.(b) Let s denote reflection of the plane about the vertical axis x = 1. Find an
isometry g such that grg−1 = s, and write s in the form taρθr.
Solution for (a). By (6.3.3),
s = t(1,1)ρπ/2t(−1,−1) = t(1,1)tρπ/2(−1,−1)ρπ/2 = t(2,0)ρπ/2.
Solution for (b). By (6.3.3), letting g = t1,0ρπ/2,
6.4 Finite Groups of Orthogonal Operators on the Plane
Exercise 6.4.2.(a) List all subgroups of the dihedral group D4, and decide which ones are normal.(b) List the proper normal subgroups N of the dihedral group D15, and identify the
quotient groups D15/N .(c) List the subgroups of D6 that do not contain x3.
Solution for (a). We claim the subgroups of D4 form the lattice diagram
{e}
〈y〉 〈x2y〉 〈x2〉 〈x3y〉 〈xy〉
〈x2, y〉 〈x〉 〈x2, xy〉
D4
By Lagrange’s theorem (Thm. 2.8.9), any subgroup H 6 D4 must have |H| ∈{1, 2, 4, 8}. {e} is the unique order 1 subgroup. The order 2 subgroups are gen-erated by one order 2 element, hence are given by the second row above. The order4 subgroups are either generated by an order 4 element or by two order 2 elements;these give the third row. Finally, D4 is the unique order 8 subgroup.
We claim the normal subgroups of D4 are
{e}, 〈x〉, 〈x2〉, 〈x2, y〉, 〈x2, xy〉, D4.
〈x2〉 is normal since x2 commutes with all other elements of D4; the other propersubgroups are normal because they are of index 2 by the counting formula (2.8.8),and by Exercise 2.8.10. Lastly, the other four subgroups in the second row of thediagram above are not normal since they are not closed under conjugation by x.
Solution for (b). Any proper subgroup H 6 D15 must have |H| ∈ {2, 3, 5, 15} byLagrange’s theorem (Thm. 2.8.9). The order 2 subgroups are of the form 〈xiy〉; theseare not normal since they are not closed under conjugation by x. Since 2 - 3, 5, 15,we know no other subgroup contains an element of the form xiy. Thus, every normalsubgroup of D15 must be of the form 〈xi〉 for i = 1, 3, 5, since any other i gives asubgroup equal to one of these three. These are all in fact normal subgroups sincethey are kernels of homomorphisms D15 → Di for i = 1, 3, 5 mapping x x, y y.
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By the first isomorphism theorem (Thm. 2.12.10), this implies the quotient groupsfor each nontrivial N are isomorphic to
D15/〈x〉 ≈ D1, D15/〈x3〉 ≈ D3, D15/〈x5〉 ≈ D5.
Solution for (c). By Lagrange’s theorem (Thm. 2.8.9), any subgroup H 6 D6 musthave |H| ∈ {1, 2, 3, 4, 6, 12}. {e} is the unique order 1 subgroup, and does not containx3. The order 2 subgroups are 〈x3〉 and 〈xiy〉 for 0 ≤ i ≤ 5; only the first containsx3. The unique order 3 subgroup is 〈x2〉, which does not contain x3. The order 4subgroups not containing x3 are of the form 〈xiy, xjy〉 for i 6= j since D6 contains noelements of order 4, but this subgroup is of order 4 if and only if xj−i = x3 = xi−j,hence there are no order 4 subgroups not containing x3. The order 6 subgroups mustbe generated by an order 2 and an order 3 element, hence those not containing x3
are of the form 〈x2, xiy〉. But different choices i, j for i give the same subgroup ifand only if i ≡ j mod 2, hence the only subgroups of order 6 not containing x3 are〈x2, y〉 and 〈x2, xy〉. The unique order 12 subgroup D16 contains x3. In summary,the subgroups that do not contain x3 are
Exercise 6.4.3.(a) Compute the left cosets of the subgroup H = {1, x5} in the dihedral group D10.(b) Prove that H is normal and that D10/H is isomorphic to D5.(c) Is D10 isomorphic to D5 ×H?
Solution for (a). The cosets are represented by e, x, x2, x3, x4, y, xy, x2y, x3y, x4y.
Solution for (b). To show that H is normal, it suffices to show x5 commutes withevery element of D10: it trivially commutes with x, and with y since yx5y = yyx5 =x5. Thus H CD10.
Now the surjective homomorphism D10 → D5 sending x x, y y has kernelH, hence D10/H ≈ D5 by the first isomorphism theorem (Thm. 2.12.10).
Solution for (c). Note D5 ↪→ D10 by having x x2, hence to show D10 ≈ D5 ×H,it suffices by Prop. 2.11.4(d) to show H
⋂D5 = {e}, HD5 = G, and H,D5CG with
this embedding of D5. The normality of both groups follows by (b) and since D5 isof index 2 in G (Exercise 2.8.10). Their intersection is e since x5 /∈ D5. HD5 = Gsince x5 commutes with every element of D5 ⊂ D10 by (b), and so we can get anyelement of the form xi, xiy for 0 ≤ i ≤ 9.
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6.5 Discrete Groups of Isometries
Exercise 6.5.5. Prove that the group of symmetries of the frieze pattern C C C CCCC is isomorphic to the direct product C2 × C∞ of a cyclic group of order 2 andan infinite cyclic group.
Proof. Let G 6M be the group of symmetries. By Thm. 6.3.2, any symmetry arisesas tvρθ or tvρθr where tv are translations, ρθ are rotations, and r are reflections acrossthe x-axis. In both cases, θ must be zero to be a symmetry, and v must be an integermultiple of the length of C. Hence the group C2 of reflections across the x-axis andthe group C∞ of translations by integer multiples of lengths of C generate G.
We claim C2 ×C∞ ≈ G. Since clearly C2 ∩C∞ = {e}, by Prop. 2.11.4(d) it onlyremains to show C2, C∞ CG. C∞ CG since it has index 2 in G by the classificationof symmetries above, and then by Exercise 2.8.10. C2 CG since our translations alllie on the x-axis, and then by using (6.3.3).
Exercise 6.5.9. Let G be a discrete subgroup of M whose translation group is nottrivial. Prove that there is a point p0 in the plane that is not fixed by any element ofG except the identity.
Proof. We first claim that the set of points fixed by any nontrivial isometry m ∈ Ghas Lebesgue measure zero. We proceed by considering each kind of isometry inThm. 6.3.4. If m is a nontrivial translation, then there are no fixed points. If m isa nontrivial rotation around a point p, then p is a fixed point. If m is a nontrivialreflection about a line `, then any point on ` is a fixed point. Ifm is a nontrivial glide,then m has no fixed points. Thus, in all cases the set of fixed points has Lebesguemeasure zero in R2.
Now for each m 6= e, let Fm be the set of fixed points of m; by the above, ithas Lebesgue measure zero. We claim there are only countably many m ∈ G. ByThm. 6.3.2, anym is of the form tvρθ or tvρθ. By Thm. 6.5.5, there are only countablymany tv. By Prop. 6.5.10, there are only finitely many ρθ and ρθr. Thus, there areonly countably many m ∈ G.
Finally, the set of all points fixed by some nontrivial element of G has Lebesguemeasure
µ
( ⋃e6=m∈G
Fm
)≤
∑e6=m∈G
µ(Fm) = 0,
hence almost all points in R2 are not fixed by any nontrivial element of G.
16
6.6 Plane Crystallographic Groups
Exercise 6.6.2. Let G be the group of symmetries of an equilateral triangular latticeL. Determine the index in G of the subgroup of translations in G.
Proof. The index of translations is given by [G : T ]. By Theorem 6.3.2, we see thatevery isometry should be given by tvρθr; we therefore want the cardinality of the setof ρθr’s. But this is exactly the set D3, and so [G : T ] = |D3| = 6.
6.7 Abstract Symmetry: Group Operations
Exercise 6.7.1. Let G = D4 be the dihedral group of symmetries of the square.(a) What is the stabilizer of a vertex? of an edge?(b) G operates on the set of two elements consisting of the diagonal lines. What
is the stabilizer of a diagonal?
Solution for (a). If the vertex lies along the axis of reflection for y, {e, y} is thestabilizer. If it does not, {e, x2y} is the stabilizer. If the given edge lies immediatelyto the +θ direction of the line of reflection, the stabilizer is {e, xy}. If it does not,{e, x3y} is the stabilizer.
Solution for (b). The stabilizer of the diagonal is {e, x2, y, x2y}.
Exercise 6.7.2. The group M of isometries of the plane operates on the set of linesin the plane. Determine the stabilizer of a line.
Solution. It suffices to consider the classes of isometries in Thm. 6.3.4. The stabilizerof a line ` consists of translations along `, rotations of an angle π about a point p ∈ `,reflections about `, and glide reflections about `.
Exercise 6.7.8. Decompose the set C2×2 of 2×2 matrices into orbits for the followingoperations of GL2(C):(a) left multiplication,(b) conjugation.
Solution (a). Left multiplication by elements of GL2(C) corresponds to products ofelementary row operations on matrices in C2×2. By row reduction, we know thatapplying elementary row operations on a given matrix in C2×2 gives a unique matrixof one of the following forms:[
1 00 1
],
[1 a0 0
],
[0 10 0
],
[0 00 0
].
17
Thus, C2×2 decomposes as(GL2(C) ·
[1 00 1
])q
(∐a∈C
GL2(C) ·[1 a0 0
])q(GL2(C) ·
[0 10 0
])q{[
0 00 0
]}.
Solution (b). Since every matrix is conjugate (similar) to a unique Jordan form (upto ordering of Jordan blocks), we have that the orbits are generated by the differentJordan forms, i.e., C2×2 decomposes as(∐
λ∈C
GL2(C) ∗[λ 10 λ
])q
( ∐λ1≥λ2,λi∈C
GL2(C) ∗[λ1 00 λ2
]).
Exercise 6.7.11. Prove that the only subgroup of order 12 of the symmetric groupS4 is the alternating group A4.
Proof. We first prove a lemma: If H is a subgroup of G, then H is normal if andonly if H is the union of conjugacy classes in G. But H is normal if and only if it isclosed under conjugation if and only if H =
⋃h∈H G ∗ h.
Now suppose H 6 S4 has order 12. Then, by the counting formula (2.8.8),|S4| = 24 = 12 · [S4 : H] = |H|[S4 : H] implies [S4 : H] = 2, and so H C S4 byExercise 2.8.10. So, we use the classification of conjugacy classes in S4 from p. 201,following Prop. 7.5.1:
Partition Element No. in Conj. Class1 + 1 + 1 + 1 e 12 + 1 + 1 (ab)
(42
)= 6
2 + 2 (ab)(cd) 12
(42
)= 3
3 + 1 (abc) 2 ·(43
)= 8
4 (abcd) 3! = 6
Now if H C S4 has order 12, it must arise from a union of conjugacy classes aboveincluding {e} (since it is a subgroup), but the only way this is possible is the sum1+3+8 = 12. The conjugacy classes associated with these terms are exactly the evenpermutations in S4, which form the subgroup A4, and so A4 is the only subgroup oforder 12 in S4.
6.8 The Operation on Cosets
Exercise 6.8.4. Let H be the stabilizer of the index 1 for the operation of the sym-metric group G = Sn on the set of indices {1, . . . ,n}. Describe the left cosets of Hin G and the map (6.8.4) in this case.
18
Solution. We see that H consists of all cycles that hold 1 fixed, i.e., permutations ofthe remaining n− 1 elements, hence is isomorphic to Sn−1. We claim
G/H = {(1i)H | i ∈ {1, . . . ,n}}
Each (1i) gives a different coset since no (1i)H contains (1j) for i 6= j, and then by(2.8.5). These are all the cosets since by Pop. 6.8.4, |G/H| = |O1| = n. Thus, themap ε : G/H → O1 is the map (1i) i.
6.9 The Counting Formula
Exercise 6.9.4. Identify the group T ′ of all symmetries of a regular tetrahedron,including orientation-reversing symmetries.
Solution. T ′ 6 O3(R) operates transitively on the set F of faces of order 4, hence isgiven by a permutation of the set {f1, f2, f3, f4}, and so there is a homomorphismT ′ → S4 by Prop. 6.11.2. This is injective since if t ∈ T ′ fixes three faces, then it fixesthe three vectors defining the centers of each face, hence is the identity matrix inO3(R). We claim this homomorphism T ′ → S4 is surjective, hence an isomorphism;it suffices to show |T ′| = 24. But the stabilizer Gf of a given face f is the groupD3 generated by a rotation by 2π/3 about the center of f and a reflection about anaxis in f ; |D3| = 6, and so the counting formula (6.9.2) gives |T ′| = 6 · 4 = 24, henceT ′ ≈ S4.
6.10 Operations on Subsets
Exercise 6.10.1. Determine the orders of the orbits for left multiplication on theset of subsets of order 3 of D3.
Solution. We know that there are(63
)= 20 subsets of order 3 of D3; we also know
by the counting formula (6.9.2) that there can only be orbits of order 1, 2, 3, 6 sinceonly these divide |D3|. We see that only the subset {e, x, x2} is a subgroup, whichproduces the orbit {e, x, x2}, {y, yx, yx2}. The other 18 subsets form 3 orbits of order6 of D3. Letting H1 = {e, x, y}, we have
We therefore have one orbit of order 2 and three orbits of order 6; this means wehave 1×2+3×6 = 20 subsets in these orbits, and so we have found all of them.
6.11 Permutation Representations
Exercise 6.11.1. Describe all the ways in which S3 can operate on a set of fourelements.
Solution. By Cor. 6.11.3, it suffices to find all homomorphisms f : S3 → S4. Recallfrom p. 42 that S3 is generated by x = (123), y = (12); a homomorphism f is thenfully determined by specifying f(x), f(y). We moreover note that x3 = e, y2 = eimplies f(x)3 = f(y)2 = e, hence f(x) = e or is one of the eight 3-cycles in S4 by theclassification in Exercise 6.7.11; similarly, f(y) = e or is one of the nine elements oforder 2 in the table from Exercise 6.7.11.
If f(x) = e, then there is no restriction on f(y), thus there are ten homomor-phisms f of this kind.
If f(x) = (abc), then since yx has order 2, f(yx)2 = f(y)2f(x)2 = e. Thusf(y) 6= e. Now suppose f(y) acts nontrivially on the fourth element d unaffected byf(x); then without loss of generality, f(y) interchanges a, d, and so f(yx) · d = a.Hence f(yx) must map a d as well, and so f(y) · b = d. But then f(y)2 · b =a 6= b, contradicting that f(y) has order 2. Hence f(y) is a permutation of order2 of the subset {a, b, c}, which are 2-cycles; assume without loss of generality thatf(y) = (ab). In the argument above, there are 4 choices for the fixed point d of f(x),3 choices for the fixed point of f(y), and two choices for a = f(x) · c. Each choicegives a homomorphism f by just renaming a, b, c as 1, 2, 3, hence realizing f as thecanonical operation of S3 on the subset {1, 2, 3} ⊂ {1, 2, 3, 4}. Thus, there are 24homomorphisms f such that f(x) 6= e.
Exercise 6.11.5. A group G operates faithfully on a set S of five elements, and thereare two orbits, one of order 3 and one of order 2. What are the possible groups?Hint: Map G to a product of symmetric groups.
Solution. Let S = {1, 2, 3, 4, 5} such that the two orbits are O3 = {1, 2, 3}, O2 ={4, 5}, respectively. G operates on O3, O2 separately, hence the action of G onO3, O2 respectively correspond to group homomorphisms f3 : G → S3 and f2 : G →
20
S2. This defines a group homomorphism f = (f3, f2) : G → S3 × S2 defined byg (f3(g), f2(g)). Since G acts faithfully on S, g is injective, hence G ≈ f(G) byCor. 2.12.11. Now the only group that acts transitively on O2 is S2 itself, and sof2(G) ≈ S2. On the other hand, there are two groups that act transitively on O3:C3 and S3. Hence G ≈ C3 × S2 or S3 × S2.
Exercise 6.11.6. Let F = F3. There are four one-dimensional subspaces of the spaceof column vectors F 2. List them. Left multiplication by an invertible matrix permutesthese subspaces. Prove that this operation defines a homomorphism ϕ : GL2(F ) → S4.Determine the kernel and the image of this homomorphism.
Proof. We have the following one-dimensional subspaces, denoting F = {0, 1, 2}:
V1 =
{[10
],
[20
]}, V2 =
{[01
],
[02
]}, V3 =
{[11
],
[22
]}, V4 =
{[21
],
[12
]}.
Call ei the first listed vector in Vi; note that Vi = {ei, 2ei}. These are all the subspacesin F 2 since there are only 3 · 3− 1 = 8 nontrivial vectors in F 2.
Left multiplication by an invertible matrix defines an action on the vectors in F 2
since left multiplication by matrices in F is associative, and since the identity matrixdefines the trivial action; this descends to a well-defined action on the subspaces Visince the span of Aei is equal to the span of 2Aei for any A ∈ GL2(F ). Hence, byCor. 6.11.3, this action defines a homomorphism ϕ : GL2(F ) → S4.
Now kerϕ = {I, 2I} since ⊃ clearly holds, and for A ∈ GL2(F ), Ae1 ∈ V1 impliesthe first column of A is either e1 or 2e1, Ae2 ∈ V2 implies the second column of A iseither e2 or 2e2, and Ae3 ∈ V3 implies that the coefficients on ei in each column haveto be the same, hence A = I or 2I.
We claim imϕ = S4. By Cor. 2.8.13, we have |GL2(F )| = |kerϕ||imϕ|, and since|S4| = 24 and |kerϕ| = 2 from above, it suffices to show |GL2(F )| = 48. Now everymatrix in GL2(F ) consists of a pair of linearly independent vectors in F2; there are8 · 6 = 48 of these pairs by the above decomposition of F2 into subspaces, hence|GL2(F )| = 48.
6.12 Finite Subgroups of the Rotation Group
Exercise 6.12.3. Let O be the group of rotations of a cube, and let S be the set offour diagonal lines connecting opposite vertices. Determine the stabilizer of one ofthe diagonals.
21
Solution. O acts on S, hence there is a homomorphism O → S4 by Prop. 6.11.2. Wefirst show this homomorphism is injective. If s ∈ O fixes all four diagonals, thens either fixes or interchanges the two endpoints of each diagonal. If s 6= e, i.e., itacts nontrivially on the vertexes, then we can pick three pairs of opposite vertexessuch that s interchanges one of the pairs of vertexes, or such that s interchanges allthree pairs of vertexes. In either case, give R3 a basis that points along the threediagonals connecting these pairs of vertexes; then the matrix for s in this basis hasan odd number of −1’s on the diagonal, and so has determinant −1, contradictingthat O 6 SO3(R). The homomorphism O → S4 is surjective, since looking at thefaces of the cube, O acts transitively on the set of faces, and each face has stabilizerC4, hence |O| = |Of ||O · f | = 4 · 6 = 24 by the counting formula (6.9.2).
Now we know O acts like S4 on the set of diagonals, hence the stabilizer of oneof the diagonals is the same as fixing one index in the set of indexes {1,2,3,4} asin Exercise 6.8.4, hence is equal to S3.
Exercise 6.12.7. The 12 points (±1,±α, 0)t, (0,±1,±α)t, (±α, 0,±1)t form thevertices of a regular icosahedron if α > 1 is chosen suitably. Verify this, and deter-mine α.
Proof. We have that the 12 points form three categories of the form above. By thedistance formula, if they are in the same category, we have three possibilities forsquare distances:
Vertex Change sign of 1 Change sign of α Change sign of bothSquare Distance 4 4α2 4(α2 + 1)
If they are in different categories, then we have two possibilities for square distances.If our coordinates are (x1, x2, x3) and (y1, y2, y3), then there is only one i such thatxi, yi 6= 0, and xi, yi must have different absolute values for each i, giving the table
Vertex xi, yi have same sign xi, yi have differing signSquare Distance 2(α2 − α + 1) 2(α2 + α + 1)
where for a given vertexes, there are four vertexes each with square distance of eachform above.
We now find α. There must be five vertexes of shortest distance from v0; bycomparing the possible distances above, this shortest square distance must be equalto both 4 and 2(α2 − α + 1). This gives the equation
2(α2 − α + 1) = 4 =⇒ α2 − α− 1 = 0 =⇒ α =1 +
√5
2
22
by choosing the root greater than 1. This shows each vertex has exactly five neighborsof distance 2 away from it; we claim that the polyhedron formed by connectingvertexes of distance 2 away from each other forms an icosahedron.
Now we show each face formed by the edges is a congruent equilateral triangle.This is true since any face is formed by neighboring vertexes, which are distance 2away from each other by the above. Each vertex moreover has the same number offaces meeting there since every vertexes has exactly five neighbors by the above.
Finally, suppose we have two neighboring vertexes v, v′ forming an edge; we claimthere are only two faces intersecting at that edge. It suffices to show there are onlytwo vertexes w that are of distance 2 from both. Suppose vi = v′i = 0; in thefollowing, we consider subscripts mod 3. Then vi+1 = −v′i+1 with absolute value1, so wi+1 = 0, |wi+2| = 1, |wi| = 3 by the table above. Next vi+2 = v′i+2, hencewi+2 must have the same sign as vi+2, v
′i+2. Finally, wi can have either sign since
vi = v′i = 0, hence there are only two vertexes that are of distance 2 from v, v′.
6.M Miscellaneous Problems
Exercise 6.M.7. Let G be a finite group operating on a finite set S. For each elementg of G, let Sg denote the subset of elements of S fixed by g: Sg = {s ∈ S | gs = s},and let Gs be the stabilizer of s.(a) We may imagine a true-false table for the assertion that gs = s, say with rows
indexed by elements of G and columns indexed by elements of S. Constructsuch a table for the action of the dihedral group D3 on the vertices of a triangle.
(b) Prove the formula∑
s∈S|Gs| =∑
g∈G|Sg|.(c) Prove Burnside’s Formula: |G| · (number of orbits) =
∑g∈G|Sg|.
Solution for (a). We construct the true-false table:
where S = {s1, s2, s3} is the set of the vertices of the triangle, x = (s1s2s3) rotations,and y is the reflection with preferred vertex s1.
23
Proof of (b). By summing over different sets, we have∑s∈S
|Gs| = |{(g, s) ∈ G× S | gs = s}| =∑g∈G
|Sg|.
Proof of (c). By the orbit-stabilizer theorem (Prop. 6.8.4) we have that there is abijection between cosets G/Gs and orbits Os. By the counting formula (2.8.8), wethen have
|Gs| · |Os| = |Gs| · |G/Gs| = |G|.Note in particular that |Gs′ | is equal for all s′ ∈ Os since Os = Os′ . Thus,
|G| · (number of orbits) =∑
orbits Os
|G| =∑
orbits Os
|Gs| · |Os| =∑s∈S
|Gs|,
and so, combining (b),
|G| · (number of orbits) =∑g∈G
|Sg|.
11 Rings
11.1 Definition of a Ring
Exercise 11.1.1. Prove that 7 + 3√2 and
√3 +
√−5 are algebraic numbers.
Proof. 7 + 3√2 is a root of (x− 7)3 − 2 = x3 − 21x2 + 147x− 345 = 0.√
3 +√−5 is a root of (x2 + 2)2 + 60 = x4 + 4x2 + 64 = 0.
Exercise 11.1.2. Prove that, for n 6= 0, cos(2π/n) is an algebraic number.
Proof. Suppose n > 0. Recall that the Chebyshev polynomials Tn(x) are defined bythe recurrence relations T0(x) = 1, T1(x) = x, and Tn(x) = 2xTn−1(x) − Tn−2(x).We claim Tn(cos θ) = cosnθ for any θ. This is clear for n = 0, 1. For arbitrary n,
Tn(cos θ) = 2 cos θ Tn−1(cos θ)− Tn−2(cos θ)
= 2 cos θ cos ((n− 1)θ)− cos ((n− 2)θ)
= cosnθ + cos ((n− 2)θ)− cos ((n− 2)θ) = cosnθ,
and so Tn(cos θ) = cosnθ for any θ as desired. Letting θ = 2π/n, we have thatTn(cos(2π/n)) = cos 2π = 1, and so cos(2π/n) satisfies the polynomial equationTn(x)− 1 = 0.
If n < 0, then using the fact that cos(2π/n) = cos(−2π/n), we see cos(2π/n)satisfies the polynomial equation T−n(x)− 1 = 0 by the above.
24
Exercise 11.1.3. Let Q[α, β] denote the smallest subring of C containing the rationalnumbers Q and the elements α =
√2 and β =
√3. Let γ = α+β. Is Q[α, β] = Q[γ]?
Is Z[α, β] = Z[γ]?Solution. Since γ ∈ Z[α, β],Q[α, β], the reverse inclusion ⊃ holds in both cases. Weclaim that the inclusion ⊂ holds for Q, but not Z.
To show Q[α, β] ⊂ Q[γ], it suffices to show α, β ∈ Q[γ]. Now γ2 = 5+2αβ ∈ Q[γ],hence αβ ∈ Q[γ], so αβγ = 3α+ 2β ∈ Q[γ]. Thus α = αβγ − 2γ ∈ Q[γ], and finallyβ = γ − α ∈ Q[γ] as well.
To show Z[α, β] 6⊂ Z[γ], we claim αβ /∈ Z[γ]. It suffices to show that γk only haseven coefficients for αβ, for then, any element
∑Nk=0 akγ
k ∈ Q[γ] will have an evencoefficient for αβ, hence αβ cannot be expressed as such a sum.
To prove this, we claim
γk =
{aα + bβ for a, b odd when k odd
c+ dαβ for c odd, d even when k even
This is clear for k = 0, k = 1. For arbitrary n, consider first when k is even. Then,by inductive hypothesis
γk = γk−1γ = (aα + bβ)(α + β) = 2a+ 3b+ (a+ b)αβ,
for a, b odd, hence 2a+ 3b is odd and a+ b is even. Likewise, when k is odd,
for c odd, d even, hence 3d+ c, c+ 2d are odd, and we are done.
11.2 Polynomial Rings
Exercise 11.2.1. For which positive integers n does x2 + x + 1 divide x4 + 3x3 +x2 + 7x+ 5 in [Z/(n)][x]?Proof. We will perform the division algorithm on x4 + 3x3 + x2 + 7x+ 5. We have
x2 + 2x− 2
x2 + x+ 1)
x4 + 3x3 + x2 + 7x + 5
− x4 − x3 − x2
2x3 + 7x−2x2 − 2x2 − 2x
−2x2 + 5x + 52x2 + 2x + 2
7x + 7
25
and so x4 + 3x3 + x2 + 7x + 5 = (x2 + 2x − 2)(x2 + x + 1) + (7x + 7), where theremainder 7x + 7 ≡ 0 mod n if and only if n ∈ {1, 7}. Hence x2 + x + 1 dividesx4 + 3x3 + x2 + 7x+ 5 in [Z/(n)][x] if and only if n ∈ {1, 7}.
Exercise 11.2.2. Let F be a field. The set of all formal power series p(t) = a0 +a1t+a2t
2+ · · · , with ai in F , forms a ring that is often denoted by F [[t]]. By formalpower series we mean that the coefficients form an arbitrary sequence of elements ofF . There is no requirement of convergence. Prove that F [[t]] is a ring, and determinethe units in this ring.
Proof. We denote p(t) =∑
i aiti, q(t) =
∑i bit
i, r(t) =∑
i citi. Let + defined as
p(t) + q(t) =∑
k(ak + bk)tk and × as p(t)× q(t) =
∑k
∑i+j=k aibjt
k.+ makes F [[t]] an abelian group because associativity and commutativity follow
since + is defined termwise, and because having ai = 0 for all i defines an identityand letting q(t) such that bi = −ai for all i defines an inverse for p(t).
× is commutative since∑
i+j=k aibj =∑
i+j=k biaj, and is associative since
(p(t)× q(t))× r(t) =∑`
∑i+j=`
aibjt` × r(t) =
∑`
∑i+j+k=`
aibjckt`
= p(t)×∑`
∑j+k=`
bjckt`= p(t)× (q(t)× r(t)).
The identity is p(t) such that a0 = 1, ai = 0 for all i > 0.It remains to show the distributive property:
(p(t) + q(t))× r(t) =∑i
(ai + bi)ti ×∑j
cjtj =
∑k
∑i+j=k
(ai + bi)cjtk
=∑k
∑i+j=k
aicjtk +
∑k
∑i+j=k
bicjtk
= p(t)× r(t) + q(t)× r(t).
Finally, we claim that the p(t) such that a0 6= 0 are the units. Any unit musthave a0 6= 0, for p(t) × q(t) = 1 =⇒ a0b0 = 1. In the other direction, suppose p(t)is such that a0 6= 0. Define q(t) such that
Exercise 11.3.3. Find generators for the kernels of the following maps:(a) R[x, y] → R defined by f(x, y) f(0, 0),(b) R[x] → C defined by f(x) f(2 + i),(c) Z[x] → R defined by f(x) f(1 +
√2),
(d) Z[x] → C defined by x √2 +
√3,
(e) C[x, y, z] → C[t] defined by x t, y t2, z t3.
Remark. We will denote each map as ϕ.
Solution for (a). We claim that (x, y) = kerϕ. By the division algorithm, any poly-nomial f ∈ R[x, y] can be written g + a0 for g ∈ (x, y), and so ϕ(f) = a0 = 0 if andonly if a0 = 0 if and only if f = g ∈ (x, y).
Solution for (b). We claim that (x2−4x+5) = kerϕ. x2−4x+5 = (x− (2+ i))(x−(2 − i)), hence (x2 − 4x + 5) ⊂ kerϕ. Conversely, let f ∈ kerϕ. By the divisionalgorithm, we can write f = g + r for g ∈ (x2 − 4x + 5), where r = a1x + a0 forai ∈ R has degree less than 2. Then, ϕ(f) = ϕ(g) + ϕ(r) = a1(2 + i) + a0, which iszero only if a1 = a0 = 0, i.e., only if f = g ∈ (x2 − 4x+ 5).
Solution for (c). We claim that (x2−2x−1) = kerϕ. x2−2x−1 = (x−(1+√2))(x−
(1 −√2)), hence (x2 − 2x − 1) ⊂ kerϕ. Conversely, let f ∈ kerϕ. By the division
algorithm, we can write f = g + r for g ∈ (x2 − 4x + 5), where r = a1x + a0 forai ∈ Z has degree less than 2, since x2−2x−1 is monic. Then, ϕ(f) = ϕ(g)+ϕ(r) =a1(1+
√2)+a0, which is zero only if a1 = a0 = 0 since 1,
√2 are linearly independent
over Z, i.e., only if f = g ∈ (x2 − 2x− 1).
Solution for (d). We claim that (x4 − 10x2 + 1) = kerϕ. We have (x4 − 10x2 + 1) ⊂kerϕ, since ϕ(x4 − 10x2 + 1) = (
√2 +
√3)4 − 10(
√2 +
√3)2 + 1 = 0. Conversely,
27
let f ∈ kerϕ. By the division algorithm, we can write f = g + r, where r =a3x
3 + a2x2 + a1x+ a0 for ai ∈ Z has degree less than 4. Then,
ϕ(f) = ϕ(g + r) = r(√2 +
√3)
= a3(√2 +
√3)3 + a2(
√2 +
√3)2 + a1(
√2 +
√3) + a0
= (11a3 + a1)√2 + (9a3 + a1)
√3 + (2a2)
√6 + (5a2 + a0).
This gives rise to the system of equations represented by the matrices1 0 5 00 0 2 00 1 0 90 1 0 11
a0a1a2a3
=
0000
,since
√2,√3,√6, 1 are linearly independent over Z. But, this only has the trivial
Conversely, we first regard f as a polynomial in z whose coefficients are in x, y,as in Corollary 11.3.8. We can apply the division algorithm to get f = g1 + r1 forg1 ∈ (y3−z2), where r1 is of degree less than 2 in z. If r1 = 0, then f ∈ (y3−z2), andso we are done. If not, then we can apply the division algorithm again with x2 − y,this time in y, on r1 to have r1 = g2 + r2 for g2 ∈ (x2 − y), where r2 is of degree 0 iny, and degree less than 2 in z. If r2 = 0, then f = g1 + g2 ∈ (y3 − z2, x2 − y), andso we are done. If not, then we can apply the division algorithm again with x3 − z,this time in x, on r2 to have r2 = g3 + r3 for g3 ∈ (x3 − z), where r3 is of degree lessthan 2 in x, degree 0 in y, and degree less than 2 in z. This means that we have
and the linear independence of tk over C implies that r3 = 0, and so f = g1+g2+g3 ∈(y3 − z2, x2 − y, x3 − z).
28
Exercise 11.3.5. The derivative of a polynomial f with coefficients in a field F isdefined by the calculus formula (anx
n + · · ·+ a1x+ a0)′ = nanx
n−1 + · · ·+ 1a1. Theinteger coefficients are interpreted in F using the unique homomorphism Z → F .(a) Prove the product rule (fg)′ = f ′g+fg′ and the chain rule (f ◦g)′ = (f ′ ◦g)g′.(b) Let α be an element of F . Prove that α is a multiple root of a polynomial f if
and only if it is a common root of f and of its derivative f ′.
Proof of (a). Let f =∑aix
i, g =∑bjx
j. Then by (11.2.7), fg =∑ckx
k, whereck =
∑i+j=k aicj, and so
f ′g =
(∑i≥0
(i+ 1)ai+1xi
)(∑j≥0
bjxj
)=∑k≥0
∑i+j=k
(i+ 1)ai+1bjxk,
and similarly
fg′ =
(∑i≥0
aixi
)(∑j≥0
(j + 1)bj+1xj
)=∑k≥0
∑i+j=k
(j + 1)aibj+1xk.
Thus,
f ′g + fg′ =∑k≥0
(∑i+j=k
(i+ 1)ai+1bj + (j + 1)aibj+1
)xk
=∑k≥0
( ∑i+j=k+1
iaibj + jaibj
)xk
=∑k≥0
(k + 1)∑
i+j=k+1
aibjxk =
∑k≥0
(k + 1)ck+1xk = (fg)′.
Proof of (b). Suppose f ∈ F [x]. Then, f = (x − α)kg for some k ≥ 0 and g ∈ F [x]such that g(α) 6= 0. By (a), f ′ = k(x − α)k−1g + (x − α)kg′ if k ≥ 1, and f ′ = g′ ifk = 0. Hence f(α) = f ′(α) = 0 if and only if k ≥ 2, i.e., α is a multiple root of f ifand only if α is a common root of f, f ′.
Exercise 11.3.7. Determine the automorphisms of the polynomial ring Z[x].
Proof. Suppose ϕ ∈ Aut(Z[x]). Then ϕ(1) = 1, hence
n = ϕ(1) + · · ·+ ϕ(1)︸ ︷︷ ︸n times
= ϕ(1 + · · ·+ 1︸ ︷︷ ︸n times
) = ϕ(n).
29
Thus, if f =∑n
j=0 bjxj ∈ Z[x], then ϕ(f) =
∑nj=0 bjϕ(x)
j, and so ϕ is uniquelydetermined by ϕ(x).
So let ϕ(x) =∑d
i=0 aixi. Then,
ϕ(f) =d∑i=0
ai
(n∑j=0
bjxj
)d
= adbnxnd + · · ·+
d∑i=0
aibdj . (1)
hence deg(ϕ(f)) = nd. Suppose f ∈ Z[x] is the unique element such that ϕ(f) = x.Then, nd = 1 and adbn = 1 imply that d = 1 and a1 ∈ {±1}. Thus ϕ(x) = ±x + afor a ∈ Z, i.e., every ϕ ∈ Aut(Z[x]) must map x ±x+a for a ∈ Z. Finally, all suchϕ give automorphisms of Z since they define ring homomorphisms by the equation(1), and since Z[±x+ a] = Z[x].
Exercise 11.3.8. Let R be a ring of prime characteristic p. Prove that the mapR → R defined by x xp is a ring homomorphism. (It is called the Frobeniusmap).
Proof. We first claim that for p prime, p |(pi
)if 1 ≤ i ≤ p− 1. By definition,(
p
i
)=p(p− 1) · · · (p− i+ 1)
i(i− 1) · · · 2 · 1.
1 ≥ i implies p appears in the numerator, and i ≤ p − 1 implies p does not appearin the denominator. Since
(pi
)∈ Z, this implies p |
(pi
), hence
(pi
)= 0 ∈ R.
Now if x, y ∈ R, the binomial theorem gives
(x+ y)p =
p∑i=0
(p
i
)xiyp−i = xp + yp +
p−1∑i=1
(p
i
)xiyp−i = xp + yp
Thus x xp respects addition. Since trivially 1 1 and (xy)p = xpyp, we thereforehave that x xp defines a ring homomorphism R → R.
Exercise 11.3.9.(a) An element x of a ring R is called nilpotent if some power is zero. Prove that
if x is nilpotent, then 1 + x is a unit.(b) Suppose that R has prime characteristic p 6= 0. Prove that if a is nilpotent
then 1 + a is unipotent, that is, some power of 1 + a is equal to 1.
Proof of (b). Suppose aN = 0. By Exercise 11.3.8, x xp is a homomorphismR → R, so (1 + a)p = 1 + ap. Iterating this map n times such that pn ≥ N gives(1 + a)p
n= 1 + ap
n= 1 + aNap
n−N = 1.
Exercise 11.3.10. Determine all ideals of the ring F [[t]] of formal power series withcoefficients in a field F (see Exercise 11.2.2).
Solution. We claim all nonzero ideals are of the form (tn) for some n. Let I be anideal and p ∈ I such the number n := min{i | ai 6= 0} is minimal. We claim I = (tn).First, p = tnq for some unit q, hence (tn) ⊂ I. Conversely, any r ∈ I has first nonzerocoefficient at degree ≥ n, hence tns for some s ∈ F [[t]], and so r ∈ (tn).
Exercise 11.3.11. Let R be a ring, and let I be an ideal of the polynomial ringR[x]. Let n be the lowest degree among nonzero elements of I. Prove or disprove: Icontains a monic polynomial of degree n if and only if it is a principal ideal.
Proof of ⇒. If I = 0, then it is principal so suppose not. Let f be a monic polynomialof lowest degree n. We claim (f) = I. f ∈ I hence (f) ⊂ I. Now suppose g ∈ I;then, by division with remainder we can write g = fq + r, where if r 6= 0, it hasdegree lower than f . But then, f, g ∈ I, hence g − fq = r ∈ I, so r = 0, andg ∈ (f).
Counterexample for ⇐. Consider I = (2x) ⊂ Z[x]. Any element in I is obtained bymultiplying 2x by a polynomial of degree ≥ 1, in which case we get an element ofdegree ≥ 2, or by multiplying by an element of Z. But then, 2 /∈ Z×, hence there isno monic polynomial of degree 1 in I.
Exercise 11.3.12. Let I and J be ideals of a ring R. Prove that the set I + J ofelements of the form x + y, with x in I and y in J , is an ideal. This ideal is calledthe sum of the ideals I and J .
Proof. Let (x + y), (x′ + y′) ∈ I + J and s ∈ R, where x, x′ ∈ I, y, y′ ∈ J . Then,s(x+ y) + (x′ + y′) = (sx+ x′) + (sy + y′) ∈ I + J , hence I + J is an ideal.
Exercise 11.3.13. Let I and J be ideals of a ring R. Prove that the intersectionI ∩ J is an ideal. Show by example that the set of products {xy | x ∈ I, y ∈ J} neednot be an ideal, but that the set of finite sums
∑xνyν of products of elements of I
and J is an ideal. This ideal is called the product ideal, and is denoted by IJ . Isthere a relation between IJ and I ∩ J?
31
Proof. Let x, y ∈ I ∩ J and s ∈ R. Then, sx + y ∈ I and sx + y ∈ J since I, J areideals, hence sx+ y ∈ I ∩ J and so I ∩ J is an ideal.
Now let R = Z[x], I = (2, x), J = (3, x). Then, 3x, 2x are in the set of products,but their difference 3x− 2x = x is not, and so the set of products is not an ideal.
Let∑xiyi,
∑x′iy
′i ∈ IJ and s ∈ R, where xi, x
′i ∈ I, yi, y
′i ∈ J . Then, s
∑xiyi+∑
x′iy′i =
∑sxiyi +
∑x′iy
′i ∈ IJ , hence IJ is an ideal.
We have in general IJ ⊂ I ∩ J since∑xiyi for xi ∈ I, yi ∈ J has xiyi ∈ I ∩ J .
IJ = I ∩ J if I + J = (1) by Exercise 11.6.8(a).I ∩ J ⊂ IJ does not hold in general, for if R = Z, I = (m), J = (n), then
I ∩ J = (lcm(m,n)) but IJ = (mn).
11.4 Quotient Rings
Exercise 11.4.2. What does the Correspondence Theorem tell us about ideals of Z[x]that contain x2 + 1?
Solution. By the Correspondence Theorem (Thm. 11.4.3), there is a bijective corre-spondence between the ideals of Z[x] that contain x2+1 and the ideals of Z[x]/(x2+1) ≈ Z[i], where these two rings are isomorphic as in Ex. 11.4.5. By Prop. 12.2.5(c),Z[i] is a Euclidean domain, hence a principal ideal domain by Prop. 12.2.7. Thus,every ideal in Z[i] is of the form (a + bi) for a, b ∈ Z. These correspond to ideals(a+ bx) in Z[x]/(x2 + 1) by the isomorphism i x from above, hence the ideals inZ[x] containing x2 + 1 are of the form (a+ bx, x2 + 1).
Solution for (a). We see that 2(x2 − 3)− (x− 2)(2x+4) = 2 ∈ (x2 − 3, 2x+4), andso (x2 − 3, 2x+ 4) = (x2 − 3, 2x+ 4, 2) = (x2 − 3, 2), since 2(x+ 2) = 2x+ 4. Then,Z[x]/(x2−3, 2x+4) = Z[x]/(x2−3, 2). But then, Z[x]/(x2−3, 2) ≈ F2[x]/(x
2−3) =F2[x]/(x
2 + 1) = F2[x]/(x+ 1)2 since x2 + 1 = (x+ 1)2 in F2.
Solution for (b). First recall Z[i] ≈ Z[x]/(x2 + 1). Thus, Z[i]/(2 + i) ≈ Z[x]/(x2 +1, 2 + x). We first consider the quotient Z[x]/(2 + x). Since (2 + x) is the kernel ofthe homomorphism Z[x] → Z[−2], f(x) f(−2), we see that this is isomorphic toZ, as in Example 11.4.5. Then, we have that the residue of g = x2 + 1 is 5, and sowe have Z[i]/(2 + i) ≈ Z/5Z = F5.
Solution for (c). We first note 6x− 3(2x− 1) = 3 ∈ (6, 2x− 1), and so (6, 2x− 1) =(3, 2x− 1). Also, 3x− (2x− 1) = x+ 1 ∈ (3, 2x− 1), and so (3, 2x− 1) = (3, x+ 1).Thus, Z[x]/(6, 2x− 1) = Z[x]/(3, x+ 1) ≈ F3[x]/(x+ 1) = F3.
Solution for (e). We see Z[x]/(x2 + 3, 5) ≈ F5[x]/(x2 + 3). But since x2 + 3 6= 0 for
any x ∈ F5, we see that we cannot reduce x2 + 3. But in F5, x
2 + 3 = x2 − 2, and sowe have F5[
√2].
Exercise 11.4.4. Are the rings Z[x]/(x2 + 7) and Z[x]/(2x2 + 7) isomorphic?
Solution. We claim they are not. Let R = Z[x]/(x2 + 7) and S = Z[x]/(2x2 + 7).Asume α : R → S is an isomorphism. α(1) = 1, hence α(2) = 2, and so if αis an isomorphism, then R/(2) ≈ S/(2). We claim this is a contradiction. For,R/(2) ≈ F2[x]/(x
2 + 1) 6= 0, whereas S/(2) ≈ F2[x]/(1) = 0.
11.5 Adjoining Elements
Exercise 11.5.1. Let f = x4+x3+x2+x+1 and let α denote the residue of x in thering R = Z[x]/(f). Express (α3 + α2 + α)(α5 + 1) in terms of the basis (1, α, α2, α3)of R.
Proof. We first have (x3 + x2 + x)(x5 + 1) = x8 + x7 + x6 + x3 + x2 + x. We thenperform the division algorithm:
x4 − x
x4 + x3 + x2 + x+ 1)
x8 + x7 + x6 + x3 + x2 + x
− x8 − x7 − x6 − x5 − x4
− x5 − x4 + x3 + x2 + xx5 + x4 + x3 + x2 + x
2x3 + 2x2 + 2x
and so (α3 + α2 + α)(α5 + 1) = 2α3 + 2α2 + 2α ∈ Z[x]/(f).
Exercise 11.5.3. Describe the ring obtained from Z/12Z by adjoining an inverse of2.
Solution. Let R = (Z/12Z)[x]/(2x−1) be our ring; it is isomorphic to Z[x]/(12, 2x−1). (12, 2x − 1) = (3, 2x − 1) since ⊂ clearly holds and 12x2 − (3 + 6x)(2x − 1) =3 ∈ (12, 2x − 1). Thus R ≈ Z[x]/(3, 2x − 1) ≈ F3[x]/(2x − 1). But 2 ∈ F×
3 , henceF3[x]/(2x− 1) ≈ F3, so R ≈ F3.
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Exercise 11.5.4. Determine the structure of the ring R′ obtained from Z by adjoin-ing an element α satisfying each set of relations.(a) 2α = 6, 6α = 15, (b) 2α− 6 = 0, α− 10 = 0, (c) α3 + α2 + 1 = 0, α2 + α = 0.
Solution for (a). Z[α] = Z[x]/(2x−6, 6x−15). But (6x−15)−3(2x−6) = 3 ∈ (2x−6, 6x−15), and 3(x−2)−(2x−6) = x ∈ (2x−6, 6x−15) imply (2x−6, 6x−15) = (x, 3).Thus, Z[α] ≈ Z[x]/(3, x) ≈ Z/3Z = F3.
Solution for (c). Z[α] = Z[x]/(x3 + x2 + 1, x2 + x). But x3 + x2 + 1 − x(x2 + x) =1 ∈ (x3 + x2 + 1, x2 + x), and so Z[α] ≈ Z[x]/(1) = 0, the zero ring.
Exercise 11.5.6. Let a be an element of a ring R, and let R′ be the ring R[x]/(ax−1)obtained by adjoining an inverse of a to R. Let α denote the residue of x (the inverseof a in R′).(a) Show that every element β of R′ can be written in the form β = αkb, with b
in R.(b) Prove that the kernel of the map R → R′ is the set of elements b of R such
that anb = 0 for some n > 0.(c) Prove that R′ is the zero ring if and only if a is nilpotent (see Exercise 11.3.9).
Proof of (a). Any element β ∈ R′ can be written as a finite sum∑bkα
k. Letting Kbe the largest k such that bk 6= 0, we see that defining b =
∑aK−kbk, α
Kb = β.
Proof of (b). Let Γa(R) = {b ∈ R | anb = 0 for some n > 0}, and call the mapdefined ϕ; it suffices to show Γa(R) = ϕ−1((ax − 1)), where (ax − 1) ⊂ R[x]. Butb ∈ R has ϕ(b) ∈ (ax − 1) if and only if (ax − 1)g = b for some g =
∑gix
i ∈ R[x].Solving recursively, we must have gi = −aib for all i. Such a g exists if and only ifb ∈ Γa(R), for otherwise g would be an infinite sum.
Proof of (c). a is nilpotent if and only if 1 ∈ Γa(R). By the proof of (b), this holdsif and only if 1 ∈ (ax − 1), and since R[x] → R′ is surjective, this holds if and onlyif R′ = 0 by the first isomorphism theorem (Thm. 11.4.2(b)).
34
11.6 Product Rings
Exercise 11.6.1. Let ϕ : R[x] → C×C be the homomorphism defined by ϕ(x) = (1, i)and ϕ(r) = (r, r) for r ∈ R. Determine the kernel and the image of ϕ.
Solution. ϕ is the evaluation map f (f(1), f(i)). So f ∈ kerϕ ⇐⇒ f(1) =f(i) = 0 ⇐⇒ f ∈ ((x− 1)(x2 + 1)), i.e., kerϕ = ((x− 1)(x2 + 1)).
We now claim imϕ = R×C. ⊂ clearly holds. Now let (c, a+ bi) ∈ R×C. Then,if f = a3x
3 + a2x2 + a1x+ a0 ∈ R[x] is of degree 3,
ϕ(f) = (a3 + a2 + a1 + a0,−ia3 − a2 + a1i+ a0)
= (a3 + a2 + a1 + a0, (a0 − a2) + (a1 − a3)i),
and the condition f (c, a+ bi) gives the system of equationsa0 + a1 + a2 + a3 = c
a0 − a2 = a
a1 − a3 = b
which has a solution by linear algebra, hence ⊃ also holds, and imϕ = R× C.
Exercise 11.6.2. Is Z/(6) isomorphic to the product ring Z/(2)× Z/(3)? Is Z/(8)isomorphic to Z/(2)× Z/(4)?
Solution. Letting R = Z/(6), I = (2), J = (3) in Exercise 11.6.8(c), we have Z/(6) ≈Z/(2)× Z/(3), since IJ = 0 in R.
Z/(8) 6≈ Z/(2)×Z/(4), for Z/(8) has an additive element of order 8 while Z/(2)×Z/(4) does not.
Exercise 11.6.3. Classify rings of order 10.
Proof. Let R be a ring of order 10. By the Sylow Theorems (Thms. 7.7.2, 7.7.4, 7.7.6),the abelian group (R,+) contains exactly one normal subgroup each of orders 2 and 5,with trivial intersection, hence by Prop. 2.11.4(d), (R,+) ≈ Z/(2)×Z/(5) ≈ Z/(10).It remains to show R ≈ Z/(10) also as a ring. Let a ∈ R be an element of order 10;a then generates R, and so 1 ∈ R implies na = 1 for some 1 ≤ n ≤ 9. If n = 2, then5a = 2a · 5a = 10a2 = 0, contradicting that a has order 10; similarly, n 6= 5. Thus,na = 1 has order 10, and generates R, hence R ≈ Z/(10).
Exercise 11.6.4. In each case, describe the ring obtained from the field F2 by ad-joining an element α satisfying the given relation:(a) α2 + α + 1 = 0, (b) α2 + 1 = 0, (c) α2 + α = 0.
35
Remark. Since each equation is of degree 2, we can write F2[α] = {0, 1, α, α + 1} byProp. 11.5.5(a) in each case.
Solution for (a). Since α(α+1) = α2+α = 1, every element of F2[α] has an inverse.Thus, we have a field of order 4, i.e., F2[α] = F4.
Solution for (b). F2[α] = F2[x]/(x2 + 1) = F2[i], the ring of Gauss integers mod 2.
We see (α + 1)2 = α2 + 1 = 0, and so α2 = 1, α(α + 1) = α2 + α = α + 1. This isnot a field, or even an integral domain, since (α + 1)2 = 0.
Solution for (c). We have α2 = α, and so by Prop. 11.6.2 we have F2[α] ≈ αF2[α]×(α+1)F2[α]. Since αF2[α] = {0, α} and (α+1)F2[α] = {0, α+1} are both isomorphicto F2, we see that F2[α] ≈ F2 × F2.
Exercise 11.6.5. Suppose we adjoin an element α satisfying the relation α2 = 1 tothe real numbers R. Prove that the resulting ring is isomorphic to the product R×R.
Proof. The resulting ring is R = R[x]/(x2 − 1). Let ϕ : R[x] → R × R be definedby f (f(1), f(−1)). kerϕ = (x2 − 1) = ((x + 1)(x − 1)), and so by the firstisomorphism theorem (Thm. 11.4.2(b)) it suffices to show ϕ is surjective, inducingan isomorphism R → R× R. Letting f = ax+ b ∈ R[x] of degree 1, we have
ϕ(ax+ b) = (a+ b, a− b),
which can be solved for arbitrary (a+ b, a− b) = (x, y) by linear algebra.
Exercise 11.6.6. Describe the ring obtained from the product ring R×R by invertingthe element (2, 0).
Solution. Using the isomorphism in Exercise 11.6.5, since x+ 1 (2, 0), the ring is
R[x, y](x2 − 1, (x+ 1)y − 1)
≈ R[x, (x+ 1)−1]
(x2 − 1)=
R[x, (x+ 1)−1]
((x+ 1)(x− 1))=
R[x, (x+ 1)−1]
(x− 1)≈ R,
since x+ 1 has residue 2 in R[x]/(x− 1), which already has an inverse in R.
Exercise 11.6.7. Prove that in the ring Z[x], the intersection (2) ∩ (x) of theprincipal ideals (2) and (x) is the principal ideal (2x), and that the quotient ringR = Z[x]/(2x) is isomorphic to the subring of the product ring F2[x] × Z of pairs(f(x), n) such that f(0) ≡ n modulo 2.
36
Proof. Clearly (2x) ⊂ (2) ∩ (x). Conversely, if f ∈ (2) ∩ (x), then f = 2g = xh forsome g, h ∈ Z[x]. But this implies g ∈ (x), hence f ∈ (2x).
Now consider the ring homomorphism ϕ : Z[x] → F2[x] × Z where∑aix
i ∑(aix
i, a0), where ai is the residue of ai in F2. We claim ϕ(Z[x]) = S := {(f(x), n) ∈F2[x] × Z | f(0) ≡ n mod 2}. By construction, ϕ(Z[x]) ⊂ S; conversely, any(f(x), n) ∈ S must have f(x) =
∑aix
i where a0 = n, and so choosing repre-sentatives ai of ai in Z such that a0 = n, we see that
∑aix
i 7→ (f(x), n), henceϕ(Z[x]) ⊃ S.
Finally we show kerϕ = (2x). (2x) ⊂ kerϕ by construction. Conversely, ϕ(f) =(0, 0) =⇒ a0 = 0 and 2 | ai for i 6= 0, hence f ∈ (2) ∩ (x) = (2x) by the above.Thus, R ≈ S by the first isomorphism theorem (Thm. 11.4.2(b)).
Exercise 11.6.8. Let I and J be ideals of a ring R such that I + J = R.(a) Prove that IJ = I ∩ J (see Exercise 11.3.13).(b) Prove the Chinese Remainder Theorem: For any pair a, b of elements of R,
there is an element x such that x ≡ a modulo I and x ≡ b modulo J . (Thenotation x ≡ a modulo I means that x− a ∈ I.)
(c) Prove that if IJ = 0, then R is isomorphic to the product ring (R/I)× (R/J).(d) Describe the idempotents corresponding to the product decomposition in (c).
Proof of (a). Clearly IJ ⊂ I∩J . In the other direction, let u+v = 1 for u ∈ I, v ∈ J .Then, if x ∈ I ∩ J , x = x(u+ v) = xu+ xv ∈ IJ .
Proof of (b). It suffices to show the ring homomorphism ϕ : R → R/I×R/J definedby x 7→ (x+I, x+J) is surjective. Let (a+I, b+J) ∈ R/I×R/J . Since I+J = R, wehave u+v = 1 for some u ∈ I, v ∈ J . Let x = bu+av. Then, ϕ(x) = (x+I, x+J) =(av + I, bu+ J) = (a(1− u) + I, b(1− v) + J) = (a+ I, b+ J).
Proof of (c). By definition in (b) kerϕ = I ∩ J , and by (a), kerϕ = IJ = 0, and sowe are done by the first isomorphism theorem (Thm. 11.4.2(b)).
Solution for (d). By (b), if u + v = 1 for u ∈ I, v ∈ J , then ϕ(v) = (1, 0) andϕ(u) = (0, 1). Hence the idempotents corresponding to the product decompositionin (c) are the images of u, v from partitions of unity u+ v = 1 for u ∈ I, v ∈ J .
11.7 Fractions
Exercise 11.7.1. Prove that a domain of finite order is a field.
37
Proof. Suppose R is a finite domain, and consider R× := R \ {0}. It suffices to showany r ∈ R× is a unit. If |R×| = n, then r · R× ⊂ R× since R is a domain, and|r · R×| = n by the cancellation law (11.7.1). Hence 1 ∈ r · R×, and so rr′ = 1 forsome r′ ∈ R×.
Exercise 11.7.2. Let R be a domain. Prove that the polynomial ring R[x] is adomain, and identify the units in R[x].
Proof. Let f, g ∈ R[x] be nonzero, and suppose fg = 0. If deg f = d, deg g = d′,then deg fg = d + d′ since the product adbd′ of their leading coefficients is nonzero,contradicting that deg fg is undefined. Hence R[x] is a domain.
If fg = 1 then deg fg = d+d′ = 0, so f and g are constant polynomials. Viewingconstant polynomials as elements of R, the units of R[x] are then the units of R.
Exercise 11.7.3. Is there a domain that contains exactly 15 elements?
Solution. Suppose R is such a domain; it is a field by Exercise 11.7.1. It suffices toshow any finite field F has order pn for some prime power pn, since 15 cannot bewritten in this way. By Lem. 3.2.10, F has characteristic p for some prime p, andso contains Fp. Then, F is of finite dimension n as a vector space over Fp, and socontains pn elements.
Exercise 11.7.4. Prove that the field of fractions of the formal power series F [[x]]over a field F can be obtained by inverting the element x. Find a neat description ofthe elements of that field (see Exercise 11.2.2).
Proof. Take an element fgin the field of fractions of F [[x]] and suppose g =
∑∞i=n aix
i
with n ≥ 0 and an 6= 0. Then, g = xng0, where g0 is a unit of F [[x]] by Exercise 11.2.2,
and so fg=
fg−10
xn. Thus, the field of fractions of F [[x]] is contained in F [[x]][x−1], and
the reverse inclusion is clear, and so the field of fractions of F [[x]] is F [[x]][x−1].This is the ring of infinite series of the form h =
∑i≥n aix
i for n ∈ Z, which iscalled the ring of formal Laurent series F ((x)).
11.8 Maximal Ideals
Exercise 11.8.1. Which principal ideals in Z[x] are maximal ideals?
Solution. We claim that none are. Suppose (f) ⊂ Z[x] is maximal. Then, deg f > 0,for otherwise (f) ( (f, x) ( Z[x], contradicting maximality of (f). Now choose psuch that p - ai for any coefficient ai of f . Then, (f) ( (p, f) since p /∈ (f), and(p, f) ( Z[x] since Z[x]/(p, f) = Fp[x]/(f) 6= 0. Thus, (f) cannot be maximal.
38
Exercise 11.8.2. Determine the maximal ideals of each of the following rings:(a) R× R, (b) R[x]/(x2), (c) R[x]/(x2 − 3x+ 2), (d) R[x]/(x2 + x+ 1).
Solution for (a). The units in R × R are elements (a, b) for a, b 6= 0, and so anymaximal ideal in R × R cannot contain a pair (a, 0) and (0, b) for a, b 6= 0. Henceevery element in a maximal ideal of R× R must have the same coordinate equal tozero, and so it is of the form R × 0 or 0 × R since 0 × 0 is not maximal; these aremaximal by Prop. 11.8.2(b) since taking quotients gives R, a field.
Solution for (b). The ideals in R[x]/(x2) are the ideals in R[x] containing (x2) by thecorrespondence theorem (Thm. 11.4.3). Since R[x] is a PID by Props. 12.2.5, 12.2.7,the only proper ideals containing (x2) are (x) and (x2), since if x2 ∈ (f), then x2 = fgimplies f | x2. We have (x2) ( (x), and so (x) is the only maximal ideal.
Solution for (c). x2 − 3x + 2 = (x− 2)(x− 1) and (x− 2) + (x− 1) = R[x] impliesR[x]/(x2 − 3x + 2) ≈ R[x]/(x − 1) × R[x]/(x − 2) ≈ R × R by Exercise 11.6.8(c).The maximal ideals of R × R are R × 0 and 0 × R as in (a), which correspond to(x− 2), (x− 1), respectively, in R[x]/(x2 − 3x+ 2).
Solution for (d). We note that x2 + x+ 1 has no real roots, and so is irreducible inR[x]. Thus, (x2+x+1) maximal and R[x]/(x2+x+1) is a field with unique maximalideal (0) by Prop. 11.8.2.
Exercise 11.8.3. Prove that the ring F2[x]/(x3+x+1) is a field, but that F3[x]/(x
3+x+ 1) is not a field.
Proof. By Prop. 11.8.2(b), it suffices to show (x3+x+1) is maximal in F2[x] but notin F3[x]. But this is true since x
3+x+1 has no roots in F2 (03+0+1 = 1 = 13+1+1),
while x3+x+1 has the root 1 in F3 (13+1+1 = 0), hence (x3+x+1) ( (x−1).
Exercise 11.8.4. Establish a bijective correspondence between maximal ideals of R[x]and points in the upper half plane.
Proof. R[x] is a PID by Props. 12.2.5, 12.2.7, and so any maximal ideal is of theform (f). But (f) is maximal if and only if f is an irreducible nonunit in R[x].Every polynomial in R[x] of degree at least 3 has a real root, and therefore is notirreducible. So, every maximal ideal is of the form (f) for f a linear polynomial oran irreducible quadratic polynomial.
Now recall we can identify the upper half plane with the subset of C of elements zwith Im(z) ≥ 0. Therefore, we can define the function H that sends every maximalideal (f) to the root of f in the upper half plane. Notice that the function is
39
well-defined: if f is linear, then it has only one solution, a real number, and if f is anirreducible quadratic polynomial, then z and z are the roots of f , for some z ∈ C\R,so only one of the roots is in the upper half-plane.
We need to show H is a bijection. We define the inverse H−1 sending z to theideal generated by x − z if z ∈ R, and (x − z)(x − z) otherwise. H−1 is clearly atwo-sided inverse of H since (f) = (cf) for any c ∈ R, and so we are done.
11.9 Algebraic Geometry
Exercise 11.9.4. Let U and V be varieties in Cn. Prove that the union U ∪ V andthe intersection U ∩ V are varieties. What does the statement U ∩ V = ∅ meanalgebraically? What about the statement U ∪ V = Cn?
Proof. Let U be defined by {f1, . . . , fr}, and V by {g1, . . . , gs}. U ∩ V is defined by{f1, . . . , fr, g1, . . . , gs} since x ∈ U ∩ V if and only if fi(x) = 0 for all i and gj(x) = 0for all j. We claim U ∪ V is equal to the variety W defined by {figj}i,j. U ∪ V ⊂ Wsince if x ∈ U (resp. V ), then fi = 0 for all i (resp. gj = 0 for all j), hence figj = 0for all i, j. Conversely, suppose x ∈ W \ U ∪ V . Then, there exists i0, j0 such thatfi0(x), gj0(x) 6= 0, and so fi0gj0 6= 0, a contradiction.
Now by Thm. 11.9.1 and Cor. 11.9.3, U ∩ V = ∅ if and only if the quotient ringC[x1, . . . , xn]/(f1, . . . , fr, g1, . . . , gs) = 0, i.e., (f1, . . . , fr, g1, . . . , gs) = (1).
Now U ∪ V = Cn if and only if any point x ∈ Cn is a common zero of all figj.But there are only finitely many figj with finitely many zeros each, and so this holdsif and only if all the figj are equal to 0, which is true if and only if all the fi or allthe gj are 0, if and only if U or V equals Cn.
Exercise 11.9.5. Prove that the variety of zeros of a set {f1, . . . , fr} of polynomialsdepends only on the ideal they generate.
Proof. For sets of polynomials {f1, . . . , fr} and {g1, . . . , gs}, let V,W be the varietiesformed by their zero sets and let I, J be the ideals they generate, respectively. Weclaim I ⊃ J implies V ⊂ W . Every gj can then be written as a linear combinationof the fi, hence if x ∈ V , fi(x) = 0 for all i, and so gj(x) =
∑aifi(x) = 0 for all j
as well, i.e., V ⊂ W .Finally, if I = J , then V ⊂ W and W ⊂ V , and so V = W . Thus, the variety
defined by a set {f1, . . . , fr} depends only on the ideal (f1, . . . , fr).
Exercise 11.9.6. Prove that every variety in C2 is the union of finitely many pointsand algebraic curves.
40
Proof. Let U be defined by {f1, . . . , fr}, and let g = gcd(f1, . . . , fr); g exists sinceC[x, y] is a UFD (Thm. 12.3.10). Then, U = V ∪W , where V is defined by g andW by {f1/g, . . . , fr/g} as in Exercise 11.9.4. W is the union of finitely many pointsby Thm. 11.9.10 since gcd(gcd(f1/g, . . . , fr−1/g), fr/g) = gcd(f1/g, . . . , fr/g) = 1.V is the union of finitely many algebraic curves since g =
∏gi for some irreducible
polynomials gi, hence V is the union of curves defined by gi as in Exercise 11.9.4.
Exercise 11.9.11. Let C1 and C2 be the zeros of quadratic polynomials f1 and f2respectively that don’t have a common linear factor.(a) Let p and q be distinct points of intersection of C1 and C2, and let L be the
(complex) line through p and q. Prove that there are constants c1 and c2, notboth zero, so that g = c1f1 + c2f2 vanishes identically on L. Prove also that gis the product of linear polynomials.
(b) Prove that C1 and C2 have at most 4 points in common.
Proof of (a). Let λ(t) = (1 − t)p + tq parametrize L ⊂ C2. Then, f1(λ(t)), f2(λ(t))are quadratic in t, and so c1f1(λ(t)) + c2f2(λ(t)) is at most quadratic in t for anyci ∈ C. Let t0 ∈ C \ {0, 1}, and choose ci such that c1f1(λ(t0)) + c2f2(λ(t0)) = 0;we can moreover choose ci to be nonzero. Then, h(t) = c1f1(λ(t)) + c2f2(λ(t)) is atmost quadratic in t hence has at most two zeros if h 6= 0 by the fundamental theoremof algebra (Thm. 15.10.1); but h(0) = h(1) = h(t0) = 0 implies h(t) = 0. Henceg = c1f1 + c2f2 vanishes identically on L.
Now since g vanishes on L, by unique factorization (Thm. 12.3.10) g = gLg′ for
some linear polynomial gL that defines L. g is at most quadratic, hence g′ is eitherlinear or constant. In either case, g is then a product of linear polynomials.
Proof of (b). Suppose C1∩C2 contains more than four points. Let g as in (a). Then,g = c1f1 + c2f2 implies that C1 ∩ C2 ⊂ {g = 0}. g is either the product of oneor two linear polynomials as in (a), and so C1 ∩ C2 is contained in either one ortwo lines. In either case, this implies at least three points in C1 ∩ C2 lie on one lineparametrized by η(t). Then, f1(η(t)), f2(η(t)) are quadratic and both vanish at threevalues of t, and so f1(η(t)) = f2(η(t)) = 0 by the fundamental theorem of algebra(Thm. 15.10.1). Thus, f1, f2 have infinitely many common zeros in C2, hence have acommon linear factor by Thm. 11.9.10, a contradiction.
Exercise 11.9.12. Prove in two ways that the three polynomials f1 = t2 + x2 − 2,f2 = tx−1, f3 = t3+5tx2+1 generate the unit ideal in C[x, t] in two ways: by showingthat they have no common zeros, and also by writing 1 as a linear combination off1, f2, and f3 with polynomial coefficients.
41
Proof. We first show that there are no common zeros. If f2(x, t) = 0, then x, t 6= 0and t = x−1. Thus,
hence if f(x, t) = 0 as well, then (x, t) = ±(1, 1). But f3(1, 1) = 7, f3(−1,−1) = −5,hence f1, f2, f3 have no common zeros, and (f1, f2, f3) = (1) by Exercise 11.9.4.
Now we write 1 as a linear combination of f1, f2, f3. 1 can be expressed as
Exercise 11.9.13. Let ϕ : C[x, y] → C[t] be a homomorphism that is the identity onC and sends x x(t), y y(t) and such that x(t) and y(t) are not both constant.Prove that the kernel of ϕ is a principal ideal.
Proof. We claim kerϕ is principal. If not, then kerϕ contains two elements f, gthat do not have a common factor. We claim they do not have a common factor inC(x)[y]. For, suppose h ∈ C(x)[y] is a common factor; then, h = a−1h0 for somea ∈ C[x], h0 ∈ C[x, y] by clearing denominator. Assuming without loss of generalitythat nothing of the form x−α divides h0, then h0 divides f, g in C(x)[y], hence alsodoes in C[x, y] by Prop. 11.9.9, contradicting that f, g do not have a common factor.
Therefore, there exist r0, s0 ∈ C(x)[y] such that r0f + s0g = 1, and clearingdenominators we get rf+sg = q ∈ C[x] for some r, s ∈ C[x]. This implies kerϕ∩C[x]is nontrivial, but this is a contradiction for any g ∈ kerϕ∩C[x] must satisfy g(x(t)) =0 for all t, hence g = 0.
11.M Miscellaneous Problems
Exercise 11.M.3. Let R denote the set of sequences a = (a1, a2, a3, . . .) of realnumbers that are eventually constant: an = an+1 = . . . for sufficiently large n.Addition and multiplication are componentwise, that is, addition is vector additionand multiplication is defined by ab = (a1b1, a2b2, . . .). Prove that R is a ring, anddetermine its maximal ideals.
Proof. Denote a to be the real number a converges to. R is a ring since sums andproducts of eventually constant sequences are eventually constant, commutativity
42
and associativity of +,× and the distributive property follow since addition andmultiplication are defined componentwise, and R has additive identity (0, 0, . . .),additive inverses −a = (−a1,−a2, . . .) for a, and multiplicative identity (1, 1, . . .).
We now want to find its maximal ideals. Consider for each i ∈ Z>0 the mapϕi : R → R defined by a ai; since +,× are defined termwise, and 0 0, thisdefines a ring homomorphism. It is moreover surjective, hence mi := kerϕi = {a |ai = 0} is maximal by Prop. 11.8.2(a). Now consider the map ϕ∞ : R → R defined bymapping a to the real number it converges to; since 0 0 and a+ b = a+b, ab = ab,this defines a ring homomorphism. It is moreover surjective, hence m∞ := kerϕ∞ ={a | a = 0} is maximal by Prop. 11.8.2(a).
We claim these are all the maximal ideals. It suffices to show if m ⊂ R is maximaland not equal to mi, it is equal to m∞. So suppose not; then, there exists a ∈ m suchthat a 6= 0. There are therefore finitely many i such that ai = 0. Since m 6= mi, foreach i there exists a sequence ai with aii 6= 0; we can moreover assume aij = 0 for allj 6= i by multiplying by the sequence in R with 1 in the ith index and 0 otherwise.
b = a+∑
{i|ai=0}
ai ∈ m
is then a unit in m, since bi 6= 0 for all i, contradicting maximality of m.
Exercise 11.M.4.(a) Classify rings R that contain C and have dimension 2 as a vector space over
C.(b) Do the same for rings that have dimension 3.
Solution for (a). Let {1, r} be a basis for R, and let ϕ : C[x] → R be defined by1 1, x r. ϕ is then surjective, and kerϕ = (f) for some f ∈ C[x] since C[x] isa PID (Props. 12.2.5, 12.2.7), giving R ≈ C[x]/(f) by the first isomorphism theorem(Thm. 11.4.2(b)). deg f > 1 since otherwise {1, r} would be linearly dependent; sincer2 = ar + b ∈ R for some a, b ∈ C, we then have f = x2 − ax− b = (x− ζ1)(x− ζ2)for some ζ1, ζ2 ∈ C. If ζ1 = ζ2 =: ζ, then
R ≈ C[x]/(f) ≈ C[x]/(x2)
by composing with the isomorphism defined by x x+ ζ. If ζ1 6= ζ2, then (x− ζ1)+(x− ζ2) = R as ideals, hence
R ≈ C[x]/(f) ≈ C[x]/(x− ζ1)× C[x]/(x− ζ2) ≈ C× C
by Exercise 11.6.8(c). Hence the two possibilities are R ≈ C[x]/(x2) or C× C.
43
Solution for (b). Suppose there exists r ∈ R such that {1, r, r2} is a basis for R.Then, defining the map ϕ : C[x] → R by x r, we again have that kerϕ = (f) fordeg f > 2, since if deg f ≤ 2 then {1, r, r2} would not be linearly independent. Also,r3 = ar2 + br + c for some a, b, c ∈ C, and so f = x3 − ax2 − bx − c. If f has onetriple root ζ, then
R ≈ C[x]/((x− ζ)3) ≈ C[x]/(x3).If f has a double root ζ and a simple root ζ ′, then since ((x− ζ)2) + (x− ζ ′) = R asideals,
R ≈ C[x]/((x− ζ)2)(x− ζ ′))
≈ C[x]/((x− ζ)2)× C[x]/(x− ζ ′)
≈ C[x]/(x2)× C
by the same argument as before. Finally, if f has three distinct roots, then
R ≈ C[x]/((x− ζ1)(x− ζ2)(x− ζ3))
≈ C[x]/(x− ζ1)× C[x]/(x− ζ2)× C[x]/(x− ζ3)
≈ C× C× C
by repeated applications of Exercise 11.6.8(c) since, for example, ((x− ζ1)(x− ζ2))+(x− ζ3) = R as ideals.
Now suppose no r ∈ R exists such that {1, r, r2} is a basis for R. Thus, if {1, r, s}is a basis for R, then
r2 = a1r + c1, s2 = b2s+ c2, rs = a3r + b3s+ c3
for some ai, bi, ci ∈ C. We claim we can assume c1 = c2 = 0. For, changingcoordinates r r + α gives
and so letting α be such that α2 + a1α − c1 = 0, we have r2 = a1r. Similarly for s,we have s2 = b2s. We then have
(r + zs)2 = r2 + 2zrs+ z2s2
= (a1 + 2a3z)r + (b2z2 + 2b3z)s+ 2c3z
= Az(r + s) +Bz
for some Az, Bz ∈ C since {1, r + zs, (r + zs)2} is linearly dependent. Thus,
Az = a1 + 2a3z = b2z2 + 2b3z, Bz = 2c3z.
44
Since the first equation must hold for all z, we have a1 = b2 = 0, hence r2 = s2 = 0,and also a3 = b3. This implies (rs)2 = r2s2 = 0, but since also
(rs)2 = (a3r + a3s+ c3)2
= 2a23rs+ 2a3c3r + 2a3c3s+ c23= 2a3(a
23 + c3)r + 2a3(a
23 + c3)s+ c3(2a
23 + c3),
we have a3(a23 + c3) = c3(2a
23 + c3) = 0. If one of a3, c3 is nonzero, then the other is
also. But this is impossible, and so we have a3 = b3 = c3 = 0.Finally, let ϕ : C[x, y] → R be defined by 1 1, x r, y s. ϕ is then sur-
jective, giving R ≈ C[x, y]/kerϕ by the first isomorphism theorem (Thm. 11.4.2(b)).We know I := (x2, y2, xy) ⊂ kerϕ, and the reverse inclusion holds since for anyf ∈ C[x, y], f ≡ αr + βs + γ mod I for some α, β, γ ∈ C, and f 0 in the compo-sition C[x, y]/I → C[x, y]/ker ϕ→ R if and only if α = β = γ = 0 since {1, r, s} is abasis for R, i.e., if and only if f ∈ I. Thus, in this case R ≈ C[x, y]/(x2, y2, xy).
In summary, there are four possibilities for R:
C[x, y]/(x2, y2, xy), C[x]/(x3), C[x]/(x2)× C, C× C× C.
12 Factoring
12.1 Factoring Integers
Exercise 12.1.4. Solve the following simultaneous congruences:(a) x ≡ 3 modulo 8, x ≡ 2 modulo 5,(b) x ≡ 3 modulo 15, x ≡ 5 modulo 8, x ≡ 2 modulo 7,(c) x ≡ 13 modulo 43, x ≡ 7 modulo 71.
Solution for (c). x = 2421 = 56 · 34 + 13 = 34 · 71 + 7.
Exercise 12.1.5. Let a and b be relatively prime integers. Prove that there areintegers m and n such that am + bn = 1 modulo ab.
Proof. Since a, b are relatively prime, am + bn = 1 mod ab if and only if am + bn =1 mod a and am+ bn = 1 mod b. Note am+ bn = bn mod a, and am+ bn = am mod b.
45
Now consider a, a2, a3, . . . mod b. There are only b equivalence classes 0, 1, . . . , b−1 mod b, hence for some i < j we have ai ≡ aj mod b. Then, ai(aj−i− 1) ≡ 0 mod b,and so b | ai(aj−i− 1). But a, b are relatively prime, hence b | aj−i− 1. Thus, lettingm := j − i, we have am ≡ 1 mod b.
By the same argument working mod a, there exists n such that bn ≡ 1 mod a,and so am + bn ≡ 1 mod a and mod b, hence am + bn ≡ 1 mod ab.
12.2 Unique Factorization Domains
Exercise 12.2.1. Factor the following polynomials into irreducible factors in Fp[x].(a) x3 + x2 + x+ 1, p = 2, (b) x2 − 3x− 3, p = 5, (c) x2 + 1, p = 7
Solution for (a). Let f(x) = x3 + x2 + x+1. Since f(1) = 0, we can then factor outx + 1 to get f(x) = (x + 1)(x2 + 1). Now since 12 + 1 = 0, x2 + 1 can be factoredinto (x+ 1)(x+ 1), and so we have f(x) = (x+ 1)3.
Solution for (b). Let f(x) = x2 − 3x − 3. Since f(1) = f(2) = 0, we claim f(x) =(x−1)(x−2); this follows since (x−1)(x−2) = x2−3x+2 ≡ x2−3x−3 mod 5.
Solution for (c). Since f(x) = x2 + 1 6= 0 for all x ∈ F7 (f(0) = 1, f(1) = 2, f(2) =5, f(3) = 3, f(4) = 3, f(5) = 5, f(6) = 2), we know it is irreducible in F7[x].
Exercise 12.2.2. Compute the greatest common divisor of the polynomials x6+x4+x3 + x2 + x+ 1 and x5 + 2x3 + x2 + x+ 1 in Q[x].
Solution. We perform the Euclidean algorithm (p. 45):
Exercise 12.2.5 (partial fractions for polynomials).(a) Prove that every element of C(x) can be written as a sum of a polynomial and
a linear combination of functions of the form 1/(x− a)i.(b) Exhibit a basis for the field C(x) of rational functions as vector space over C.
46
Proof of (a). We first show that f(x)/(x − a)i ∈ C(x) can be so written, where(x − a)i - f(x) ∈ C[x]. We apply the Euclidean algorithm (p. 45) considering f(x)as an element in C[x] i times to obtain
for some rj ∈ C, and qi(x) ∈ C(x). Dividing this out by (x− a)i, we have
f(x)
(x− a)i= qi(x) +
ri−1
(x− a)+ · · ·+ r2
(x− a)i−1+
r1(x− a)i
.
Now suppose we have arbitrary f(x)/g(x) ∈ C(x), where we can assume withoutloss of generality that gcd(f, g) = 1. We can first write g(x) =
∏ni=1(x − ai)
αi forai ∈ C, αi ∈ N, by the fundamental theorem of algebra (Thm. 15.10.1) (and movingany unit factor into f(x)). If n = 1, we are done by the above, and so we proceed byinduction on n. If g(x) has n distinct roots, we can write g(x) = (x−a)jh(x) for somea ∈ C, j ∈ N, h(x) ∈ C[x] where h(x) has n− 1 distinct roots. Since (x − a)j, h(x)share no roots, they are relatively prime and so 1 = r(x)(x−a)j + s(x)h(x) for somer(x), s(x) ∈ C[x]. Multiplying throughout by f(x)/g(x) gives
f(x)
g(x)=r(x)
h(x)+
s(x)
(x− a)j,
where the first term can be written as in the statement by inductive hypothesis, andthe second by the above paragraph.
Proof of (b). We claim that
B = {xi : i ∈ N}⋃{
1
(x− a)i: a ∈ C, i ∈ N
}is a basis for C(x) over C. We see that this set spans C(x) by (a); it suffices to showthat any finite subset of this set is linearly independent to show this is a basis. LetS ( B be finite, and suppose it is linearly dependent. Then,∑
j∈J
cjxj +∑i∈I
bi(x− ai)αi
= 0,
47
where J indexes the xj ∈ S and I the 1/(x− ai)αi ∈ S. Let D(x) be the product of
all the denominators appearing in the sum on the right; then,
D(x)
(∑j∈J
cjxj +∑i∈I
bi(x− ai)αi
)= 0,
and so cj = 0 for all j since the powers of x that come from the sum over J havehigher degree than those that come from the sum over I. Thus, we have
D(x)∑i∈I
bi(x− ai)αi
= 0.
But then, all the bi must equal zero, for suppose not. Then, we can write
D(x)∑
{i∈I|a1 6=ai}
bi(x− ai)αi
= −D(x)∑
{i∈I|a1=ai}
bi(x− ai)αi
Note that the left side has all terms divisible by x − a1, while the right side has noterms divisible by x− a1. Suppose the right side is zero; then the left side is nonzeromod x− a′ for any a′ 6= a1, while the right side is zero mod x− a′. If the right sideis nonzero, then the left side is zero mod x − a1 and the right side is nonzero modx− a1. In either case this is a contradiction, and so cj = bi = 0 for all i, j, and B islinearly independent.
Exercise 12.2.6. Prove that the following are Euclidean domains:(a) Z[ω], ω = e2πi/3; (b) Z[
√−2].
Proof. Let α = ω or√−2. In either case Z[α] = {x+ yα | x, y ∈ Z}. We claim that
σ(x+ yα) := |x+ yα|2 is an appropriate size function. This simplifies to
Multiplying throughout by |a|2 and letting r = b− aq, we either get r = 0 or
σ(r) = |b− aq|2 < |a|2 = σ(a),
and so our function is an appropriate size function.
Exercise 12.2.9. Let F be a field. Prove that the ring F [x, x−1] of Laurent polyno-mials (Chapter 11, Exercise 5.7) is a principal ideal domain.
Proof. It suffices by Prop. 12.2.7 to show that F [x, x−1] is a Euclidean domain. Wedefine the size function σ(f) as deg+(f)−deg−(f), where deg+(f) denotes the largestpower of x in f and deg−(f) denotes the smallest power of x in f , both includingnegative powers. σ(f) ∈ Z≥0 by considering possible values for the degree function.Now consider the Euclidean algorithm on f where we divide by g. Let p = x− deg−(f)
and q = x− deg−(g); note they are both units in F [x, x−1]. Then, fp, gq have allnon-negative exponents and
deg(gq) = deg+(gq) = deg+(g)− deg−(g) = σ(g).
Thus, fp, gq ∈ F [x]. We can then apply the Euclidean algorithm (p. 45) on fp, gqin F [x] to get (fp) = r(gq) + s, where r, s ∈ F [x], and s = 0 or deg(s) < deg(gq)by the Euclidean algorithm in F [x]. Then, dividing by p throughout, we get f =g(rqp−1) + sp−1, and (12.2.4) holds since either sp−1 = 0, or
σ(sp−1) = σ(s) = deg(s) < deg(gq) = σ(g).
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12.3 Gauss’s Lemma
Exercise 12.3.1. Let ϕ denote the homomorphism Z[x] → R defined by(a) ϕ(x) = 1 +
√2, (b) ϕ(x) = 1
2+√2.
Is the kernel of ϕ a principal ideal? If so, find a generator.
Solution for (a). We want to find the polynomials f(x) such that f(1 +√2) = 0.
Using the canonical embedding of Z[x] into R[x], we see that (x− (1 +√2)) would
generate the kernel of ϕ in R[x]. But then, since (x − (1 +√2))(x − (1 −
√2)) =
x2 − 2x− 1, and (x2 − 2x− 1) ⊆ Z[x], we obtain (x2 − 2x− 1) ⊆ kerϕ, as an idealin Z[x]. But since x2 − 2x − 1 is primitive and irreducible in Q[x], since its rootsare (1 ±
√2), x2 − 2x − 1 is irreducible in Z[x] by Prop. 12.3.7(a). Now suppose
that f(x) ∈ kerϕ \ (x2 − 2x − 1). By the Euclidean algorithm dividing f(x) byx2 − 2x− 1, we can assume without loss of generality that f(x) = ax+ b. But then,ϕ(f) = a+a
√2+b 6= 0 unless a = b = 0, and so we see that (x2−2x−1) = kerϕ.
Solution for (b). We proceed similarly to (a). Considering the embedding of Z[x]into R[x], we first see that (x− (1
2+√2)) would generate the kernel in R[x], which
contains (x − (12+
√2))(x − (1
2−
√2)) = x2 − x − 7
4. x2 − x − 7
4is in Q[x], and is
irreducible since the roots are in the extension R[x] as in (a). But then since Q isa field, (4x2 − 4x− 7) =
(x2 − x− 7
4
)as ideals. This former ideal is generated by a
primitive and irreducible element in Q[x], which as an element of Z[x] is irreducibleas well by Prop. 12.3.7(a). We thus have (4x2 − 4x− 7) ⊆ kerϕ. Now suppose thatf(x) ∈ kerϕ \ (4x2 − 4x− 7). But taking the canonical embedding of Z[x] into Q[x],and seeing that the latter is a principal ideal domain since Q is a field, we obtain thatf(x) ∈ (4x2−4x−7) as an ideal in Q[x]. But then, 4x2−4x−7 | f(x) in Q[x], and byThm. 12.3.6, this implies 4x2−4x−7 | f(x) in Z[x]; thus (4x2−4x−7) = kerϕ.
Exercise 12.3.2. Prove that two integer polynomials are relatively prime elementsof Q[x] if and only if the ideal they generate in Z[x] contains an integer.
Proof. f, g ∈ Z[x] are relatively prime in Q[x] if and only if af + bg = 1 for somea, b ∈ Q[x]. By clearing denominators of a, b, this is true if and only if a′f + b′g = dfor some a′, b′ ∈ Z[x], d ∈ Z, which is equivalent to saying (f, g) ∩ Z 6= ∅.
Exercise 12.3.4. Let x, y, z, w be variables. Prove that xy− zw, the determinant ofa variable 2× 2 matrix, is an irreducible element of the polynomial ring C[x, y, z, w].
Proof. Suppose xy − zw factors as pq for p, q /∈ C, where p = a0 + a1 + a2 and q =b0+ b1+ b2 for ai, bi homogeneous of degree i or zero. We claim p, q are homogeneousof degree one; we proceed by comparing degrees. a0b0 = 0 hence without loss of
50
generality a0 = 0. a1b0 = 0 as well; if a1 = 0, then b1 = b2 = 0, contradicting thatq /∈ C, hence a1 6= 0, b0 = 0. This implies b2 = 0, hence p, q are homogeneous ofdegree one.
Thus, since xy− zw = x2(y/x− z/x ·w/x) = x2(p/x)(q/x), xy− zw is reducibleonly if y/x − z/x · w/x is reducible in C[y/x, z/x, w/x]. Relabeling variables andtaking the contrapositive, it suffices to show t − rs is irreducible in C[r, s, t]. ButC[r, s, t]/(t − rs) ≈ C[r, s], a domain, hence t − rs is prime (Prop. 13.5.1(a)) andthus irreducible since C[r, s, t] is a domain (Lem. 12.2.10).
Exercise 12.3.6. Let α be a complex number. Prove that the kernel of the substitu-tion map Z[x] → C that sends x α is a principal ideal, and describe its generator.
Proof. Let ϕ be this map. By the mapping property of fractions (Thm. 11.7.2(c),which is still true if ϕ is not injective), we have the commutative diagram
Z[x] C
Q[x]
ϕ
ψ
Since Q[x] is a Euclidean domain, kerψ is principal. We claim it is generated by thepolynomial f of minimal degree which has α as a root, or zero if α is transcendental.For, in the former case, if g ∈ kerϕ, we can write g = fq + r by the Euclideanalgorithm, and r 6= 0 implies deg r < deg f , contradicting the minimality of f .
We claim if f = cf0 as in Lem. 12.3.5 for f0 primitive, then kerϕ = (f0).(f0) ⊂ kerϕ by the commutativity of the diagram, so it suffices to show the op-posite inclusion. Let g ∈ kerϕ. Then, f | g in Q[x] by the above, hence f0 | g in Z[x]by Thm. 12.3.6(a). Thus kerϕ is generated by f0, the primitive polynomial obtainedfrom f as above, or is zero if α is transcendental.
12.4 Factoring Integer Polynomials
Exercise 12.4.4. Factor the integer polynomial x5 + 2x4 + 3x3 + 3x + 5 mod 2,mod 3, and over Q.
Remark. We denote f(x) = x5 + 2x4 + 3x3 + 3x+ 5.
Solution for F2. f(x) = x5 + x3 + x+ 1 over F2. Then, f(1) = 0, hence f(x) = (x+1)(x4+x3+1). Now x4+x3+1 has no roots in F2 hence has no linear factors; x
2+x+1is the only irreducible quadratic in F2[x] and has (x2+x+1)2 = x4+x2+1 6= x4+x3+1,hence we cannot factor further.
51
Solution for F3. f(x) = x5 + 2x4 + 2 over F3. Then, f(2) = 0, hence f(x) =(x + 1)(x4 + x3 + 2x2 + x + 2). But 2 is also a root of x4 + x3 + 2x2 + x + 2,hence f(x) = (x + 1)2(x3 + 2x + 2), and we cannot factor further since x3 + 2x + 2has no roots in F3.
Solution for Q. From (a), (b), we suspect that −1 is a root. Indeed, f(x) = (x +1)(x4 + x3 + 2x2 − 2x + 5). The residue x4 + x3 + 1 of the second factor in F2[x] isirreducible as in (a), and so x4+x3+2x2−2x+5 is irreducible in Q[x] by Prop. 12.4.3.Thus we cannot factor further.
Exercise 12.4.6. Factor x5 + 5x+ 5 into irreducible factors in Q[x] and in F2[x].
Solution for Q[x]. x5 + 5x + 5 is irreducible in Q[x] by the Eisenstein criterion(Prop. 12.4.6) since 5 doesn’t divide the leading coefficient 1 while it divides allother coefficients, and 52 = 25 does not divide the constant term 5.
Solution for F2[x]. The residue in F2[x] is x5+x+1, which is reducible in F2[x] since
x5+x+1 = (x3+x2+1)(x2+x+1); each of these factors have no roots in F2 hencewe cannot factor further.
Exercise 12.4.7. Factor x3 + x+ 1 in Fp[x], when p = 2, 3, and 5.
Remark. We denote f(x) = x3 + x+ 1.
Solution for p = 2. Since f(0) = 1, f(1) = 1, we see that there are no roots in F2,and so x3 + x+ 1 is irreducible.
Solution for p = 3. Since f(1) = 0, we can factor out (x + 2) to get f(x) = (x +2)(x2 + x+ 2). We cannot factor further since x2 + x+ 2 has no roots in F3.
Solution for p = 5. Since f(0) = 1, f(1) = 3, f(2) = 1, f(3) = 1, f(4) = 4, we seethat there are no factors of f(x) in F5 and so it is irreducible.
Exercise 12.4.12. Determine:(a) the monic irreducible polynomials of degree 3 over F3,(b) the monic irreducible polynomials of degree 2 over F5,(c) the number of irreducible polynomials of degree 3 over the field F5.
Remark. It suffices to make sure our polynomials have no roots, for any factorizationof a quadratic or a cubic will produce a linear factor.
52
Solution for (a). We consider all polynomials of the form x3 + ax2 + bx + c in F3.c = 1, 2 since otherwise x = 0 would be a root. Moreover, 1 + a + b + c 6≡ 0 mod 3and 2 + a+ 2b+ c 6≡ 0 mod 3 must be true for irreducibility. Suppose c = 1. Then,1+ a+ b+1 6≡ 0 implies a+ b 6≡ 1. Likewise, 2+ a+2b+1 ≡ a+2b 6≡ 0, and so theonly ordered triples that work for c = 1 are (0, 2, 1), (1, 2, 1), (2, 0, 1), (2, 1, 1), i.e.,
are monic irreducible polynomials of degree 3 over F3.
Solution for (b). We consider polynomials of the form x2+ax+ b in F5. b = 1, 2, 3, 4since otherwise x = 0 would be a root. Moreover, we have the system of equations
1 + a+ b 6≡ 0
4 + 2a+ b 6≡ 0
4 + 3a+ b 6≡ 0
1 + 4a+ b 6≡ 0
=⇒
{a, 4a 6≡ 4− b,
2a, 3a 6≡ 1− b.
We can consider this system for each b value. For b = 1, we have{a, 4a 6≡ 3
2a, 3a 6≡ 0=⇒ a = 1, 4.
For b = 2, we have {a, 4a 6≡ 2
2a, 3a 6≡ 4=⇒ a = 0, 1, 4.
For b = 3, we have {a, 4a 6≡ 1
2a, 3a 6≡ 3=⇒ a = 0, 2, 3.
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For b = 4, we have {a, 4a 6≡ 0
2a, 3a 6≡ 2=⇒ a = 2, 3.
Thus, our monic irreducible polynomials of degree 2 over F5 are
Solution for (c). There are 5 monic polynomials of degree 1, and from (b) there are10 monic irreducible polynomials of degree 2. The irreducible monic polynomials ofdegree 3 are those that cannot be obtained by multiplying together polynomials oflower degree. We see that any choice of factors gives a unique polynomial of degree3 by Prop. 12.2.14(c), and so it suffices to count all monic polynomials of degree 3 inF5[x], and subtract those that can be obtained by multiplying together polynomialsof lower degree.
We see that reducible polynomials of degree 3 must be a product of 3 polynomialsof degree 1 or a product of 2 polynomials of degree 1 and 2 respectively. The numberof polynomials in the former category is(
5
3
)+
(5
1
)(4
1
)+
(5
1
)= 35,
where the terms describe all different, 2 same, and all same factors. The numberof polynomials in the latter category is 5 × 10 = 50. Since the number of degree 3monic polynomials is 53 = 125, we see that there are 125− (35 + 50) = 40 degree 3monic irreducible polynomials in F5.
Since we want to find the number of (not necessarily monic) irreducible polyno-mials of degree 3 in F5[x], we see that we can multiply our 40 monic polynomials byany element of F5 \ {0}; this results in 160 irreducible polynomials of degree 3.
Exercise 12.4.13 (Lagrange interpolation formula).(a) Let a0, . . . , an be distinct complex numbers. Determine a polynomial p(x) of
degree n, which has a1, . . . , an as roots, and such that p(a0) = 1.(b) Let a0, . . . , ad and b0, . . . , bd be complex numbers, and suppose that the ai are
distinct. There is a unique polynomial g of degree ≤ d such that g(ai) = bi foreach i = 0, . . . , d. Determine the polynomial g explicitly in terms of ai and bi.
54
Solution for (a). We let
pi(x) =∏j 6=i
x− ajai − aj
.
Any aj with j 6= i is a root since pi(x) = 0 ⇐⇒ x−aj = 0 for some j, and p(ai) = 1since each term in the product would equal 1. Letting p(x) = p0(x), we are done.
Solution for (b). We let
g(x) =d∑i=0
bipi(x) =d∑i=0
bi∏j 6=i
x− ajai − aj
,
where pi(x) is as in (a). g(ai) = bi for all i since pi(aj) = δij. deg g ≤ d by the factthat each term in the sum has degree d.
We now prove uniqueness. Suppose h(x) is another such polynomial. Then,f(x) = g(x) − h(x) has degree ≤ d and therefore has ≤ d roots; however, h(ai) =g(ai)− h(ai) = bi − bi = 0 for all 0 ≤ i ≤ d. But then, f(x) has d+ 1 distinct roots,which contradicts Prop. 12.2.20 unless g(x) = h(x).
Exercise 12.4.15. With reference to the Eisenstein criterion, what can one saywhen(a) f is constant, (b) f = xn + bxn−1?
Claim (a). Let f(x) = anxn + · · · + a0 ∈ Z[x] and let p ∈ Z be prime. Then f is
irreducible in Q[x] if (i) p - a0; (ii) p | ai for 1 ≤ i ≤ n; (iii) p2 - an. Note that (i)and (ii) hold if and only if f is a nonzero constant.
Proof of Claim (a). Suppose not; then f = gh for g = brxr + · · · + b0, h = csx
s +· · ·+c0, r+s = n. Since then f = gh = a0 6= 0 by (i) and (ii), we see that g = b0 andh = c0 since they must divide a0. But then p | br and p | cs imply p2 | brcs = an.
Claim (b). Let f(x) = anxn + · · · + a0 ∈ Z[x] and let p ∈ Z be prime. Then f is
irreducible in Q[x] if (i) an ≡ 1 mod p; (ii) p - an−1; (iii) p | ai for 0 ≤ i ≤ n − 2;(iv) p - a0. Note that (i), (ii), and (iii) hold if and only if f = xn + bxn−1.
Proof of Claim (b). Suppose not; then f = gh for g = brxr + · · · + b0, h = csx
s +· · ·+ c0, r+ s = n. Since then f = gh = xn+ an−1x
n−1 = xn−1(x+ an−1) by (i), (ii),and (iii), without loss of generality x + an−1 | g, and so g = brx
r−1(x + an−1) andh = csx
s. But then, p | c0 implies p | b0c0 = a0.
Exercise 12.4.19. Factor x5 − x4 − x2 − 1 mod 2, mod 16, and over Q.
55
Remark. We denote f(x) = x5 − x4 − x2 − 1.
Solution for F2. f(x) = x5 + x4 + x2 + 1 over F2. Then, f(1) = 0, and so f(x) =(x+1)(x4 + x+1). x4 + x+1 has no roots in F2, and the only irreducible quadraticin F2[x] is x
Solution for Z/16Z. f(−5) ≡ 0 mod 16, and so f(x) = (x+5)(x4−6x3−2x2+9x+3).This does not factor further, for a factorization of x4−6x3−2x2+9x+3 over Z/16Zwould produce a factorization of x4 + x+ 1 over F2, contradicting (a).
Solution for Q. The only possible factorization is as in (b). But the constant term ofthe linear factor in (b) has constant term 5 6≡ ±1 mod 16, which would be necessaryfor f to have constant term −1, and so f is irreducible.
12.5 Gauss Primes
Exercise 12.5.2. Find the greatest common divisor in Z[i] of (a) 11+7i, 4+7i, (b)11 + 7i, 8 + i, (c) 3 + 4i, 18− i.
Proof of (a). First, (11+7i)(11−7i) = 112+72 = 170 = 2×5×17 = (1+i)(1−i)(2+i)(2− i)(4 + i)(4− i), and so 11 + 7i = (1 + i)(2− i)(4 + i) is a prime factorization.Likewise, (4+7i)(4− 7i) = 42+72 = 65 = 5× 13 = (2+ i)(2− i)(3+2i)(3− 2i), andso 4+7i = (2+ i)(3+2i) is a prime factorization. Thus, gcd(11+7i, 4+7i) = 1.
Proof of (b). 11 + 7i = (1 + i)(2 − i)(4 + i) is a prime factorization by (a). Then,(8 + i)(8 − i) = 82 + 12 = 65 = 5 × 13 = (2 + i)(2 − i)(3 + 2i)(3 − 2i), and so8 + i = (2− i)(3 + 2i). Thus, gcd(11 + 7i, 8 + i) = 2− i.
Proof of (c). Since (18− i)/(3 + 4i) = 2− 3i, gcd(3 + 4i, 18− i) = 3 + 4i.
Exercise 12.5.3. Find a generator for the ideal of Z[i] generated by 3+4i and 4+7i.
Proof. Z[i] is a PID (Props. 12.2.5(c), 12.2.7), so it suffices to find gcd(3 + 4i, 4 +7i). Since (3 + 4i)(3 − 4i) = 25 = (2 + i)2(2 − i)2, (3 + 4i) = (2 + i)(2 + i) is aprime factorization, and from Exercise 12.5.2(a), 4 + 7i = (2 + i)(3 + 2i) is a primefactorization. Thus, 2 + i is a generator of the ideal.
Exercise 12.5.5. Let π be a Gauss prime. Prove that π and π are associates if andonly if π is an associate of an integer prime, or ππ = 2.
56
Proof. Suppose π and π are associates. Then, e(a + bi) = a − bi for some unite ∈ {±1,±i}. Thus, ππ = e(a+ bi)2 = e(a2− b2+2bi). First suppose e = ±1. Then,b = 0 since ππ ∈ Z, and so ππ = ea2 = a2, since ππ > 0. By Theorem 12.5.2(a), πprime implies a is an integer prime, and so π must be an associate of a, an integerprime. Now suppose e = ±i. Then, ππ = ±i(a2 − b2 + 2bi) = ∓2b ± i(a2 − b2).Then, a2 − b2 = 0 since ππ ∈ Z, and so ππ = ∓2b. By Theorem 12.5.2(a), we seeb ∈ {±1,±2}. But since a2 − b2 = 0 and b = ±2 imply a = ±2, we have that 2 | π,which contradicts its primeness. Thus, b = ±1, and so ππ = ∓2b = 2.
Now if π is an associate of an integer prime, then π = ep for some e ∈ {±1,±i},and so π = ep; since then π/π ∈ {±1,±i}, π and π are associates. Now supposeππ = 2. Then, (a− bi)(a+ bi) = a2 + b2 = 2 = (1+ i)(1− i). By the fact that Z[i] isa UFD, this prime factorization is unique (up to a unit), and so since 1+ i = i(1− i),we see that π and π are associates.
Exercise 12.5.6. Let R be the ring Z[√−3]. Prove that an integer prime p is a
prime element of R if and only if the polynomial x2 + 3 is irreducible in Fp[x].
Proof. We consider the ring R = Z[√−3]/(p). Since R ≈ Z[x]/(x2 + 3), we can take
the quotient with respect to (p) and (x2 + 3) in different orders:
Z[x] Fp[x]
R R
killp
killx2 + 3
killx2 + 3
killp
Now R is a domain if and only if p is prime in R by Prop. 13.5.1. But R is a domainif and only if x2+3 is prime in Fp[x] as well, and since Fp[x] is a domain, this is trueif and only if x2 + 3 is irreducible in Fp[x] by Lem. 12.2.10.
Exercise 12.5.7. Describe the residue ring Z[i]/(p) for each prime p.
Solution. Recall that Z[i] ≈ Z[x]/(x2 + 1), and so Z[i]/(p) ≈ (Z[x]/(p))/(x2 + 1) ≈Fp[x]/(x2 + 1).
When p ≡ 3 mod 4, p is a Gauss prime and so Fp[x]/(x2 + 1) ≈ Fp[i] is a field byLem. 12.5.3, and so is isomorphic to Fp2 by Thm. 15.7.3(d).
When p ≡ 1 mod 4 or p = 2, p is not a Gauss prime and so x2 + 1 is reducible inFp[x] with roots α, α−1 by Lem 12.5.3. Then, (x− α) + (x− α−1) = Fp[x] as ideals,
57
and (x − α)(x − α−1) = (x2 + 1) = 0 as ideals in Fp[x]/(x2 + 1), hence by Exercise11.6.8(c) we have Fp[x]/(x2 + 1) ≈ Fp[x]/(x− α)× Fp[x]/(x− α−1) ≈ Fp × Fp.
Exercise 12.5.9. Let R = Z[ω] with ω = e2πi/3. Let p be a prime integer 6= 3. Adaptthe proof of Theorem 12.5.2 to prove the following:(a) The polynomial x2 + x+ 1 has a root in Fp if and only if p = 1 modulo 3.(b) (p) is a maximal ideal of R if and only if p = −1 modulo 3.(c) p factors in R if and only if it can be written in the form p = a2 + ab+ b2, for
some integers a, b.
Proof of (a). x2 + x + 1 has a root in Fp if and only if x3 − 1 has a nontrivial rootin Fp, i.e., F×
p has an element of order 3. This is true if and only if 3 | |F×p | =
p− 1 by corollaries to Lagrange’s theorem (Cor. 2.8.10) and the first Sylow theorem(Cor. 7.7.3). This is equivalent to having p ≡ 1 mod 3.
Proof of (b). We consider the ring R = Z[ω]/(p). Since R ≈ Z[x]/(x2 + x + 1), wecan take the quotient with respect to (p) and (x2 + x+ 1) in different orders:
Z[x] Fp[x]
R R
killp
killx2 + x+ 1
killx2 + x+ 1
killp
Now R is a field if and only if (p) is maximal in R by Prop. 11.8.2(b). But R is afield if and only if (x2 + x+ 1) is maximal in Fp[x] as well, and since Fp[x] is a PID(Props. 12.2.5(b), 12.2.7), this is true if and only if x2 + x + 1 is irreducible in Fp[x]by Cor. 12.2.9(C). But this holds if and only if p ≡ −1 mod 3 by (a) since all primeintegers 6= 3 are equivalent to ±1 mod 3.
Proof of (c). If p = a2 + ab+ b2, then p factors as (a− bω)(a− bω2) in R.Conversely, suppose p factors in R. p is not a unit in R hence is divisible by a
prime π ∈ R. Then, π | p = p, so ππ | p2 in R and Z, and so ππ = p or p2. Thelatter case is impossible for otherwise p would be an associate of π, hence irreducible.Thus, p = (a− bω)(a− bω)(a− bω)(a− bω2) = a2 + ab+ b2 for some a, b ∈ Z.
Exercise 12.5.10.(a) Let α be a Gauss integer. Assume that α has no integer factor, and that αα
is a square integer. Prove that α is a square in Z[i].
58
(b) Let a, b, c be integers such that a and b are relatively prime and a2 + b2 = c2.Prove that there are integers m and n such that a = m2 − n2, b = 2mn, andc = m2 + n2.
Proof of (a). We can decompose α = uπk11 · · · πknn for πj prime and u a unit, suchthat πi, πj are not associates for any i 6= j; we want to show 2 | kj for each j. Then,
αα = u2(π1π1)k1 · · · (πnπn)kn = c2 for some c ∈ Z.
By Theorem 12.5.2(a), each πjπj is an integer prime or a square of an integer prime.For each j, πjπj integer prime implies 2 | kj, since otherwise c2 would not be asquare integer. Now suppose πjπj is a square of an integer prime for some j. Then,p2 = πjπj implies πj is an associate of p, since this prime decomposition is unique upto units; this implies p | α, which contradicts that α has no integer factor. Finally,c2 > 0 hence u2 = 1, and so u = ±1. In either case, α is a square since i2 = −1.
Proof of (b). Not both of a, b are odd: otherwise, c2 = a2 + b2 ≡ 1 + 1 = 2 mod 4,a contradiction since all even square integers must be 0 mod 4. So assume withoutloss of generality that a is odd. Let α = a + bi. Then, αα = a2 + b2 = c2, hence by(a) letting α = (m+ ni)2 we have a = m2 − n2, b = 2mn, and c = m2 + n2.
12.M Miscellaneous Problems
Exercise 12.M.5. For which integers n does the circle x2 + y2 = n contain a pointwith integer coordinates?
Solution. We claim that x2 + y2 = n has integer solutions if and only if every prime≡ 3 mod 4 has even exponent in the prime factorization of n.
⇐ The prime factorization can be written n = ab2, where a is square free. Then,by Thm. 12.5.2(d), every p that divides a can be written p = πpπp. Letting x+ iy =b ·∏
p|a πp, we have x2 + y2 = (x− iy)(x+ iy) = ab2 = n.
⇒ Let n = x2 + y2 = (x − iy)(x + iy). If p ≡ 3 mod 4 divides a, it is a Gaussprime by Thm. 12.5.2(b), and so p | x − iy or x + iy. But then, p | x, y, hencep | x ± iy. Thus, if pk is the power of p that appears in the prime factorizationfor x − iy, then p2k is the power of p that appears in the prime factorization for(x− iy)(x+ iy) = n.
Exercise 12.M.6. Let R be a domain and let I be an ideal that is the product ofdistinct maximal ideals in two ways, say I = P1 · · ·Pr = Q1 · · ·Qs. Prove that thetwo factorizations are the same, except for the ordering of the terms.
59
Proof. Suppose there is an i such that Qj 6⊂ Pi for all j. Let ϕi : R → R/Pi be thecanonical quotient map. Then, for each j there is qj ∈ Qj \ Pi, and so∏
j
ϕi(qj) = ϕi
(∏j
qj
)6= 0 ∈ R/Pi,
which is a contradiction since∏
j qj ∈ Q1 · · ·Qs ⊂ Pi. Thus, for every i, there is a jsuch that Qj ⊂ Pi; since both ideals are maximal, we moreover have Qj = Pi. Thus,{P1, . . . , Pr} ⊂ {Q1, . . . , Qs} as sets; interchanging the role of the Pi and Qj givesthe opposite inclusion, so the two factorizations are equal.
Exercise 12.M.7. Let R = Z[x].(a) Prove that every maximal ideal in R has the form (p, f), where p is an integer
prime and f is a primitive integer polynomial that is irreducible modulo p.(b) Let I be an ideal of R generated by two polynomials f and g that have no
common factors other than ±1. Prove that R/I is finite.
Proof of (a). Let M ⊂ R be a maximal ideal. It is not principal by Exercise 11.8.1,and so there exist f1, f2 ∈M that do not share a common factor. f1, f2 do not sharea common factor in Q[x], either by Thm. 12.3.6(b). Thus, r0f1 + s0f2 = 1 for somer, s ∈ Q[x], and so clearing denominators, rf1 + sf2 = q ∈ Q for r, s ∈ Z[x], i.e.,M ∩ Z 6= (0).
Now M ∩ Z is prime, since if ab ∈ M ∩ Z, then, say, a ∈ M and moreovera ∈ Z for otherwise ab /∈ Z. Thus, M ∩ Z = (0) or (p) for some prime p; theformer case is impossible by the above. Now consider the image M ′ of M in Fp[x];then, Z[x]/M ≈ Fp[x]/M ′ is a field, and so M ′ is maximal by Prop. 11.8.2(b), andis generated by some irreducible f0 ∈ Fp[x] by Cor. 12.2.9(c) since Fp[x] is a PID byProps. 12.2.5(b), 12.2.7. Now if f ∈ Z[x] is a lift of f0, then (p, f) ⊂M since f = 0 inFp[x]/M ′; M ⊂ (p, f) since if g ∈M \ (p, f), then g 6= 0 ∈ Fp[x]/M ′ ≈ Z[x]/M .
Proof of (b). As in (a), if I = (f, g) and f, g have no common factors, then I ∩ Z 6=(0); since Z is a PID, M ∩ Z = (n) for some n ∈ Z such that n 6= ±1, for otherwiseR/I = 0 and we are done. So, letting n =
∏pkii be a prime factorization, we claim
R
I≈ (Z/nZ)[x]
(f, g)≈∏
i(Z/pkii Z)[x]
(f, g)≈∏i
(Z/pkii Z)[x](f, g)
. (3)
The first isomorphism is clear, and the second follows by applying Exercise 11.6.8(c)repeatedly. For the third, consider the canonical surjection
π :∏i
(Z/pkii Z)[x] →∏i
(Z/pkii Z)[x](f, g)
.
60
kerπ = (f, g) since an element on the left maps to zero on the right if and onlyif each direct factor is in (f, g), and so (3) holds by the first isomorphism theorem(Thm. 11.4.2(b)). Let each direct factor on the right be called Ri.
It suffices to show each Ri is finite. Since f, g do not share a common factor,without loss of generality pi - f . Then, separating out the terms in f , we can writef = f1 − f2 where pi - f1 while pi | f2. Then, pkii | fki2 , hence in (Z/pkii Z)[x],
fki1 = fki1 − fki2 = (f1 − f2)h = fh ∈ (f, g)
for some h ∈ (Z/pkii Z)[x]. If a is the leading coefficient of f1, then fki1 has leading
coefficient aki . This is a unit since if not, then (aki) is contained in a maximal idealof Z/pkii Z by Thm. 11.9.2, but (pi) is the unique maximal ideal of this ring by thecorrespondence theorem (Thm. 11.4.3), contradicting that pi - a. Thus, m = a−kifki1is a monic polynomial in (Z/pkii Z)[x], hence each Ri is finite since any polynomialwith degree ≥ degm can be reduced to a polynomial of lower degree using thatm = 0, and there are only finitely many polynomials in (Z/pkii Z)[x] with degree< degm since Z/pkii Z is finite.
Exercise 12.M.8. Let u and v be relatively prime integers, and let R′ be the ringobtained from Z by adjoining an element α with the relation vα = u. Prove that R′
is isomorphic to Z[uv
]and also to Z
[1v
].
Proof. We see that R′ = Z[x]/(vx − u) ≈ Z[α]. Now since vα = u by construction,we see that having α u
vgives an isomorphism Z[α] ≈ Z
[uv
], and so R′ ≈ Z
[uv
].
To show R′ ≈ Z[1v
], we first see that Z
[1v
]⊂ Z
[uv
]as a subring, since u · 1
v= u
v.
Since u, v are relatively prime, au+ bv = 1 for some a, b ∈ Z. Then, dividing out byv gives a · u
v+ b = 1
v. Thus, 1
v∈ Z
[uv
], and so Z
[1v
]= Z
[uv
].
13 Quadratic Number Fields
13.1 Algebraic Integers
Exercise 13.1.4. Let d and d′ be integers. When are the fields Q(√d) and Q(
√d′)
distinct?
Solution. Write d = s2e and d′ = s′2e′ such that e, e′ are square-free. We claimQ(
√d) ≈ Q(
√d′) if and only if e = e′. ⇐ clearly holds by p. 384, so it suffices to
show ⇒. So suppose e 6= e′, and without loss of generality e 6= 1. We claim thereis no x + y
√e′ with x, y ∈ Q such that e = (x + y
√e′)2 = x2 + 2xy
√e′ + y2e′. This
implies 2xy = 0 and x2+y2e′ = e. If x = 0, then y2e′ = e, so if y = a/b for (a, b) = 1,
61
we have a2e = b2e′; since e, e′ are square-free, then a = b = 1, so e = e′. On theother hand, if y = 0, then x2 = e, contradicting that e 6= 1 is square-free.
13.2 Factoring Algebraic Integers
Exercise 13.2.2. For which negative integers d = 2 modulo 4 is the ring of integersin Q[
√d] a unique factorization domain?
Solution. Suppose d ≡ 2 mod 4. The integers R in Q[√d] are of the form a + b
√d
for a, b ∈ Z (Prop. 13.1.6). We claim R is a UFD if and only if d = −2. ⇐. Z[√−2]
is a Euclidean domain (Exercise 12.2.6), hence a UFD by Props. 12.2.7, 12.2.14(b).⇒. Suppose d 6= −2. Consider 2(2 − d/2) = 4 − d = (2 −
√d)(2 +
√d). We claim
2 is irreducible in R; as on p. 386 it suffices to show there is no a + b√d such that
N(a + b√d) = a2 − b2d = 2. For, this would imply 2 | d hence 2 | a2, and so 2 | a,
which implies a = 0, b = 1, d = −2 is the only possibility, a contradiction. Thus, forR to be a UFD we must have either 2 | 2−
√d or 2 +
√d, but 1±
√d/2 /∈ R when
d ≡ 2 mod 4 by Prop. 13.1.6.
13.3 Ideals in Z[√−5]
Exercise 13.3.2. Let δ :=√−5. Decide whether or not the lattice of integer
combinations of the given vectors is an ideal: (a) (5, 1 + δ), (b) (7, 1 + δ), (c)(4− 2δ, 2 + 2δ, 6 + 4δ).
Remark. We denote A to be the lattice spanned by the given vectors.
Solution for (a). A is not an ideal. For, suppose it were; then, δ(1+δ) = −5+δ ∈ A,and so −5 + δ = 5a+ (1 + δ)b for some a, b ∈ Z. Thus, b = 1, but −5 = 5a+ 1 hasno solution a ∈ Z.
Solution for (b). A is not an ideal, for if it were, then as in (a), δ(1 + δ) = −5+ δ =7a+ (1 + δ)b ∈ A implies b = 1, but −5 = 7a+ 1 has no solution a ∈ Z.
Solution for (c). We claim A is an ideal. A is closed under addition since it is alattice, and so it suffices to show A is closed under external multiplication. By thedistributive property, it is moreover sufficient to show δ times any of the generatorsis in A, for A is closed under integer combinations. But this follows since
Exercise 13.3.3. Let A be an ideal of the ring of integers R in an imaginaryquadratic field. Prove that there is a lattice basis for A, one of whose elementsis an ordinary positive integer.
Proof. Let a′ ∈ A; then a′a′ ∈ A is a positive integer hence A∩N is nonempty. Choosea ∈ A ∩ N that is minimal, and choose b′ ∈ A which is Z-linearly independent froma′. Consider the parallelogram Π(a, b′) = {ra + sb′ | 0 ≤ r, s ≤ 1}. A ∩ Π(a, b′) isfinite, and so we can choose b ∈ A ∩ Π(a, b′) with minimal positive imaginary part.We claim a, b form a lattice basis for A. First, A ∩ Π(a, b) = {0, a, b, a + b}, for ifc ∈ (A∩Π(a, b))\{0, a, b, a+ b}, then either c ∈ Z but c < a, violating minimality ofa, or 0 < Im(c) < Im(b), violating minimality of b. Finally, we can move Π(a, b) byZ-linear combinations of a, b, and none of these parallelograms contain anything inA other than linear combinations of a, b for otherwise we can move that point intothe parallelogram Π(a, b) with an appropriate Z-linear combination of a, b. Thus,a, b form a lattice basis for A.
13.4 Ideal Multiplication
Exercise 13.4.3. Let R be the ring Z[δ], where δ =√−5, and let A and B be ideals
of the form A = (α, 12(α + αδ)), B = (β, 1
2(β + βδ)). Prove that AB is a principal
ideal by finding its generator.
Solution. By Prop. 13.1.6, we know 2 | α and 2 | β. By Prop. 13.4.3(c), we haveA = α
2(2, 1 + δ) and B = β
2(2, 1 + δ). Thus, by (13.4.2) and Prop. 13.4.3(a), we get
AB =αβ
4(2, 1 + δ)2 =
αβ
4(4, 2 + 2δ,−4 + 2δ) =
αβ
4(2) =
(αβ
2
).
13.5 Factoring Ideals
Exercise 13.5.2. Let δ =√−3 and R = Z[δ]. This is not the ring of integers in
the imaginary quadratic number field Q[δ]. Let A be the ideal (2, 1 + δ).(a) Prove that A is a maximal ideal, and identify the quotient ring R/A.(b) Prove that AA is not a principal ideal, and that the Main Lemma is not true
for this ring.(c) Prove that A contains the principal ideal (2) but that A does not divide (2).
Proof of (a). It suffices to show R/A is a field by Prop. 11.8.2(b). But R/A =Z[δ]/(2, 1 + δ) ≈ Z/(2) = F2.
Now if the Main Lemma holds and AA is a principal ideal, then by the cancellationlaw (Cor. 13.4.9(a)) we have A = (2), a contradiction since 1 + δ ∈ A \ (2).
Proof of (c). (2) ⊂ A since 2 is a generator of A. Now suppose AB = (2) for someideal B ( R. B 6⊂ A for otherwise
which implies 1 ∈ A, contradicting that A is maximal from (a). Then, 2A = AAB =2AB by (b), and so A = AB. Now let C be a maximal ideal containing B as existsby Thm. 11.9.2; note C 6= A since B 6⊂ A. Then, A = AB ⊂ AC ⊂ A, and so wehave equalities throughout. But this contradicts Exercise 12.M.6, for A and AC aretwo distinct factorizations of A into a product of distinct maximal ideals.
14 Linear Algebra in a Ring
14.1 Modules
Exercise 14.1.3. Let R = Z[α] be the ring generated over Z by an algebraic integerα. Prove that for any integer m, R/mR is finite.
Proof. Let f(x) ∈ Z[x] be the (monic) irreducible polynomial for α. Then, I = (m, f)is an ideal in Z[x] generated by two polynomials that have no common factors, andso R/mR ≈ Z[x]/(m, f) is finite by Exercise 12.M.7(b).
Exercise 14.1.4. A module is called simple if it is not the zero module and if ithas no proper submodule.(a) Prove that any simple R module is isomorphic to an R module of the form
R/M where M is a maximal ideal.(b) Prove Schur’s Lemma: Let ϕ : S → S ′ be a homomorphism of simple modules.
Prove that ϕ is either zero, or an isomorphism.
Proof of (a). Let S be simple and let s ∈ S be nonzero. Define ψ : R → S by r 7→ rs;this is a homomorphism since S is a module. Then, imϕ is a submodule of S, henceequal to S since S is simple, and so ψ is surjective. By the first isomorphism theorem(Thm. 14.1.6(c)), R/ker ϕ ≈ S. Now M := kerϕ is a submodule of R, then by thecorrespondence theorem (Thm. 14.1.6(d)) M is a maximal submodule of R, hence amaximal ideal of R by Prop. 14.1.3.
64
Proof of (b). kerϕ is a submodule of S, and imϕ is a submodule of S ′. Since S ′ issimple, imϕ is 0 or S ′. If imϕ = 0, then ϕ = 0. If imϕ = S ′, then kerϕ ( S ′, henceequals 0, and so S ≈ S ′ by the first isomorphism theorem (Thm. 14.1.6(c)).
14.2 Free Modules
Exercise 14.2.3. Let A be the matrix of a homomorphism ϕ : Zn → Zm of freeZ-modules.(a) Prove that ϕ is injective if and only if the rank of A, as a real matrix, is n.(b) Prove that ϕ is surjective if and only if the greatest common divisor of the
determinants of the m×m minors of A is 1.
Remark. By Thm. 14.4.6, there exist matrices P ∈ GLn(Z), Q ∈ GLm(Z) such thatA′ = Q−1AP is diagonal of the form
d1. . .
dk0
(4)
with di ∈ Z>0 and d1 | d2 | · · · | dk. Note A defines an injective (resp. surjective) mapϕ : Zn → Zm if and only if A′ defines an injective (resp. surjective) map ϕ′ : Zn → Zmsince P,Q are invertible as integer matrices.
Proof of (a). ϕ is injective if and only if A′ above has k = n ≤ m, and this isequivalent to A having rank n as a real matrix since rank is preserved by columnand row operations P,Q.
Proof of (b). ϕ is surjective if and only if A′ above has d1 = · · · = dk = 1 andk = m ≥ n, i.e., the greatest commmon divisor of the determinants of the m ×mminors of A′ is 1. It suffices to show that this property is preserved by column androw operations P,Q.
Recall (14.4.2): the operations are (i) adding an integer multiple of a row (resp.column) to another, (ii) interchanging two rows (resp. columns), and (iii) multi-plying a row (resp. column) by −1. But (i) simply adds integer multiples of somedeterminants to others, (ii) switches some pairs of determinants and multiplies someby −1, and (iii) multiplies some determinants by −1. So we are done.
65
14.4 Diagonalizing Integer Matrices
Exercise 14.4.6. Let ϕ : Zk → Zk be a homomorphism given by multiplication byan integer matrix A. Show that the image of ϕ is of finite index if and only if A isnon-singular that it if so, then the index is equal to |detA|.
Proof. Since Zk is abelian, the index of imϕ 6 Zk is given by |Zk/ imϕ|. ByThm. 14.4.6, there exist P,Q ∈ GLk(Z) such that A′ = Q−1AP is diagonal of theform (4), with diagonal entries d1 | d2 | · · · | dr followed by k − r zeros, and so
Zk/ imϕ = Zk/AZk ≈ Zk/A′Zk ≈r∏i=1
(Z/d1Z)× Zk−r. (5)
Thus, |Zk/ imϕ| <∞ if and only if r = k. But this is true if and only if
detA = detQ detA′ detP = detQ detPk∏i=1
di 6= 0, (6)
i.e., detA is nonsingular. If |Zk/ imϕ| < ∞, then (5) gives |Zk/ imϕ| =∏k
i=1 di,which equals |detA| by (6).
14.5 Generators and Relations
Exercise 14.5.1. Let R = Z[δ] where δ =√−5. Determine a presentation matrix
as an R-module for the ideal (2, 1 + δ).
Proof. Let ϕ : R2 → I that sends (x, y) 2x+ (1 + δ)y. Now
kerϕ =
{(x, y) ∈ R2
∣∣∣∣ x = −1 + δ
2y
}↔{y ∈ R2
∣∣∣∣ 1 + δ
2y ∈ R
}where ↔ denotes bijection. But y = 2, 1−δ are the elements in R with smallest normsatisfying this condition, and moreover the corresponding vectors (−3, 1− δ), (−1−δ, 2) ∈ R are independent over R since 2r 6= 1 − δ for any r ∈ R. Thus, kerϕ isgenerated by these two vectors, and by pp. 424–425 we have a presentation matrix
A =
(−3 −1− δ1− δ 2
).
66
14.7 Structure of Abelian Groups
Exercise 14.7.7. Let R = Z[i] and let V be the R-module given by the elements v1and v2 with relations (1 + i)v1 + (2− i)v2 = 0 and 3v1 +5iv2 = 0. Write this moduleas a direct sum of cyclic modules.
Proof. We use the relations to find a presentation matrix A and then diagonalize itfollowing the proof of Thm. 14.4.6:(
1 + i 32− i 5i
) (1 + i 3−i −3 + 8i
) (−i −3 + 8i1 + i 3
)
(1 −8− 3i
1 + i 3
) (1 −8− 3i0 8 + 11i
)
(1 0
−1+i 1
)·
(0 11 0
)·
(i 01 0
)·
(1 0
−1−i 1)·
This reduces to the 1×1 matrix (8+11i) by Prop. 14.5.7(iv), and so V ≈ R/(8+11i).To decompose further, we need to factor 8 + 11i. But
and i(2− i)(6− i) = i(12− 8i− 1) = 8 + 11i, hence V ≈ R/(8 + 11i) ≈ R/(2− i)⊕R/(6− i) by Exercise 11.6.8(c), where we note the map ϕ in Exercise 11.6.8 is alsoan R-module isomorphism.
14.8 Applications to Linear Operators
Exercise 14.8.2. Let M be a C[t]-module of the form C[t]/(t−α)n. Show that thereis a C-basis for M , such that the matrix of the corresponding linear operator is aJordan block.
Proof. By Prop. 11.5.5, 1, t, . . . , tn−1 is a basis for M over C, and following p. 435,we have the following matrix for multiplication by t in this basis:
T =
0 −a01 0 −a1
1. . .
.... . . 0 −an−2
1 −an−1
, an−i = (−α)i(n
i
)
67
by using the binomial formula. This has characteristic polynomial f(t) = (t−α)n ason p. 435, hence by the Jordan normal form it suffices to show dimker(αI − T ) = 1.By rank-nullity it suffices to show rk(αI−T ) ≥ n−1, for an eigenvalue has geometricmultiplicity at least one. For α = 0, we are done, so suppose α 6= 0. Then, the matrix
αI − T =
α a0−1 α a1
−1. . .
.... . . α an−2
−1 α + an−1
turns into the matrix
1 a0/α1 a1/α + a0/α
2
. . ....
1 an−2/α + · · ·+ a0/αn−1
1 + an−1/α + an−2/α2 + · · ·+ a0/α
n
by Gaussian elimination, and so rk(αI − T ) ≥ n− 1.
Exercise 14.8.4. Let V be an F [t]-module, and let B = (v1, . . . , vn) be a basis forV as F -vector space. Let B be the matrix of T with respect to this basis. Prove thatA = tI −B is a presentation matrix for the module.
Proof. We want to show that imA = ker(p : F [t]n → V ) where p is given by (fi)ni=1 ∑n
i=1 fivi. The inclusion ⊂ is clear, for B is multiplication by t on V .For the other direction, suppose (fi)
ni=1 ∈ ker p, i.e.,
∑ni=1 fivi = 0. If each
fi =∑aijt
j, we then have
n∑i=1
fivi =n∑i=1
∑j≥0
aijtjvi
=n∑i=1
[ai0vi +
∑j≥1
aijtj−1((t−B)vi +Bvi)
]
= (p ◦ A)
(∑j≥0
aijtj−1
)n
i=1
+n∑i=1
[ai0vi +
∑j≥1
aijtj−1Bvi
]
=n∑i=1
[ai0vi +
∑j≥1
aijtj−1Bvi
],
68
since p ◦ A = 0 by the above. Then, repeating this process, we get
n∑i=1
fivi =n∑i=1
∑j≥0
aijBjvi =
n∑i=1
bivi = 0
for some coefficients bi ∈ F , i.e., each contribution from the fi above could be put inthe form (p ◦ A)(w) for some w ∈ F [t]n, and so imA ⊃ ker p.
14.M Miscellaneous Problems
Exercise 14.M.10.(a) Prove that the multiplicative group Q× of rational numbers is isomorphic to the
direct sum of a cyclic group of order 2 and a free abelian group with countablymany generators.
(b) Prove that the additive group Q+ of rational numbers is not a direct sum oftwo proper subgroups.
(c) Prove that the quotient group Q+/Z+ is not a direct sum of cyclic groups.
Proof of (a). Let 〈−1〉 = {−1, 1} with the obvious group structure making it iso-morphic to Z/2Z. Let 〈p〉 = {pk | k ∈ Z} with the obvious group structure makingit isomorphic to Z. Then, define the map
〈−1〉 ⊕⊕
p prime
〈p〉 → Q×
σ ⊕⊕
p prime
pkp σ∏
p prime
pkp
Note that⊕
〈p〉 is a free abelian group. This map is a group homomorphism since(σ, k2, . . .) · (τ, `2, . . .) gets mapped to στ
∏pkp+`p = σ
∏pkp · τ
∏p`p . This map
is clearly surjective since any rational number r/s has prime decompositions for rand s, which we use to find a preimage on the left. Similarly, this is injective sincesomething on the right is equal to 1 if and only if σ = 1 and kp = 0 for all p.
Proof of (b). Let G,H be proper nontrivial subgroups of Q+; it suffices to show theyhave nontrivial intersection by Prop. 2.11.4(d). If p/q ∈ G, s/t ∈ H for some p, s 6= 0,then qs(p/q) = pt(s/t) = ps ∈ G ∩H, while ps 6= 0 by assumption.
Proof of (c). Suppose Q+/Z+ ≈⊕
kHk for Hk cyclic. Let (the image of) pk/qk +Z+ ∈ Hk generate Hk for (pk, qk) = 1. Let spk + tqk = 1 for some integers s, t; then,
1
qk+ Z+ =
spkqk
+tqkqk
+ Z+ = spkqk
+ Z+ ∈ Hk,
69
hence we can assume that (the image of) 1/qk generates Hk. Now consider theelement 1/q21 + Z+ ∈ Q+/Z+. 1/q21 + Z+ then has a decomposition
1
q21+ Z+ =
∑k
rkqk
+ Z+,
where qk | rk for all but finitely many values of k. Adding it to itself q1 times gives
1
q1+ Z+ =
∑k
q1rkqk
+ Z+
and since we had a direct sum decomposition, qk | q1rk for all k 6= 1, and q1 | q1r1−1.But this last property implies q1 | 1, a contradiction.
15 Fields
15.2 Algebraic and Transcendental Elements
Exercise 15.2.1. Let α be a complex root of the polynomial x3 − 3x + 4. Find theinverse of α2 + α + 1 in the form a+ bα + cα2 with a, b, c ∈ QProof. Suppose 1 = (a+ bα + cα2)(1 + α + α2). We can rewrite this as
1 = a+ (a+ b)α + (a+ b+ c)α2 + (b+ c)α3 + cα4
We then note that α3 = 3α− 4 and α4 = 3α2 − 4α, and so we get the equation
Since 1, α, and α2 are linearly independent over Q, we get a system of linear equa-tions, with solutionab
c
=
1 −4 −41 4 −11 1 4
−1100
=1
49
17 12 20−5 8 −3−3 −5 8
100
=1
49
17−5−3
and so (1 + α + α2)−1 = 1
49(17− 5α− 3α2).
Exercise 15.2.3. Let β = ω 3√2 where ω = e2πi/3, and let K = Q(β). Prove that the
equation x21 + · · ·+ x2k = −1 has no solution with xi in K.
Proof. β has minimal polynomial x3−2, which has roots 3√2 = ω2β, β, and ωβ. Thus,
by Prop. 15.2.8, there is an isomorphism Q(β) ≈ Q( 3√2). But then, if x21+ · · ·+x2k =
−1 has a solution in Q(β), it has a solution xi = ai + bi3√2 ∈ Q( 3
√2) ⊂ R by
Prop. 15.2.10, contradicting that this equation has no real solutions.
70
15.3 The Degree of a Field Extension
Exercise 15.3.2. Prove that the polynomial x4 + 3x+ 3 is irreducible over the fieldQ[ 3
√2].
Proof. Since f := x4 + 3x+ 3 ≡ x4 + x+ 1 mod 2 is irreducible over F2 by Exercise12.4.19, f is irreducible over Q as well by Prop. 12.4.3.
Now let α be a root of f ; then [Q(α) : Q] = 4 while [Q( 3√2) : Q] = 3, and so
[Q(α, 3√2)] = 12 by Cor. 15.3.8. But 12 = [Q(α, 3
√2) : Q( 3
√2)][Q( 3
√2) : Q] implies
[Q(α, 3√2) : Q( 3
√2)] = 4 by the multiplicative property of the degree (Thm. 15.3.5).
Thus, the minimal polynomial of α over Q( 3√2) has degree 4, and so x4 +3x+3 has
no factors in Q( 3√2)[x] by Lem. 15.3.2(b).
Exercise 15.3.5. Determine the values of n such that ζn has degree at most 3 overQ.
Proof. If p is a prime dividing n, then ζnp = 1, hence Q(ζn) ⊃ Q(ζp). Now [Q(ζp) :Q] = p − 1 by Thm. 12.4.9, and so [Q(ζn) : Q] ≤ 3 implies that if p | n, thenp ∈ {2, 3}. Thus, n = 2i3j.
We claim i ≤ 2, j ≤ 1. [Q(ζ2) : Q] = 1 and [Q(ζ4) : Q] = 2 because the minimalpolynomial of ζ4 is x2 + 1. Similarly, [Q(ζ3) : Q] = 2. But any extension of Q(ζ4) orQ(ζ3) will have degree ≥ 4 over Q by Thm. 15.3.5, hence i ≤ 2, j ≤ 1.
By the above, n ∈ {1, 2, 3, 4, 6, 12}, and also [Q(ζn) : Q] ≤ 3 for n ∈ {1, 2, 3, 4},hence it suffices to check whether n ∈ {6, 12} are also possible values of n. ζ6 = −ζ3hence Q(ζ6) = Q(ζ3). On the other hand, Q(ζ12) ⊃ Q(ζ3, ζ4), which has degree ≥ 6over Q by Thm. 15.3.5, and so n 6= 12. Thus, n ∈ {1, 2, 3, 4, 5, }.
Exercise 15.3.7. (a) Is i in the field Q( 4√−2)? (b) Is 3
√5 in the field Q( 3
√2)?
Proof of (a). Suppose i ∈ Q( 4√−2). Then, there exists a, b ∈ Q such that (a +
b 4√−2)2 = −1. But then,
(a+ b 4√−2)2 = a2 + 2ab 4
√−2 + b2 2
√−2 = −1,
and since {1, 4√−2, 2
√−2} are linearly independent over Q by Prop. 15.2.7, a2 = −1,
which is impossible since a ∈ Q.
Proof of (b). If 3√5 ∈ Q( 3
√2), then α := 3
√2 + 3
√5 ∈ Q( 3
√2) must have degree at
most 3 over Q. Let f(x) = x9 − 21x6 − 123x3 − 343; then, f(α) = 0. But f(x) isirreducible by the Eisenstein criterion (Prop. 12.4.6) since 3 | 21, 123 while 9 - 343,hence f is irreducible and so α has degree 9 over Q, a contradiction.
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Exercise 15.3.9. Let α and β be complex roots of irreducible polynomials f(x) andg(x) in Q[x]. Let K = Q(α) and L = Q(β). Prove that f(x) is irreducible in L[x]iff g(x) is irreducible in K[x].
Proof. Let m = deg f, n = deg g. By the multiplicative property of the degree(Thm. 15.3.5), we have [Q(α, β) : Q] = [L(α) : L][L : Q] = [K(β) : K][K : Q].Hence, f(x) is irreducible in L[x] if and only if [Q(α, β) : Q] = mn if and only if g(x)is irreducible in K[x].
Exercise 15.3.10. A field extension K/F is an algebraic extension if every elementof K is algebraic over F . Let K/F and L/K be algebraic field extensions. Provethat L/F is an algebraic extension.
Proof. Let α ∈ L; we want to show α is algebraic over F . Let α =∑m
i=1 biαi forbi ∈ K, αi ∈ L linearly independent and algebraic over K. Write αi =
∑nj=1 cjβj
for cj ∈ F , βj ∈ K linearly independent and algebraic over F . Thus, we have α ∈F (β1, . . . , βn)(α1, . . . , αm). Now if α is transcendental over F , then by Thm. 15.3.5,
Now Thm. 15.3.5 implies [K : Q] = [K : Q(γ)][Q(γ) : Q] = 3 and so [Q(γ) : Q] = 3since γ /∈ Q. Thus, x3 − 5x2 + 8x− 5 is the irreducible polynomial of γ over Q.
Exercise 15.4.2. Determine the irreducible polynomial for α =√3 +
√5 over the
following fields. (a) Q, (b) Q(√5), (c) Q(
√10), (d) Q(
√15).
Solution for (a). Let f(x) = (x2 − 8)2 − 60 = x4 − 16x2 + 4. Then, f(α) = 0; weclaim f is irreducible. f cannot have linear factors since a root must be an integerdividing 4 by the rational root test, and these are not roots. Now suppose f hadquadratic factors; then by Prop. 12.3.7(a), for some a, b, c, d ∈ Z we have
Now a+ c = 0 implies ad+ bc = a(d− b) = 0. a 6= 0 since otherwise b+d = −16 andbd = 4, a contradiction. So, d = b = 2, but then 2+ ac+2 = −16 implies ac = −20,which contradicts a+ c = 0. Hence f is the irreducible polynomial for α over Q.
Solution for (b). Let f(x) = (x−√5)2 − 3 = x2 − 2
√5 + 2; then f(α) = 0. f splits
if and only if the two roots√5±
√3 are in Q(
√5). But
√3 /∈ Q(
√5), for otherwise√
3 = a+ b√5 implies 3 = a2 + 2ab
√5 + 5b2, and solving for
√5 gives that
√5 ∈ Q,
a contradiction. Hence f is the irreducible polynomial for α over Q(√5).
Solution for (c). Let f(x) = x4 − 16x2 + 4. By Cor. 15.3.8 and (a), Q(α,√10) must
have degree either 4 or 8 over Q, hence by the multiplicative property of the degree(Thm. 15.3.5),
√10 has degree 2 or 1 over Q(α). To show
√10 does not have degree
1 over Q(α), it suffices to show it is not in Q(α) by Lem. 15.3.2(b). But this is clearsince if
√10 = a+ bα for a, b ∈ Q, then
10 = a2 + 2abα + b2α2 = a2 + 8b2 + 2ab√3 + 2ab
√5 + 2b2
√15
implies b = 0, contradicting that√10 /∈ Q. Now α has degree 4 over Q(
√10) since
[Q(α,√10) : Q(
√10)][Q(
√10) : Q] = [Q(α,
√10) : Q(
√10)] · 2 = 8
and so f is the irreducible polynomial for α over Q(√10).
Solution for (d). Let f(x) = x2 − 8− 2√15; then f(α) = 0. Now
4 = [Q(α,√15) : Q] = [Q(α,
√15) : Q(
√15)][Q(
√15) : Q]
by (a) since√15 = (α2 − 8)/2 ∈ Q(α), hence α has degree 2 over Q(
√15), and so f
is the irreducible polynomial for α over Q(√15).
15.6 Adjoining Roots
Exercise 15.6.1. Let F be a field of characteristic 0, let f ′ be the derivative off ∈ F [x], and let g be an irreducible polynomial that is a common divisor of f andf ′. Prove that g2 divides f .
Proof. f = gh for some h ∈ F [x], hence f ′ = gh′ + g′h by Exercise 11.3.5(a). Ifg | f ′, then g | g′h. g is prime since it is irreducible, hence g | g′ or g | h. But g - g′since deg g′ < deg g, hence h = gr for some r ∈ F [x]. Thus, f = g2r, i.e., g2 | f .
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15.7 Finite Fields
Exercise 15.7.4. Determine the number of irreducible polynomials of degree 3 overF3 and over F5.
Proof. By Thm. 15.7.3(b), for prime p the irreducible factors of xp3 −x are the monic
irreducible polynomials in Fp[x] with degree 1 or 3. x, x−1, x−2, . . . , x− (p−1) arethe linear irreducible polynomials, hence there are p(p−1)(p+1)/3 monic irreduciblepolynomials of degree 3 in Fp[x]. Since by Thm. 15.7.3(c) there are p−1 units in Fp,we have that there are p(p−1)2(p+1)/3 irreducible polynomials of degree 3 over Fp.Thus, for F3 we have 16, and for F5 we have 160 irreducible polynomials of degree3. Note this matches our result from Exercise 12.4.12.
Exercise 15.7.6. Factor the polynomial x16 − x over the fields F4 and F8.
Solution for F4. We claim x16 − x factors as the product of all monic irreduciblepolynomials over F4 of degree 1 and 2. First, x4 − x | x16 − x by Thm. 15.7.3(a), (e),hence every monic irreducible polynomial of degree 1, x − a | x16 − x for a ∈ F4
divides x16 − x. Now let g be a monic irreducible polynomial of degree 2 over F4. Ifβ is a root of g in F16 then [F4(β) : F4] = 2, hence F4(β) ≈ F16 by Thm. 15.7.3(d).Thus, x−β | x16−x, so g | x16−x by considering the other root of g and proceedingin the same way. Thus, x16 − x is divisible by every monic irreducible polynomialover F4 of degree 1 or 2. Now there are 4 irreducible monic polynomials of degree 1and 42−
(42
)− 4 = 6 irreducible monic polynomials of degree 2 over F4, which means
the total degree of their product is 16 as desired. Thus, we have that
It suffices to show the degree 4 factors are irreducible over F8. Since there are nointermediate fields between F2 and F8 by Thm. 15.7.3(e), we know there are no newlinear factors of x16−x, since its roots are the elements in F16. Let α ∈ F8 such thatF8 ≈ F2(α). Suppose one of the degree 4 factors has a degree 2 factor. Let β be oneof its roots; by Thm. 15.7.3(a) it is an element of F16, and F16 ≈ F2(β) as in thesolution for F4. Then, [F8(β) : F8] = 2, hence F8(β) ≈ F64 by Thm. 15.7.3(d). ButF64 ≈ F8(β) ≈ F2(α, β) ≈ F2r where r = 12 by Cor. 15.3.8, a contradiction.
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Exercise 15.7.7. Let K be a finite field. Prove that the product of the nonzeroelements of K is −1.
Proof. For every nonzero a ∈ K there is a−1 ∈ K such that aa−1 = 1 since K is afield. In the product Π of nonzero elements of K, we can pair off each a, a−1 suchthat a 6= a−1 so that Π is the product of all nonzero elements of K such that a−1 = a.We claim this is true if and only if a = ±1. If a = a−1, then a2 = 1 and so a is aroot of x2 − 1 = (x+ 1)(x− 1). Since K[x] is a UFD (Prop. 12.2.14(c)), this impliesa = ±1. Thus, Π = 1 · (−1) = −1.
Exercise 15.7.8. The polynomials f(x) = x3 + x + 1 and g(x) = x3 + x2 + 1 areirreducible over F2. Let K be the field extension obtained by adjoining a root of f ,and let L be the extension obtained by adjoining a root of g. Describe explicitly anisomorphism from K to L, and determine the number of such isomorphisms.
Proof. Let α be a root of f , and β one of g. Then, α3 = α+1 and β3 = β2+1. Anyfield homomorphism ϕ : K → L is specified by ϕ(α), satisfying that ϕ(α3) = ϕ(α+1)by the mapping property of quotient rings (Thm. 11.4.2). So let ϕ(α) = a+bβ+cβ2.
must then be equal to ϕ(α + 1).We claim ϕ is an isomorphism if and only if (a, b, c) ∈ {(1, 1, 0), (1, 0, 1), (0, 1, 1)}.
By Prop. 11.8.4(b), it suffices to show ϕ is surjective if and only if this is true. Supposeϕ is surjective; then not both b, c are zero by the above. Matching 1, β, β2 terms,we have that if b = 0, then a = c = 1; if c = 0, then a = b = 1; and if b = c = 1,then a = 0. Conversely, since ϕ(α + 1) = β in the first case, ϕ(α2 + α + 1) = β inthe second case, and ϕ(α2 + 1) = β in the third case, we see in each case that ϕ issurjective by using that ϕ is an homomorphism.
Exercise 15.7.10. Let F be a finite field, and let f(x) be a nonconstant polynomialwhose derivative is the zero polynomial. Prove that f cannot be irreducible over F .
Lemma 15.7.10a. In a finite field of order pr, every element is a pth power.
Proof of Lemma 15.7.10a. Let a ∈ F ; if a = 0 we are done so suppose not. |F×| =pr − 1 by Thm. 15.7.3(c), hence ap
r−1 = 1 and apr= a. Thus, b := ap
r−1satisfies
bp = a.
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Main Proof. Let f(x) =∑n
i=0 aixi. Then, if f ′(x) =
∑ni=1 iaix
i−1 = 0, we haveiai = 0 for all i ≥ 1. Since F is a domain, this implies either ai = 0 or i ≡ 0 mod p.Thus, f(x) =
∑mj=0 ajx
pj for some aj ∈ F where p = charF . There exist bj such
that bpj = aj by Lemma 15.7.10a. Thus, f(x) = g(x)p where g(x) =∑m
j=0 bjxj by
Exercise 11.3.8, and so f(x) is not irreducible.
Exercise 15.7.11. Let f = ax2 + bx+ c with a, b, c in a ring R. Show that the idealof the polynomial ring R[x] that is generated by f and f ′ contains the discriminant,the constant polynomial b2 − 4ac.
Proof. f ′ = 2ax+ b. Proceeding as in the Euclidean algorithm (p. 45), we have
2ax2 + 2bx+ 2c = x(2ax+ b) + (bx+ 2c), b(2ax+ b) = 2a(bx+ 2c) + (b2 − 4ac),
hence b2 − 4ac = −4af + f ′2 ∈ (f, f ′) ⊂ R[x].
Exercise 15.7.13. Prove that a finite subgroup of the multiplicative group of anyfield F is a cyclic group.
Proof. A finite subgroup H 6 F× is a finite abelian group. Thus, H ≈ C1⊕· · ·⊕Cnwhere the Ci are cyclic groups, and |Ci| divides |Ci+1| by Thm. 14.7.3. Let d := |Cn|.Then, xd = 1 for every x ∈ H. This means that every x ∈ H is a root of xd − 1.Now, xd − 1 has at most d roots, so d ≥ |H|. On the other hand, d ≤ |H| by thedecomposition above, so d = |H|, and n = 1, i.e., H is cyclic.
15.8 Primitive Elements
Exercise 15.8.2. Determine all primitive elements for the extension K = Q(√2,√3)
of Q.
Proof. We claim γ = a+ b√2 + c
√3 + d
√6 is a primitive element for the extension
K/Q if and only if at least two of b, c, d are nonzero. Recall that
Q(√2,√3) = Q(
√2,√6) = Q(
√3,√6) = {a+ b
√2 + c
√3 + d
√6 | a, b, c, d ∈ Q}.
We can assume without loss of generality that a = 0 since Q(γ) = Q(γ − a). ⇒ isclear since if only one of b, c, d is nonzero, then K = Q(
√2), Q(
√3), or Q(
√6).
Conversely, suppose at least two of b, c, d are nonzero. If exactly two of b, c, dare nonzero, then relabel γ = kα + jβ for α, β ∈ {
√2,√3,√6}; we can moreover
assume k = 1 by dividing by k. Note K = Q(α, β) as above, and so by Lem. 15.8.2γ = α+ jβ is a primitive element for K/Q for all but finitely many j. In particular,
76
as in the proof of Lem. 15.8.2, γ fails to be primitive when at least two of ±α ± jβare equal, but this occurs if and only if j = 0, a contradiction.
Now suppose that all three of b, c, d are nonzero. Letting
γ′ = γ2 − 2b2 − 3c2 − 6d2 = 6cd√2 + 4bd
√3 + 2bc
√6,
we have that
bγ′ − 6cdγ = 2d(2b2 − 3c2)√3 + 2c(b2 − 3d2)
√6
cγ′ − 4bdγ = 2d(3c2 − 2b2)√2 + 2b(c2 − 2d2)
√6
dγ′ − 2bcγ = 2c(3d2 − b2)√2 + 2b(2d2 − c2)
√3
are all in Q(γ). But each of their coefficients have no nonzero solutions over Q byclearing denominators and checking for integer solutions, hence any three of theseproduces a primitive element forK/Q by the paragraph above, and so γ is a primitiveelement for K/Q as well.
15.9 Function Fields
Exercise 15.9.1. Let f(x) be a polynomial with coefficients in a field F . Prove thatif there is a rational function r(x) such that r2 = f , then r is a polynomial.
Proof. Suppose f 6= 0 for otherwise r is necessarily 0. Let r = p(x)/q(x) for p, qcoprime; then p2 = fq2. Now using that F [x] is a UFD (Prop. 12.2.14(c)), since p - qwe have p2 | f , and so f = p2s for some s ∈ F [x]. But then, fq2 = p2sq2 = p2, andso sq2 = 1 and s, q ∈ F [x]; since the units in F [x] are the constant polynomials, wehave s, q ∈ F and so r ∈ F [x].
15.10 The Fundamental Theorem of Algebra
Exercise 15.10.1. Prove that A ⊂ C is algebraically closed, where A is the subsetof C consisting of the algebraic numbers.
Remark. We prove that a root of a polynomial with coefficients that are algebraicover a field F is also algebraic.
Proof. Let f ∈ A[x] be non-constant, and let α be a root of f . It suffices to show αis algebraic over F . Suppose not, and let a0, . . . , an be the coefficients of f . Then,
[F (α) : F ] ≤ [F (α, a0, . . . , an) : F ] ≤ n ·n∏i=0
mi <∞
77
by Cor. 15.3.8, where mi is the degree of ai over F . Thus, there exists a polynomialin F [x] of degree ≤ n ·
∏ni=0mi with α as a root.
Exercise 15.10.2. Construct an algebraically closed field that contains the primefield Fp.
Remark. We prove that any field K is contained in an algebraically closed field L.
Proof. We first claim the monic irreducible polynomials of degree ≥ 1 in K[x] can bewell-ordered. This is a consequence of the well-ordering theorem, but in the case ofFp (or any countable field) it is possible to well-order these polynomials without theaxiom of choice since there are only countably many of them. Then, letting (fi) bea well-ordering of these polynomials, and letting L0 = K and Li be the extension ofLi−1 such that fi splits completely which exists by Prop. 15.6.3, we claim L :=
⋃i Li
is an algebraically closed field containing K. It suffices to show it is algebraicallyclosed, but this is true since if f(x) ∈ L[x] has a root α, then the proof of Exercise15.10.1 shows α is algebraic over K, hence in L by construction.
15.M Miscellaneous Problems
Exercise 15.M.1. Let K = F (α) be a field extension generated by a transcendentalelement α, and let β be an element of K that is not in F . Prove that α is algebraicover F (β).
Proof. Suppose β ∈ K \ F ; then β = p(α)/q(α) for p(x), q(x) ∈ F [x], and soβq(α) − p(α) = 0. Thus, α is a root of the polynomial βq(x) − p(x) ∈ F (β)[x],i.e., α is algebraic over F (β).
Exercise 15.M.2. Factor x7 + x+ 1 in F7[x].
Proof. First substitute x x + 3, giving x7 + x. Then, x7 + x = x(x6 + 1). Nowx6 +1 has solutions x2 = 3, 5, 6 ∈ F7, hence x
7 + x = x(x2 +1)(x2 +2)(x2 +4). Now1, 2, 4 are the squares in F7, none of which are 3, 5, 6, hence we cannot factor further.Substituting back x x− 3, since (x− 3)2 = x2 + x+ 2,
Exercise 15.M.3. Let f(x) be an irreducible polynomial of degree 6 over a field F ,and let K be a quadratic extension of F . What can be said about the degrees of theirreducible factors of f in K[x]?
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Proof. K = F (α), where α is the root of some irreducible quadratic polynomialg ∈ F [x]. Let β be a root of f ; then, 2 = [K : F ] and 6 = [F (β) : F ] divide[K(β) : F ] ≤ 12 by Cor. 15.3.8, and so [K(β) : F ] ∈ {6, 12}. In either case, 3 divides[K(β) : K], and so the minimal polynomial for β over K is of degree 3 or 6. In theformer case, f splits into a product of polynomials of degree 3 over K. In the latter,f remains irreducible over K. These are the only possibilities for the irreduciblefactors of f in K, for the only possibility is that the factors of degree 3 factor furtherto include linear factors, which is impossible since β is of degree 6 over F .
Exercise 15.M.4.(a) Let p be an odd prime. Prove that exactly half of the elements of F×
p are squaresand that if α and β are nonsquares, then αβ is a square.
(b) Prove the same assertion for any finite field of odd order.(c) Prove that in a finite field of even order, every element is a square.(d) Prove that the irreducible polynomial for γ =
√2 +
√3 over Q is reducible
modulo p for every prime p.
Remark. (b) implies (a) and Lemma 15.7.10a implies (c), so it suffices to show (b)and (d).
Proof of (b). Let F be our finite field of odd order, and consider the group homo-morphism ϕ : F× → F× defined by α α2. If α ∈ kerϕ then α2 = 1, hence(x−α) | x2−1 in F [x]. Since F [x] is a UFD (Prop. 12.2.14(c)), this implies α = ±1.Thus |kerϕ| = 2 since charF 6= 2 implies 1 6= −1, and so the set of squares imϕ inF× has order 1
2|F×| by the counting formula (Cor. 2.8.13).
Now let β be a nonsquare. Consider the map µ : F× → F× of sets defined byα αβ. This is a bijection, and so to show µ sends non-squares to squares, it sufficesto show µ sends squares to non-squares. But this is true since if α is a square andαβ is also a square, then α−1 is also a square, hence β is also, a contradiction.
Proof of (d). First, f := x4 − 10x2 + 1 is the irreducible polynomial for γ sincef(γ) = 0 and Q(γ) = Q(α, β) by Exercise 11.1.3, and so [Q(γ) : Q] = 4. In this field,we have the factorization
f = (x− (√2 +
√3))(x− (
√2−
√3))(x− (−
√2 +
√3))(x− (−
√2−
√3)).
If f reduces over Fp, then it must reduce to quadratic factors since γ /∈ Fp; thus, fcan factor in one of the three following ways by pairing up the factors above:
(x2 − 1− 2√2)(x2 − 1 + 2
√2), (x2 + 1− 2
√3)(x2 + 1 + 2
√3),
(x2 − 5− 2√6)(x2 − 5 + 2
√6).
79
The first factorization can occur if 2 is a square mod p, the second can occur if 3 isa square mod p, and the last can occur if 6 is a square mod p. By (a), (c), at leastone of 2, 3, 6 is a square in Fp, so at least one of these factorizations is possible andf is reducible mod p for all p.
Exercise 15.M.6.(a) Prove that a rational function f(t) that generates the field C(t) of all rational
functions defines a bijective map T ′ → T ′.(b) Prove a rational function f(x) generates the field of rational functions C(x) if
and only if it is of the form (ax+ b)/(cx+ d), with ad− bc 6= 0.(c) Identify the group of automorphisms of C(x) that are the identity on C.
Proof of (a). Let f(t) ∈ F = C(t), and f = p/q for some coprime p, q ∈ C[t].C[t](f) is then isomorphic to F , which is isomorphic to F [x]/(p − qx); note thisis still isomorphic to F . Likewise, consider F [y]/(y); this is also isomorphic to F .By Prop. 15.9.5, we have isomorphic field extensions, hence the Riemann surfaces{p− qx = 0} and {y = 0} are isomorphic branched coverings of T ′. Since {y = 0} isthe complex t-plane T ′, the isomorphism of coverings is an isomorphism T ′ → T ′.
Proof of (b). Any function (ax + b)/(cx + d) with ad − bc 6= 0 generates C(x) byExercise 12.2.5, since any element in C(x) can be expressed as a C-linear combinationof elements 1/(cx+ d)i, and then by using the Euclidean algorithm (p. 45) to get
ax+ b
cx+ d=q(cx+ d) + r
cx+ d= q +
r
cx+ d, q, r ∈ C
which implies 1/(cx + d)i for any i can be expressed as a C-linear combination of(ax + b)i/(cx + d)i. Conversely, suppose a rational function f(x) generates C(x).By (a), it induces an automorphism of the Riemann surface T ′; it suffices to showthat the automorphisms of T ′ are of the stated form. Suppose f = p/q definesan automorphism of T ′; then, if either p, q had degree larger than 1, the inducedmorphism would not be bijective since then p has two zeros; likewise, if q had degreelarger than 1, considering 1/f gives us the same argument. Moreover, one of p, qmust be non-constant also to induce a bijection. ad− bc 6= 0 corresponds to havingp, q coprime, which was necessary in (a).
Solution for (c). By (a), (b), Aut(C(x)) consists of maps x (ax + b)/(cx + d) forad − bc 6= 0, which is a group under composition. We claim there is a surjectivehomomorphism ϕ : GL2(C) → Aut(C(x)) defined by ( a bc d ) (ax+ b)/(cx+d). Thismap clearly maps ( 1 0
0 1 ) x, is surjective, and is a homomorphism since(a′ b′
c′ d′
)(a bc d
)=
(aa′ + b′c a′b+ b′dac′ + cd′ bc′ + dd′
),
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and both sides get mapped to(a′x+ b′
c′x+ d′
)◦(ax+ b
cx+ d
)=a′(ax+ b) + b′(cx+ d)
c′(ax+ b) + d′(cx+ d)=
(aa′ + b′c)x+ (a′b+ b′d)
(ac′ + cd′)x+ (bc′ + dd′).
It remains to find the kernel of ϕ. (ax + b)/(cx + d) = x ∈ C(x) if and only ifax+ b = x(cx+d) if and only if c = b = 0 and a = d. Hence kerϕ = C · ( 1 0
0 1 ), and soby the first isomorphism theorem (Thm. 11.4.2(b)), Aut(C(x)) ≈ GL2(C)/ kerϕ =:PGL2(C), the projective general linear group of order 2.
Exercise 15.M.7. Prove that the homomorphism SL2(Z) → SL2(Fp) obtained byreducing the matrix entries modulo p is surjective.
Proof. Let π : Z → Fp be the quotient map reducing mod p. We first show that ifu, v ∈ Fp are not both zero, then there exist c, d ∈ Z such that π(c) = u, π(d) = v,and gcd(c, d) = 1. So suppose v 6= 0, and let 0 ≤ u, v < p be lifts of u, v in Z. Nowsince p - v, there exist x, y ∈ Z such that xv + yp = 1 since (p) is maximal in Z, andso let c = xuv + yp and d = v. Then, π(d) = v, π(c) = u since xv ≡ 1 mod p, andgcd(c, d) = 1 since 1 = c+ xd(1− u). If v = 0, then necessarily u 6= 0, and switchingthe roles of u, v gives the desired result.
Now suppose we have ( s tu v ) ∈ SL2(Fp); let 0 ≤ s, t < p be lifts of s, t in Z. Then,
letting c, d as constructed above, we have sd − tc = 1 + Np for some N ∈ Z. So,letting a = s+mp and b = t+ np where m,n ∈ Z such that N = cn− dm, which ispossible since gcd(c, d) = 1,
ad− bc = (s+mp)d− (t+ np)c
= sd− tc+ (dm− cn)p = 1 + (N + dm− cn)p = 1,
hence ( a bc d ) ∈ SL2(Z) maps to ( s tu v ) ∈ SL2(Fp).
16 Galois Theory
16.1 Symmetric Functions
Exercise 16.1.1. Determine the orbit of the polynomial below. If the polynomial issymmetric, write it in terms of the elementary symmetric functions.(a) u21u2 + u22u3 + u23u1 (n = 3),(b) (u1 + u2)(u2 + u3)(u1 + u3) (n = 3),(c) (u1 − u2)(u2 − u3)(u1 − u3) (n = 3),
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(d) u31u2 + u32u3 + u33u1 − u1u32 − u2u
33 − u3u
31 (n = 3),
(e) u31 + u32 + · · ·+ u3n.
Solution for (a). The orbit consists of u21u2 + u22u3 + u23u1 and u23u2 + u22u1 + u21u3,corresponding to odd and even permutations respectively in S3.
Solution for (b). The orbit consists of only (u1 + u2)(u2 + u3)(u1 + u3), i.e., thispolynomial is symmetric. Since this polynomial is homogeneous of degree 3, write
Substituting (1, 0, 0), we have c1 = 0. Substituting (1, 1, 0), we have 2 = 2c2 hencec2 = 1. Lastly, substituting (1, 1, 1), we have 8 = 9 + c3 hence c3 = −1. Thus,
(u1 + u2)(u2 + u3)(u1 + u3) = s1s2 − s3.
Solution for (c). The orbit consists of (u1−u2)(u2−u3)(u1−u3) and −(u1−u2)(u2−u3)(u1 − u3), corresponding to odd and even permutations respectively in S3.
Solution for (d). The orbit consists of u31u2 + u32u3 + u33u1 − u1u32 − u2u
33 − u3u
31 and
u33u2+u32u1+u
31u3−u3u32−u2u31−u1u33, corresponding to odd and even permutations
respectively in S3.
Solution for (e). The orbit consists of only u31 + u32 + · · · + u3n, i.e., this polynomialis symmetric. Since this polynomial is homogeneous of degree 3, write
u31 + u32 + · · ·+ u3n = c1s31 + c2s1s2 + c3s3.
Substituting (1, 0, . . . , 0), we have 1 = c1. Substituting (1, 1, 0, . . . , 0), we have 2 =8+2c2 hence c2 = −3. Lastly, substituting (1, 1, 1, 0 . . . , 0), we have 3 = 27−27+c3,hence c3 = 3. Thus,
u31 + u32 + · · ·+ u3n = s31 − 3s1s2 + 3s3.
Exercise 16.1.3. Let wk = uk1 + · · ·+ ukn.(a) Prove Newton’s identities: wk − s1wk−1 + · · · ± sk−1w1 ∓ ksk = 0.(b) Do w1, . . . , wn generate the ring of symmetric functions?
Proof of (a). For 1 < i ≤ k, let ri be the sum of all distinct monomials of degree kwhere each monomial is the product of one variable raised to the power i and k − idistinct other variables; note rk = wk. We claim sk−iwi = ri + ri+1 for 1 < i < k.This is true since each product of terms with distinct variables on the left contributes
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to ri, while each product which has the term from wi occurring in the term fromsk−i contributes to ri+1, and since all terms on the right are obtained exactly oncein this way. For i = k we recall s0 = 1, hence s0wk = wk = rk. Finally, for i = 1 wehave sk−1w1 = ksk + r2, since the terms contributing to r2 arise in the same way asbefore, while the remaining terms produce k times each monomial in sk. The desiredidentity is formed by taking the alternating sum of these equations.
Claim (b). w1, . . . , wn generates the ring of symmetric functions if and only if m ∈R× for all 1 ≤ m ≤ n.
Proof of Claim (b). ⇐. By Thm. 16.1.6, it suffices to show sk ∈ R[w1, . . . , wn] forall k; we proceed by induction. If k = 1 then we are done, for s1 = w1. Then, ifsj ∈ R[w1, . . . , wn] for all j < k, by (a) we have
⇒. We show the contrapositive. Suppose m /∈ R× for some 0 < m ≤ n; we claimsm /∈ R[w1, . . . , wn]. Suppose not; then, since sm is homogeneous of degree m, write
sm =∑
I⊂{1,...,n}∑i∈I i=m
rI∏i∈I
wi, rI ∈ R.
Then, substituting u1 = · · · = um = 1, um+1 = · · · = un = 0, we get wi = m for all i,while sm = 1. But this is a contradiction, for the right side is in the ideal mR whilethe left side = 1 /∈ mR.
16.2 The Discriminant
Exercise 16.2.2.(a) Prove that the discriminant of a real cubic is non-negative if and only if the
cubic has three real roots.(b) Suppose that a real quartic polynomial has positive discriminant. What can
you say about the number of real roots?
Proof of (a). Recall that the discriminant D is the product of squares of the differ-ences of roots. So, if f has 3 roots in R, then the discriminant is trivially non-negative.Conversely, if f does not have 3 roots in R, we know it must have one root a ∈ R,and the other two roots must be a conjugate pair z, z since otherwise the constantterm in f would not be real. Then, we have
Now (a− z)(a− z) > 0, and so it suffices to show (z− z)2 < 0. But this is true sincez − z = 2 Im z.
Solution for (b). We claim that a real quartic polynomial with positive discriminantcan have 0 or 4 real roots; note in particular this implies all roots are distinct. Thediscriminant is given by
D = (α1 − α2)2(α1 − α3)
2(α1 − α4)2(α2 − α3)
2(α2 − α4)2(α3 − α4)
2,
where αi are our roots; recall that complex roots must come in conjugate pairs, andso we can have 0, 2, or 4 real roots in general. If all the roots are real D is clearlynon-negative. If none of the roots are real and α2 = α1, α4 = α3, we get
D = (α1 − α1)2(α1 − α3)
2(α1 − α3)2(α1 − α3)
2(α1 − α3)2(α3 − α3)
2
= (2 Imα1)2(2 Imα3)
2(α1 − α3)2(α1 − α3)
2(α1 − α3)
2(α1 − α3)2,
which is positive since each consecutive pair of factors multiply to be positive. How-ever, 2 real roots is impossible since if α2 = α1 and α3, α4 are real,
D = (2 Imα1)2(α1 − α3)
2(α1 − α4)2(α1 − α3)
2(α1 − α4)2(α3 − α4)
2
= (2 Imα1)2(α1 − α3)
2(α1 − α3)2(α1 − α4)
2(α1 − α4)2(α3 − α4)
2,
which is negative since the first factor is negative and the rest of the product ispositive.
Exercise 16.2.4. Use undetermined coefficients to determine the discriminant ofthe polynomial(a) x3 + px+ q, (b) x4 + px+ q, (c) x5 + px+ q.
Lemma 16.2.4a. The discriminant Dn of xn + px+ q for n ≥ 2 is given by
(−1)(n−1)(n−2)
2 (n− 1)n−1pn + (−1)n(n−1)
2 nnqn−1.
Proof of Lemma 16.2.4a. Let Dn(p, q) be the determinant of xn+px+q. If u1, . . . , unare the roots of xn + px+ q, by the first equation in §16.2 we have sn(u1, . . . , un) =(−1)nq, sn−1(u1, . . . , un) = (−1)n−1p, and sk(u1, . . . , un) = 0 for all 0 < k < n − 1.Note the discriminant is a homogeneous symmetric polynomial of degree n(n − 1);we claim the only monomials in the si appearing in Dn(p, q) are p
n, qn−1. For, by theabove sn−1, sn are the only nonzero elementary symmetric polynomials that wouldappear in Dn(p, q), and moreover if pkqj had degree n(n − 1), then the equation
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k(n− 1) + jn = n(n− 1) must hold, for which k = 0, j = n− 1 and k = n, j = 0 arethe only non-negative solutions. Thus, we have Dn(p, q) = r(n)pn + s(n)qn−1.
Now consider the discriminant Dn(p, 0) for xn+ px; since this has the same roots
as xn−1 + p in addition to the root un = 0, we have
Dn(p, 0) = Dn−1(0, p)n−1∏i=1
u2i = Dn−1(0, p)p2
and so r(n)pn = s(n−1)pn, i.e., r(n) = s(n−1), andDn(p, q) = s(n−1)pn+s(n)qn−1.It remains to find s(n). Calculating the discriminant Dn(0,−1) for f := xn − 1,
Dn(0,−1) = (−1)n−1g(n) = (−1)n(n−1)
2
n∏i=1
∏i 6=j
(ξi − ξj)
where ξi are the roots of xn − 1. Then, we have the equation xn − 1 =∏
i(x − ξi),and so deriving both sides and substituting in x = ξi, we get nξn−1
i =∏
i 6=j(ξi − ξj),
and so Dn(0,−1) = (−1)n(n−1)
2
∏ni=1 nξ
n−1i . Since ξi satisfies f , ξ
n−1i = ξ−1
i , and so
Dn(0,−1) = (−1)n(n−1)
2 nnn∏i=1
ξ−1i = (−1)
n(n−1)2 (−1)n−1nn,
hence s(n) = (−1)n(n−1)
2 nn, and finally
(−1)(n−1)(n−2)
2 (n− 1)n−1pn + (−1)n(n−1)
2 nnqn−1.
Solution for (a). D3 = −4p3 − 27q2 by Lemma 16.2.4a.
Solution for (b). D4 = −27p4 + 256q3 by Lemma 16.2.4a.
Solution for (c). D5 = 256p5 + 3125q4 by Lemma 16.2.4a.
Exercise 16.2.7. There are n variables. Let m = u1u22u
33 · · ·un−1
n−1 and let p(u) =∑σ∈An
σ(m). The Sn-orbit of p(u) contains two elements, p and another polynomial q.
Prove that (p− q)2 = D(u).
Proof. Let R[u1, . . . , un] be our polynomial ring, and let δ(u) =∏
i<j(ui − uj). Wefirst claim ui − uj | p − q for all i < j. This is equivalent to showing p − q = 0 ifui − uj = 0.
85
So suppose ui = uj for i < j, and let τ = (ij) ∈ Sn. Then, p(u) = p(τ(u)) andq(u) = q(τ(u)), and we have
p(τ(u)) =∑σ∈An
σ(τ(m)) =∑
σ′∈An·τ
σ′(m) = q(u).
Similarly, q(τ(u)) = p(u), since An has exactly two cosets in Sn. Thus, p(u)− q(u) =q(u)− p(u), and so p(u)− q(u) = 0.
Now since δ(u) is homogeneous of degree n(n − 1)/2, as are p, q, ui − uj | p − qfor all i < j implies p− q = a
∏i<j(ui−uj) for some a ∈ R. But a = ±1 since in the
equation p−q =∏
σ∈Snsgn(σ)σ(m), the coefficient on m is 1, while in the expression
δ(u) =∏
i<j(ui−uj), the coefficient on m is ±1. Thus, (p− q)2 = δ(u)2 = D(u).
16.3 Splitting Fields
Exercise 16.3.1. Let f be a polynomial of degree n with coefficients in F and let Kbe a splitting field for f over F . Prove that [K : F ] divides n!.
Proof. We use induction on the degree of f . The claim is trivial if n = 1, so supposen > 1. Suppose f is reducible, and let g be an irreducible factor of f of degreem. Then, we can choose a subfield F1 ⊂ K such that g splits completely, hence[F1 : F ] | m! by inductive hypothesis. Similarly, [K : F1] | (n − m)! by inductivehypothesis since K is a splitting field for f/g over F1. Hence by the multiplicativeproperty of the degree (Thm. 15.3.5), [K : F ] | m!(n−m)! | n!.
Now suppose f is irreducible. Then, F1 := F [x]/(f) is a field extension of degreen such that f has a root α. Thus, [K : F1] | (n− 1)! by inductive hypothesis since Kis a splitting field for f/(x−α) over F1, and so [K : F ] | n(n−1)! = n! as above.
16.4 Isomorphisms of Field Extensions
Exercise 16.4.1.(a) Determine all automorphisms of the field Q( 3
√2), and of the field Q( 3
√2, ω),
where ω = e2πi/3.(b) Let K be the splitting field over Q of f(x) = (x2 − 2x − 1)(x2 − 2x − 7).
Determine all automorphisms of K.
Solution for (a). We first claim that any automorphism ϕ of either field must leaveQ fixed. Any integer is held fixed by Exercise 2.6.2 and since ϕ(1) = 1. Any rationalp/q must then be fixed since multiplicative inverses are preserved through ϕ.
86
Now let ϕ ∈ AutQ( 3√2). ϕ is then determined by ϕ( 3
√2), and it is necessary that
ϕ( 3√2)3 = 2. Thus, ϕ( 3
√2) = ωj 3
√2 for some j ∈ {0, 1, 2}; however, Q( 3
√2) ⊂ R, and
ω ∈ C \ R, hence j = 0 and ϕ = id.Now let ϕ ∈ AutQ( 3
√2, ω). ϕ is then determined by ϕ( 3
√2) and ϕ(ω), and it is
necessary that ϕ( 3√2)3 = 2 and ϕ(ω)3 = 1. Thus, ϕ( 3
√2) = ωj 3
√2 and ϕ(ω) = ωk for
some j, k ∈ {0, 1, 2}. k 6= 0 for otherwise ϕ is not surjective. Any of the remainingchoices gives an automorphism, hence Q( 3
√2, ω) has six automorphisms.
Solution for (b). The roots of f are 1±√2, 1±2
√2 by the quadratic formula, hence
K = Q(√2), which only has the trivial automorphism and the automorphism sending√
2 −√2 as in Ex. 16.4.1.
16.5 Fixed Fields
Exercise 16.5.2. Show that the automorphisms σ(t) =t+ i
t− iand τ(t) =
it− i
t+ 1of
C(t) generate a group isomorphic to the alternating group A4, and determine thefixed field of this group.
Solution. We first calculate products of σ, τ :
id = t, στ = −t, στ 2σ =1
t, τσ = −1
t,
σ =t+ i
t− i, σ2τ =
t− i
t+ i, τσ2 =
it+ 1
−it+ 1, τ 2 =
−it+ 1
it+ 1,
σ2 =t+ 1
−it+ i, τ =
it− i
t+ 1, τ 2σ =
t+ 1
it− i, στ 2 =
−it+ 1
t+ 1,
(7)
where we have used the relations σ3 = τ 3 = στστ = id, hence there are 12 elementsin this group G. Now consider the four pairs of 3 points defining circles in CP1:
Labeling these 1, 2, 3, 4 respectively, we have that σ permutes these pairs as (123)and τ does as (234), hence G is isomorphic to a subgroup of S4. By the above, thishas order 12; by Exercise 6.7.11 we then know that G ≈ A4.
To find the fixed field, the first row in (7) gives a subgroup of A4 isomorphic tothe Klein 4-group C2 × C2. We then note by Thm. 16.5.2(a) that the irreduciblepolynomial for t over the fixed field is the polynomial whose roots form its orbit:
Thus letting u = t2+ t−2, we have that [C(t) : C(u)] ≤ 4. Note u is fixed by C2×C2,hence C(u) ⊂ C(t)C2×C2 , and that since by the fixed field theorem (Thm. 16.5.4),[C(t) : C(t)C2×C2 ] = 4, it follows that C(u) = C(t)C2×C2 .
We now want to find the subfield of C(u) that is fixed by all of A4; by (7) since σand C2 ×C2 generate A4, it suffices to find the subfield of C(u) fixed by σ. We have
σ(u) = σ(t)2 +1
σ(t)2=
(t+ i
t− i
)2
+
(t− i
t+ i
)2
=2(t4 − 6t2 + 1)
(t2 + 1)2=
2(u− 6)
u+ 2
σ2(u) = σ2(t)2 +1
σ2(t)2=
(it+ 1
t− 1
)2
+
(it− 1
t+ 1
)2
=−2(t4 + 6t2 + 1)
(t2 − 1)2=
−2(u+ 6)
u− 2
hence
(x− u)(x− σ(u))(x− σ2(u)) = x3 − u(u2 − 36)
u2 − 4x2 − 36x+
4u(u2 − 36)
u2 − 4.
Now letting v = u(u2−36)u2−4
, v is fixed by σ and [C(u) : C(v)] ≤ 3, hence by the fixed
Exercise 16.6.1. Let α be a complex root of x3 + x + 1 over Q, and let K be asplitting field of this polynomial over Q. Is
√−31 in the field Q(α)? Is it in K?
Solution. By the multiplicative property of degree (Thm. 15.3.5),√−31 /∈ Q(α)
since α has degree 3 while√−31 has degree 2 over Q. However,
√−31 ∈ K because
the discriminant of x3 + x + 1 is −31 by Exercise 16.2.4(a), and the square root ofthe discriminant is a product of differences of elements in K.
16.7 The Main Theorem
Exercise 16.7.4. Let F = Q, and let K = Q(√2,√3,√5). Determine all interme-
diate fields.
Proof. K is the splitting field of the polynomial (x2−2)(x2−3)(x2−5), and so F ⊂ Kis a Galois extension of order 8 with Galois group G = C2 × C2 × C2 = 〈σ2, σ3, σ5〉,where σk is the field automorphism over Q such that
√k −
√k.
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The lattice diagram for subgroups of C2 × C2 × C2 is given by
which corresponds to the lattice diagram of intermediate fields
Q(√2,√3,√5)
Q(√2,√3) Q(
√3,√5) Q(
√5,√6) Q(
√6,√10) Q(
√2,√15) Q(
√3,√10) Q(
√2,√5)
Q(√2) Q(
√3) Q(
√5) Q(
√6) Q(
√15) Q(
√30) Q(
√10)
Q
by computing fixed fields according to Thm. 16.7.1.
Exercise 16.7.6. Let K/F be a Galois extension whose Galois group is S3. Is Kthe splitting field of an irreducible cubic polynomial over F?
89
Solution. We claim it is. Let S2 6 S3 be the subgroup fixing 3, and consider thechain F ⊂ KS2 ⊂ K. By Thm. 16.7.5, the first extension in this chain is not Galoissince S2 6 S3 is not normal, while the second extension is Galois with Galois groupS2. By Cor. 16.7.2(c), [KS2 : F ] = [S3 : S2] = 3. Now if α ∈ KS2 \F , let f ∈ F [x] beits irreducible polynomial; note that deg f = 3, and that KS2 = F (α). Since KS2/Fis not Galois, f does not split in KS2 by Thm. 16.6.4, but it does split in K by thesplitting theorem (Thm. 16.3.2). Since [K : KS2 ] = 2, there are no intermediatefields between KS2 , K, and so K/F must be the splitting field of f .
Exercise 16.7.10. Let K/F be a Galois extension with Galois group G, and let Hbe a subgroup of G. Prove that there exists an element β ∈ K whose stabilizer isequal to H.
Proof. Recall from p. 484 that we assume K,F have characteristic zero, and thatK/F is a finite extension. So consider the chain of extensions F ⊂ KH ⊂ K;by the primitive element theorem (Thm. 15.8.1) there exists β ∈ KH such thatKH = F (β). We claim Gβ = H. Clearly H ⊂ Gβ, so suppose σ ∈ Gβ \ H. Then,H ′ := 〈H, σ〉 ) H, and KH′ ⊂ KH by Cor. 16.7.2(a). Every element in KH isfixed by H ′, hence KH′
= KH . But then [K : KH ] = |H| = |H ′| by Cor. 16.7.2(c),contradicting that H ( H ′.
Exercise 16.7.11. Let α = 3√2, β =
√3, and γ = α+β. Let L be the field Q(α, β),
and let K be the splitting field of the polynomial (x3 − 2)(x2 − 3) over Q.(a) Determine the irreducible polynomial f for γ over Q, and its roots in C.(b) Determine the Galois group of K/Q.
and so Q(γ) = Q(α, β). Moreover, since α has degree 3 and β has degree 2 over Q,by Cor. 15.3.8 [L : Q] = 6. Thus, the irreducible polynomial for γ has degree 6. Nowletting f(x) = ((x− β)3 − 2)((x+ β)3 − 2), f(γ) = 0 by construction, and
Since deg f(x) = 6, it is then the irreducible polynomial for γ. Now z ∈ C is a rootof f(x) if and only if z± β = ωjα where ω = e2πi/3 and j ∈ {0, 1, 2}, hence the rootsof f(x) are ±β + ωjα.
90
Solution for (b). Denote G = G(K/Q). f(x) from (a) splits completely in K by thesplitting theorem (Thm. 16.3.2); similarly, (x3 − 2)(x2 − 3) splits completely in thesplitting field for f(x) since α, β can be obtained as linear combinations of the rootsof f(x). Thus, K is the splitting field for f(x). If j ∈ {1, 2},
±β + ωjα = ±β +−1− (−1)jiβ
2α = ±β − 1
2α− (−1)j
2iβ,
hence K = Q(γ, i) = Q(α, β, i), and so [K : Q] = [K : Q(γ)][Q(γ) : Q] = 12 by themultiplicative property of the degree (Thm. 15.3.5).
Now let K1 be the splitting field for x3 − 2 and let K2 be the splitting field forx2 − 3. We have the lattice diagram of intermediate fields
K
K1
K2
Q
splits x2 − 3
splits x3 − 2
splits x3 − 2
splits x2 − 3
Each field extension is Galois by Thm. 16.6.4(c), and so in particular the subgroupsG1 := G(K/K1) and G2 := G(K/K2) of G fixing K1, K2 respectively are normalsubgroups of G by Thm. 16.7.5. The extension K/K1 is quadratic, hence G1 ≈ C2;the extension K/K2 is cubic with [K : K2] = 6, hence G2 ≈ S3 as on p. 492.
We therefore claim G ≈ G1 × G2 ≈ C2 × S3. Consider the multiplication mapµ : G1×G2 → G. We first claim µ is injective. By Thm. 16.6.6(b), it suffices to showany σ ∈ G1 ∩G2 operates as the identity on ±β +ωjα. But ωjα ∈ K1 and ±β ∈ K2
by definition, hence G1 ∩ G2 = {1}. Now since also |G1 × G2| = 12 = |G|, µ is alsosurjective, and is therefore an isomorphism.
16.8 Cubic Equations
Exercise 16.8.4. Let K = Q(α), where α is a root of the polynomial x3 + 2x + 1,and let g(x) = x3 + x+ 1. Does g(x) have a root in K?
Solution. Let f(x) = x3 + 2x + 1. By the rational root test, f(x), g(x) are bothirreducible in Q[x]. By (16.2.8), their discriminants are −59,−31 respectively, sosince both are not squares, by Thm. 16.8.5 their splitting fields Lf , Lg have degree 6over Q, with Galois group S3. Now if g(x) has a root in K, then it splits completely
91
in K by the splitting theorem (Thm. 16.3.2), hence Lf = Lg since both are of degree6 over Q. Since the Galois group is S3, there should be one intermediate field ofdegree 2 over Q, but we have two, Q(
√−59) and Q(
√−31), a contradiction.
16.9 Quartic Equations
Exercise 16.9.6. Compute the discriminant of the quartic polynomial x4 + 1, anddetermine its Galois group over Q.
Solution. By Exercise 16.2.4(c), the discriminant is 256. By Prop. 16.9.5, this impliesthe Galois group G is A4 or D2. Now the roots of x4 + 1 are
α1 =1√2(1 + i), α2 = − 1√
2(1 + i), α3 =
1√2(1− i), α4 = − 1√
2(1− i).
β1 = 0, β2 = 2, β3 = −2 gives g(x) = x(x− 2)(x+2), so G = D2 by Prop. 16.9.8.
Exercise 16.9.13. Let K be the splitting field over Q of the polynomial x4−2x2−1.Determine the Galois group G of K/Q, find all intermediate fields, and match themup with the subgroups of G.
Solution. We find the roots of x4 − 2x2 − 1. We first apply the quadratic equationto get x2 = 1±
√2; then,
α1 =
√1 +
√2, α2 =
√1−
√2, α3 = −α1, α4 = −α2.
β1 = 2i, β2 = −2, β3 = −2i, so g(x) = (x + 2)(x2 + 4). As in Ex. 16.9.2(a), G is asubgroup of D4 = 〈σ, τ〉 where σ = (1234), τ = (24). By Prop. 16.9.8, G = D4 or C4.
As in Ex. 16.9.2(a), ρ = σ2 = (13)(24) corresponds to an automorphismK/Q; callN the normal subgroup of order 2 generated by ρ. Now α2
1 = 1+√2 and α1α2 = i are
both fixed by ρ, hence Q(√2, i) ⊂ KN . The chain of fields Q ⊂ Q(
√2, i) ⊂ KN ⊂ K
has [Q(√2, i) : Q] = 4, [K : KN ] = 2 by Thm. 16.5.4, and so [K : Q] = 8, so G = D4.
The lattice diagram for subgroups of D4 is as given in Exercise 6.4.2:
{e}
{e, τ} {e, τσ2} {e, σ2} {e, τσ} {e, τσ3}
{e, σ2, τ, τσ2} {e, σ, σ2, σ3} {e, σ2, τσ, τσ3}
D4
92
which corresponds to the lattice diagram of intermediate fields
K
Q(α1) Q(α2) Q(√2, i) Q(α1 − α2) Q(α1 + α2)
Q(i) Q(i√2) Q(
√2)
Q
by computing fixed fields according to Thm. 16.7.1.
Exercise 16.9.14. Let F = Q(ω), where ω = e2πi/3. Determine the Galois group
over F of the splitting field of (a)3√
2 +√2, (b)
√2 + 3
√2.
Remark. We note there are analogues of results from §§12.3–12.4 for checking ir-reducibility of polynomials in F [x]: Gauss’s lemma (Prop. 12.3.4(b)) only needs Fto be the fraction field of a UFD, and Z[ω] is a UFD by Exercise 12.2.6(a) andProps. 12.2.7 and 12.2.14(b). Moreover, Eisensten’s criterion generalizes to Z[ω][x]by looking at prime elements p ∈ Z[ω].
Solution for (a). Let α =3√
2 +√2, with splitting field K/F . α satisfies the poly-
nomial f(x) = (x3 − 2)2 − 2 = x6 − 4x3 + 2. The quadratic equation gives that
Thus if α1 αi, then α3 ωαi, α5 ω2αi. The permutations with this propertyare generated by σ = (123456), τ = (246) in S6; these form what is called thesemidirect product C6 o C3, and so G(K/F ) 6 C6 o C3.
Now f is irreducible over F by Eisenstein’s criterion using p = 2 which is primeby Exercise 12.5.9(b), hence [F (α) : F ] = 6. Also,
√2 ∈ F (α), hence α′ has degree 3
or 1 over F (α). Thus, [K : F ] = 6 or 18.We claim [K : F ] = 18. Consider σ2 = (135)(246); it is in all subgroups of C6oC3
of order 6, hence extends to an F -automorphism of K. Let N be the subgroup of
order 3 generated by σ2. The fixed field KN contains α2α′ = 3
√2(2 +
√2); let L
be the field generated by this element over F . The chain F ⊂ L ⊂ KN ⊂ K has[K : KN ] = 3 by the fixed field theorem (Thm. 16.5.4), and [L : F ] ≥ 3, hence[K : F ] = 18, and so G(K/F ) = C6 o C3.
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Solution for (b). Let α =√
2 + 3√2, with splitting field K/F . α satisfies the poly-
nomial f(x) = (x2 − 2)3 − 2 = x6 − 6x4 + 12x2 − 10. The roots of f(x) are
α1 = α, α2 =
√2 + ω
3√2, α2 =
√2 + ω2 3
√2, α4 = −α1, α5 = −α2, α6 = −α3.
Thus if αi αj, then αi+3 αj+3, where we treat subscripts mod 6. The groupG of permutations satisfying these is generated by the subgroup of S6 isomorphic toS3 permuting {{1, 4}, {2, 5}, {3, 6}} isomorphic to S3 and C2 ≈ 〈(14)(25)(36)〉 6 S6.Now S3 ∩ C2 = {1}, and if σ ∈ S3, τ ∈ C2, στ = τσ, hence S3, C2 are normal in G,G ≈ S3 × C2 by Prop. 2.11.4(d), and G(K/F ) 6 G.
Now f is irreducible over F by Eisenstein’s criterion, again using that 2 is prime,hence [F (α) : F ] = 6. Also, ωj 3
√2 ∈ F (α) for j ∈ {0, 1, 2}, hence α2, α3 have degree
2 or 1 over F (α). Thus, [K : F ] ∈ {6, 12, 24}. But |G| = 12, hence [K : F ] = 24 isimpossible.
We claim [K : F ] = 12. Consider ρ = ({1, 4}{2, 5}{3, 6}) ∈ G; it is contained inevery subgroup of G of order 6, hence extends to an F -automorphism of K. Let N bethe subgroup of order 3 generated by ρ. The fixed field KN contains α1α2α3 =
√10;
let L be the field generated by this element over F . The chain F ⊂ L ⊂ KN ⊂ Khas [K : KN ] = 3 by the fixed field theorem (Thm. 16.5.4), and [L : F ] = 2.But [KN : L] > 1 since α1 + α2 + α3 ∈ KN \ L, hence [K : F ] = 12, and soG(K/F ) = S3 × C2.
16.10 Roots of Unity
Exercise 16.10.3. Let ζ = ζ7. Determine the degree of the following elements overQ.(a) ζ + ζ5, (b) ζ3 + ζ4, (c) ζ3 + ζ5 + ζ6.
Remark. Suppose we want to find the degree of α ∈ Q(ζ). By Prop. 16.10.2,G(Q(ζ)/Q) = C6 consisting of σi given by σi(ζ) = ζ i for 1 ≤ i ≤ 6. If H 6 C6
is the subgroup fixing α, then by the main theorem (Thm. 16.7.1) Q(α) = Q(ζ)H .Thus, the multiplicative property of degree (Thm. 15.3.5) gives us [Q(α) : Q] =[Q(ζ) : Q]/[Q(ζ) : Q(ζ)H ] = 6/|H| by the fixed field theorem (Thm. 16.5.4).
Solution for (a). {σ1} fixes ζ + ζ5, hence ζ + ζ5 has degree 6.
Solution for (b). {σ1, σ6} fixes ζ3 + ζ4, hence ζ3 + ζ4 has degree 3.
Exercise 16.11.5.(a) How does Cardano’s formula (16.11.5) express the roots of the polynomials
x3 + 3x, x3 + 2 and x3 − 3x+ 2?(b) What are the correct choices of roots in Cardano’s formula?
Solution for (a). For x3 + 3x, we get3√
0 +√1 +
3√
0−√1.
For x3 + 2 we get3√
−1 +√1 +
3√
−1−√1.
For x3 − 3x+ 2 we get3√
−1 +√0 +
3√
−1−√0 = 3
√−1 + 3
√−1.
Solution for (b). For x3 + 3x, the solutions are 0,±i√3. To get 0, we choose
√1 to
have the same sign to be ±1 in both terms, and then choose 3√±1 = ±e2πik/3 and
3√∓1 = ∓e2πik/3 for k ∈ {1, 2}. For i
√3, we choose
√1 to have different signs ±1,∓1
so that3√
0 +√1 +
3√
0−√1 = 2 3
√±1 = ±2 3
√1, and then choose 3
√1 = e2πik/3 for
k = 1 if we chose ± to be +, and k = 2 if we chose ± to be −. For −i√3 we would
switch our choice for k.For x3 + 2, we choose the same square roots, giving
3√
−1 +√1 +
3√
−1−√1 =
3√−2, and each choice of cube root gives us a solution.For x3 − 3x+ 2, x3 − 3x+ 2 = (x− 1)2(x+ 2), so 1,−2 are the solutions. −2 is
obtained by choosing −1 for both. −1 is obtained by choosing a primitive cube rootof unity α for one root, and α for the other.
16.12 Quintic Equations
Exercise 16.12.7. Find a polynomial of degree 7 over Q whose Galois group is S7.
Solution. We first claim Sp is generated by a p-cycle and a transposition for p prime.Let (ab · · · cd), (ef) be our p-cycle and transposition. By renaming e = 1, we can as-sume they are (ab · · · 1 · · · cd) = (1 · · · cdab · · · ), (1f). Now (1 · · · cdab · · · )k = (1f · · · )for some 1 ≤ k ≤ p − 1 since p is prime, hence we can assume our generators are(1f · · · ), (1f); by renaming f = 2 and the rest of the elements in (1f · · · ) accordingly,we can assume our generators are α = (12 · · · p), β = (12). Now
so we can generate all permutations of the form ((k − 1)k). These generate Sp.Now let f(x) = (x3−2)(x2−4)(x2−32)+2 = x7−36x5−2x4+128x3+72x2−254.
This is irreducible by the Eisenstein criterion; also it has 5 real roots and 2 complexroots by looking at its graph. The only permutations of its roots that fix the real
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roots is conjugation, which corresponds to a transposition in G; moreover, G operatestransitively on the roots, hence contains a 7-cycle, and the Galois group is thereforeS7 by the above.
16.M Miscellaneous Problems
Exercise 16.M.4.(a) The non-negative real numbers are those having a real square root. Use this
fact to prove that the field R has no automorphism except the identity.(b) Prove that C has no continuous automorphisms other than complex conjuga-
tion and the identity.
Proof of (a). Any automorphism ϕ is clearly the identity on the integers and ratio-nals, since ϕ(n) = ϕ(1+· · ·+1) = ϕ(1)+· · ·+ϕ(1) = n, and then qϕ(p/q) = ϕ(p) = p.
Let a be such that ϕ(a) = b 6= 0. Then b < a after possibly multiplying by −1.Subtracting a suitable rational from each, we get b < 0 < a. But then a = α2, soϕ(α)2 = ϕ(α2) = ϕ(a) = b, contradicting b < 0.
Proof of (b). By (a), any automorphism ϕ is the identity on Q. Now letting i besuch that i2 = −1, we see ϕ(i)2 = −1, but since only ±i square to get −1 in C, onQ[i] we have that ϕ is either the identity or complex conjugation.
Finally, Q[i] is dense in C, hence any automorphism of C must also be either theidentity or complex conjugation by continuity.
Exercise 16.M.7. A polynomial f in F [u1, . . . , un] is12-symmetric if f(uσ1, . . . , uσn)
= f(u1, . . . , un) for every even permutation σ, and skew-symmetric if f(uσ1, . . . , uσn)= (sign σ)f(u1, . . . , un) for every permutation σ.(a) Prove that the square root of the discriminant δ :=
∏i<j(xi − xj) is skew-
symmetric.(b) Prove that every 1
2-symmetric polynomial has the form f + gδ, where f , g are
symmetric polynomials.
Proof of (a). It suffices to show that for any permutation of two elements contributesa negative sign, for any permutation is generated by permutations of two elements,and the number of permutation of two elements needed to generate a given permu-tation is equal to its sign.
So suppose we are interchanging ui, uj; assume without loss of generality thati < j. Then the terms in δ involving ui, uj have the product
(ui−uj)i−1∏k=1
(uk−ui)j−1∏k=1
(uk−uj)j−1∏k=i+1
(uk−uj)j−1∏k=i+1
(ui−uk)n∏
k=j+1
(ui−uk)n∏
k=j+1
(uj−uk)
96
Now interchanging ui, uj causes a sign change in each factor; the sign change inui−uj is the only one that remains since each two adjacent factors in the rest of theproduct have canceling sign changes. Hence δ also changes sign when interchangingui and uj, i.e., δ is skew-symmetric.
Proof of (b). Let h be our 12-symmetric polynomial. Assume that charF 6= 2. If h is
symmetric, we are done by letting f = h, g = 0 so suppose not. The action of S = Snon h has orbit {h, h′} for some other polynomial h′, since by the orbit-stabilizertheorem |Sh||Sh| = |S|, but Sh = An so |Sh| = 2. This implies f = 1
2(h + h′) is
symmetric, for h, h′ will just be interchanged. g = h−f = 12(h−h′) is antisymmetric
for the same reason.It now suffices to show that any antisymmetric polynomial g is divisible by δ,
i.e., any binomial ui−uj divides it. For, if ϕ is the substitution map letting uj = ui,then ϕ(g) = −ϕ(g) implies ϕ(g) = 0. So letting g = g/δ suffices, since g cannot beanti-symmetric otherwise g would be symmetric by (a).
Exercise 16.M.10. Let K be a finite extension of a field F , and let f(x) be in K[x].Prove that there is a nonzero g(x) in K[x] such that the product f(x)g(x) is in F [x].
Proof. We can assume f(x) is irreducible by working with irreducible factors indi-vidually; suppose moreover that it is monic. Any root α of f(x) is algebraic over Khence algebraic over F by Exercise 15.10.1. Thus we have h(x) ∈ F [x] the minimalpolynomial for α over F . Then f | h since f is defined over K; thus h = fg for someg ∈ K[x].
