Class Notes - Dec. 10

7/31/2019 Class Notes - Dec. 10

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Class Notes for Math 5820

Fall 2010


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Contents

Preface v

Chapter 1. Rings and Modules 11. Review of prime ideals and maximal ideals 12. Review of homomorphisms of rings 53. Modules over a ring 6

4. Free modules 105. Noetherian Rings 14

Chapter 2. Types of domains 191. Factorization in domains 192. Euclidean domains 223. Integral extensions of rings. 264. A quick look at field extensions 295. Quadratic number fields 316. Linear Algebra 337. The Trace and Norm functions for a finite field extension 368. First properties of the ring of integers in a number field 39

Chapter 3. Dedekind domains 431. Definitions and first properties 432. The Ideal Class Group of a Dedekind domain 473. Unique Factorization of Ideals in a Dedekind domain 50

Chapter 4. Modules over a PID 551. Split short exact sequences 552. Structure of f.g. modules over a PID 573. Linear algebra and the Jordan Canonical Form 604. Presentation of a f.g. module over a PID. 625. Norms of ideals in a ring of algebraic integers 66

Appendix. Bibliography 71

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Preface

Students should have taken an undergraduate course in Abstract Algebra.Hence results on groups should be familiar to you. In addition, some ring the-ory should have been covered in the course as well. At UMSL, textbooks used forMath 4400 are Rotman (A First Course in Abstract Algebra) or Lang (Undergrad-uate Algebra). The following prerequisite topics can be found in Ch 3 of Rotmanor Ch 3 of Lang.

Definition of a commutative ring (with unity). Subrings, ideals. Homomorphisms of rings, kernel and image. Quotient rings R/I, First Isomorphism Theorem. Definition of (integral) domains, fields. Sums, products and intersections of ideals in a ring. Polynomial ring R[X].

The idea of these notes is that the class will fill in details of proofs as part ofthe course work. For example, I may state a theorem or an exercise (an examplefrom undergraduate algebra: a commutative ring R is a field iff the only ideals inR are (0) and R.) I will assign a student to write up the details of the proof (wemay have discussed the proof in class).

I quote from Rotman ([2] page 221):It is unusual to give such a detailed proof, for it tends to makea simple idea look difficult. You should regard a proof as anexplanation why a statement is true. Depending on who you aretalking to, you would probably give one explanation to a highschool student, another to one of your classmates, and yet an-other to your professor. As a rule of thumb, your proofs shouldbe directed toward your peers, one of whom is yourself. Makeyour proof as clear as possible, not too short, not too long. Ifyour proof is challenged, you must be prepared to explain fur-ther, so try to anticipate challenges by giving enough details inyour original proof.

In view of this, I will expect the student to consult with at least one other studentand the professor while writing up the answer.The final version in LaTeX form willbe inserted into the class notes under the students by-line. It will be importantthat this be done in a timely fashion, so that all students have access to the proofsof results from earlier classes.

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CHAPTER 1

Rings and Modules

1. Review of prime ideals and maximal ideals

In this course, all rings will be commutative rings with unity; ie. for elementsa, b in a ring R, a b = b a, and there will be an element in R called 1, (or 1R whena distinction is needed) such that a 1 = 1 a = a for any a R.

Such a ring R will be called an integral domain (or often, just a domain) if

for any a, b R, ab = 0 a = 0 or b = 0. In any ring, if two non-zero elementsmultiply to give 0, they are called zero-divisors, hence a domain is a ring withoutany zero-divisors.

One of our rings will be called a field if every non-zero element has a mul-tiplicative inverse; ie. if a = 0, then there exists an element (usually calledthe multiplicative inverse or reciprocal and denoted a1) in the ring such thata a1 = 1. It is easily seen that the multiplicative inverse of a is unique if it exists,and (ab)1 = b1a1 = a1b1.

Conventionally, letters used for a field include F , k, K, L, while letters for a ringinclude R,S,A,B,D,O.

Proposition 1.1. A field F is always a domain. A domain R with a finitenumber of elements is a field.

Proof. Jaime Perkins.

Example 1.2. Jesse Bassett

A domain that is not a field:Given the general understanding of integers, we know that a, b Z, ifab = 0 then a = 0 or b = 0 . So Z is an integral domain.However, Z is not a field because for the number 2 Z, there is no integerwhich can take the role of a1 in the equation 2 a1 = 1. Indeed, themultiplicative inverse of 2 would be a rational number which is not aninteger.

A field:The set of rational numbers Q is a field. Any nonzero rational number

likeab which is not zero must have both a, b as non-zero integers. Then

ba

is the multiplicative inverse of ab . A ring which is not a domain:Let R = Z/(10), be a ring whose elements are as follows:

{0 + (10), 1 + (10), ..., 9 + (10)}Then, 2+(10)5+(10) = 10+(10) = 0+(10). Therefore, Z/(10) is not adomain because there exists divisors of zero that are not zero themselves

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2 1. RINGS AND MODULES

The notion of a ring being a sub-ring of another is quite easy. Recall that theunit elements of each have to coincide.

The following construction shows that any domain can be viewed as a subringof a field. We say that a domain D can be embedded in a field [D] called the fieldof fractions of D.

What is a fraction? It can be defined as a formal expression ab , where a, b Dand where b = 0.We have to agree that if c = 0, then ab and acbc are the same(note that in a domain bc cannot become 0!) Fractions defined thus can be added,subtracted and multiplied. Any element a of D can be viewed as a fraction a1 . Inthis way D is contained in the set of fractions as a sub-ring. Non zero fractions areones that look like ab , where both a, b are non-zero in D, hence

ba is its multiplicative

inverse.This is of course what we teach school children when we make them work

with rational numbers. Another common example from calculus is: if elements of

R[X] are called real polynomials in X, then expressions of the form f(X)g(X) are called

rational functions. So the field of rational functions is just the field of fractions ofthe ring (domain, in fact) of real polynomials.

Definition 1.3. Ideals

(i) A subset I of a ring R is called an ideal if it is an additive subgroup of R suchthat a I, r R ar I. The set {0} is called the zero ideal, denoted (0).The set R is called the unit ideal.

(ii) Let a R. The set (a) = {ra|r R} of all multiples of a is an ideal in Rcalled the principal ideal generated by a.

(iii) Let a1, a2, . . . , am R. We denote by (a1, a2, . . . , am) the set {r1a1 + r2a2 + + rmam|r1, r2, . . . , rm R}. This is an ideal in R and is called the idealof R generated by a1, a2, . . . , am.

(iv) Let I be an ideal in R. I is called finitely generated (f.g.) if you can

find a1, a2, . . . , am I such that I = (a1, a2, . . . , am). We may also say I ism-generated.(v) Let A be a (possibly infinite) subset of R. Consider the subset ofR described

as {nk=1 rkak|n N, rk R, ak A}. This is an ideal of R, called the idealof R generated by the subset A.

(vi) Suppose R is a subring of the ring S, and I is an ideal of R. The subset ISof S (see last definition) is an ideal of S which contains I.

Proposition 1.4. LetI, J be ideals in a ring R. Then

(i) I + J, defined as {i + j|i I, j J}, is an ideal in R containing both I andJ.

(ii) I J is an ideal contained in both I and J.(iii) IJ, defined as {

nk=1 rkikjk|n N, rk R, ik I, jk J}, is an ideal

contained in both I and J.Proof. Steve Burkett.

In general, to show that I R is an ideal we must show that a I, r R ar Iand that I is an additive subgroup of R.

IfG is a group, recall that S G is a subgroup iffS = and a, b S ab S. In our case, since a I, r R ra I, then in particular (1R)a = a I.So to show that I is a subgroup, it is enough to show that I is nonempty and that


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1. REVIEW OF PRIME IDEALS AND MAXIMAL IDEALS 3

it is closed under addition. Note that since 0R I,J, nonemptiness is triviallyverified in each case.

(i) Ifa, b I+J then we can write a = i+j, b = i+j, so a+b = i+j +i+j =(i + i) + (j + j) I + J, since I, J are closed under addition. If r R anda = i +j I+ J, then ra = r(i +j) = ri + ij I+ J, since I, J are closed undermultiplication by R.

(ii) Clearly I J is a subset of both I and J. For a, b I J, we havea, b I a + b I and a, b J a + b J, so a + b IJ. For r R, a IJ,we have ra I and ra J, so ra I J.

(iii) For a IJ, we can write a = nk=1 rkikjk. Then IJ I since a =nk=1(rkjk)ik, so each summand is in I, and I is closed under addition. A similar

argument gives IJ J.

Closure under external multiplication is immediate: if r0 R then r0a =r0n

k=1 rkikjk =n

k=1(r0rk)ikjk IJ, since r0rk R

It is not too hard to see that the sum a + b is also of the desired form. Sinceprecise notation is tricky, let us reword the situation as follows. Let A be the setA = {ij|i I, j J}. Then any element in IJ is obtained as follows: first choosefinitely many elements from A, then take a linear combination of these elements. Sothe element a is obtained from one finite subset ofA, the element b is obtained fromanother possibly different finite subset. Hence a + b can be seen to be obtained asfollows: consider the union of the finitely many elements from A corresponding toa and to b. a + b is a linear combination of these elements from A. Thus a + b IJ.

Example 1.5. Ideals I and J in a ring R for which IJ IJ, but IJ = IJ.Charles Macaulay Let R be the ring of integers, and let I = (4) and J =

(6). Then I J = (12) But, the odd multiples of 12 are not in IJ since IJ ={nk=1 rkikjk | rk R, ik (4), jk (6)} and no multiple of 4 times a multiple of6 will ever equal an odd multiple of 12. Hence, IJ = (24) and IJ I J.

Definition 1.6. Let R be a ring and I an ideal not equal to R. corrected fromearlier.

(i) I is called a prime ideal if ab I a Ior b I.(ii) I is called a maximal ideal if the only ideals J in R such that J contains I

are J = I and J = R.

Proposition 1.7. Let P be an ideal in the ring R. Then the following are

equivalent:

(i) P is a prime ideal.(ii) R/P is an integral domain.

Proof. Ashley Neal. An ideal P of R is prime if and only if for all r, s R,rs P implies r P or s P.

This is equivalent to saying for all (r + P), (s + P) R/P, (r + P)(s + P) = 0 R/P implies r + P = 0 R/P or s + P = 0 R/P, which is equivalent to saying


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that R/P has no zero divisors. Since all rings by our definition are commutativewith unity, this is equivalent to R/P being an integral domain.

Proposition 1.8. Let M be an ideal in the ring R. Then the following areequivalent:

(i) M is a maximal ideal.(ii) Whenever a R and a / M, then M + (a) is the unit ideal.

(iii) R/M is a field.

Proof. Steve Burkett.(i)(ii). M+ (a) is an ideal in R, by Proposition 1.4. We also have M M+ (a),since a = 0 + 1a M+ (a), but a / M by hypothesis. Since M+ (a) is an ideal inR that properly contains M, it must be all of R, because M is maximal.

(ii)(iii). For any a + M R/M, if a + M = 0R/M then a / M, so by(i) above, M + (a) = R. Since 1

R, there is some m

M and b

R, such

that m + ba = 1 ba 1 = m ba 1 M (since M is an ideal). Thusba + M = 1 + M, so (b + M)(a + M) = 1 + M, and b + M = (a + M)1. Thus anynon-zero element of R/M has an inverse, so R/M is a field.

(iii)(i). Suppose N is an ideal of R with M N, and let n N. Thena R such that (a + M)(n + M) = an + M = 1 + M, since R/M is a field. Thenan 1 M, that is, m M such that an m = 1. But M N m N, andN is an ideal, so 1 = an m N, thus N = R, and M is a maximal ideal.

Corollary 1.9. A maximal ideal is always a prime ideal.

Proof. This is obvious from 1.8 and 1.1.

Remark 1.10. A field F has only two ideals: (0) and F itself. These ideals,the zero ideal and the unit ideal are often called the trivial ideals. ie. a non-trivialideal should be interesting.

1.1. Construction ofR from Q. We finish this section with a complicatedbut important example that every math graduate student should know.

A sequence of rational numbers an is called a Cauchy sequence in Q if it satisfiesthe following condition:

Given Q>0, there exists an integerN > 0 such that m,n > N |aman| < .Let C denote the set of all Cauchy sequences in Q. It is easy to see (any course in

real analysis) that given two Cauchy sequences, their sum, difference and product is

also a Cauchy sequence. The sequence 0, 0, 0 . . . and the sequence 1, 1, 1, . . . standfor the zero element and the unit element, making C a commutative ring (with unitelement).

Let M denote the subset of all null Cauchy sequences where an is callednull if given Q>0, there exists an integer N > 0 such that n > N |an| < .(ie. it converges to 0). It is easy to see that M is an ideal in C. It is also easy toprove that M is a maximal ideal in C.

Hence C/M is a field. This is the field that we usually call R.


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2. REVIEW OF HOMOMORPHISMS OF RINGS 5

2. Review of homomorphisms of rings

Let : R

S be a homomorphism of rings. In particular, we require (1R) =

1S. The following are important properties to be remembered.(i) ker , defined as {r R|(r) = 0S} is an ideal in R.

(ii) image is a subring of S.(iii) is monomorphism (often denoted R S) iff ker = (0R).(iv) The map : R/ ker image given by (r + ker ) = (r) is well-defined

and is an isomorphism. (This is the First Isomorphism Theorem.)

One of the most important homomorphisms will be the map to the quotientring: let I be an ideal of a ring R. Form the quotient ring R/I = {r + I|r R}.Very often, the coset r + I will be just denoted by r, if there is no confusion aboutwhat the I is. The canonical homomorphism from R to R/I sends r R tor R/I. It is a surjective homomorphism with kernel equal to I. In view of theFirst Isomorphism Theorem, any surjective homomorphism (denoted by ) can be

thought of as a canonical homomorphism because we have a commuting diagram

R

can## ##G

GGGG

GGGG

// // S

R/ ker

=;;wwwwwwwww

Proposition 2.1. . Let : R S be a ring homomorphism.(i) LetJ be an ideal in S. Then 1(J) is an ideal in R.

(ii) Suppose (1(J)) = J. (Always true if is onto). Then 1(J) is the onlyideal I in R with the two properties:(a) I ker (b) (I) = J.

(iii) LetI be an ideal in R. Then (I) is an ideal in S, provided is onto.(iv) When is onto, there is a one-to-one correspondence between the set of ideals

of R satisfying the two conditions above and the set of ideals ofS.

Proof. Ryan SemanProof of i: Let J be an ideal in S. To show that 1(J) = I is an ideal in R.(0R) = 0S J. Therefore 0R I. Take a, a I. Then (a) = b, (a) = b,where b, b J. Then (a a) = (a) (a) = (b b) J. Therefore a a I.Take r R. Then (r) = s S. Then (ra) = (r)(a) = sb J. Thereforera I. And so 1(J) = I is an ideal in R.Proof of ii: Given that (1(J)) = J. To show that (1(J)) = I is the onlyideal in R that satisfies:a) ker()

I.

b) (I) = J.First to show I satisfies (a) and (b). Take a R such that a ker(). Then(a) = 0S J, hence a I. Therefore ker() I. (1(J)) = J as given, andI = 1(J), so (I) = J. Therefore I satisfies (a) and (b).Next to show if K is an ideal in R that satisfies (a) and (b), then K = I. LetK be such an ideal. Take k K. Let (k) = j J. So k 1(J), but1(J) = I. So k I. Therefore K I. Take x I. Let (x) = y J.Since (K) = J, y = (x) for some x K. Then (x) = y = (x). Then


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(x x) = (x) (x) = b b = 0S. Therefore (x x) ker(). ker() K asgiven, therefore (x x) K. Since x K, it must be that (x x + x) K, andso x

K. So I

K. Therefore I = K.

And so 1(J) = I is the only ideal in R such that ker() I, and (I) = J.Proof of iii: Given that : R S is an epimorphism. Suppose I is an ideal inR. Let (I) = J. To show J is an ideal.0R I, and (0R) = 0S . Therefore 0S J. Take x J and y J. Then thereis some x I such that (x) = x. and there is some y I such that (y) = y.So (x y) = (x) (y) = (x y) J. Therefore x, y J (x y) J.Take any s S. Because is onto, there is an r R such that phi(r) = s. So(rx) = (r)(x) = sx S. Therefore s S, x J sx J. Therefore J is anideal in S.Proof of iv: Given that : R S is an epimorphism. Let Ibe the set of allideals of R containing ker(). Let J be the set of all ideals of S. Need to showthere is a one-to-one correspondence between Iand J.Let : I J be given by (I) = (I) (ponder and digest the reasons for suchstrange notation!).We need to show that is one-to-one and onto. Let I1, I2 I, and J J suchthat (I1) = J = (I2). As given above, (I1) = (I1), and (I2) = (I2). Butby proof of part (ii), if (I1) = J = (I2), then I1 = I2. Therefore, is one-to-one.Let K J, then K is an ideal in S. By proof of part (ii), there is an ideal I inR, such that (I) = K with the property ker() I, so I is an element in I. Weknow from above that (I) = (I), so there is an ideal I in Isuch that (I) = J.So is onto, and therefore is an bijection from I J.

Example 2.2. Ashley Neal A situation : R S, I an ideal in R where (I)is not an ideal in S.Consider : Z

C, where (z) = z. Clearly, I = (2) is an ideal in Z; it is the even

integers. In order for (I) to be an ideal in C, each element in I multiplied by eachelement in C must be in (I). But 2 12 = 1 / (I), so (I) is not an ideal in C.

Remark 2.3. Let I be an ideal in R. Construct the quotient ringR/I. OftenR/I is denoted R, when I is understood. Let H be another ideal in R. Its imagein R is an ideal. This image is denoted by the notation H and also HR. If Hcontains I, then H is the same as H/I. In general, H = (H+ I)/I.

3. Modules over a ring

Definition 3.1. Let R be a commutative ring with unit element. A set M iscalled a (left) module over R ifM is an abelian group (under an operation +), andthere is a map (called multiplication by scalars) R M M, which is denotedby (r, m)

rm or r

m, with the following properties: for any r, r1, r2

R and

m, m1, m2 M,(a) r(m1 + m2) = rm1 + rm2(b) (r1 + r2)m = r1m + r2m(c) (r1r2)m = r1(r2m)(d) 1R m = m

There is also the notion of a right R-module, but for a commutative ring R,the concepts coincide. We will always strive to keep scalars on the left and module


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3. MODULES OVER A RING 7

elements on the right, but sometimes for typographical reasons it may be hard andif you see an expression mr, you should take it for rm. It is quite easy to verifythat for any r, r

0M

= 0M

and (

r)m =

(rm). (Private exercise for students.)

Example 3.2. Some modules:

When our ring is a field F, an F-module is traditionally called a vectorspace over F.In otherwords, working with a vector space is a special caseof working with modules.

When R is a subring of a ring S, then S is an R-module. In particular,R is an R-module.

Any ideal I in R is an R-module. Any abelian group G is a Z-module.

A submodule H of an R-module M is an abelian subgroup of M which isclosed under scalar multiplication. Given a submodule H of M, we can constructthe quotient module M/H.

Proposition 3.3. LetH be a submodule of the R-module M. Then the abeliangroup M/H has a natural scalar multiplication r (m + H) = (rm) + H (or oftenwritten r m = rm) which is well-defined, making M/H also an R-module.

Proof. Michael YoungLet H be a submodule of the R-module M. Since H is an abelian subgroup ofM,closed under scalar multiplication, the quotient group M/H = {m + H|m M}can be defined.

Let us define scalar multiplication as r(m+H) = (rm)+H, r R, m M. Thisscalar multiplication in the quotient module is well-defined if, for m + H = m + H,r(m + H) = r(m + H).Since m, m are in the same coset, m m H and hence r(m m) H, since His an R-module. Thus, (rm

rm)

H or rm + H = rm + H and

r(m + H) = (rm) + H = (rm) + H = r(m + H).Thus scalar multiplication in the quotient module M/H is well-defined.

We need to prove various axioms for an R-module, like:For any r1, r2 R and m M it follows that(r1 + r2)(m + H) = r1(m + H) + r2(m + H).RHS: r1(m + H) + r2(m + H) = (r1m + H) + (r2m + H) by the definition of scalarmultiplication.This equals (r1m + r2m) + H by adding of cosets. Using the distributive propertyon M, this in turn equals ((r1 + r2)m) + H = (r1 + r2)(m + H) :LHS, again by thedefinition of scalarmultiplication. The rest are proven similarly

The following results will look similar to the earlier results on ideals.

Proposition 3.4. LetM be an R-module.

(i) Let m M. The set Rm = {rm|r R} of all multiples of m is a submoduleof M called the cyclic submodule generated by m.

(ii) Letm1, m2, . . . , mh M. We denote byRm1 +Rm2+. . . R mh the set{r1m1+r2m2 + + rhmh|r1, r2, . . . , rh R}. This is a submodule ofM and is calledthe submodule generated by m1, m2, . . . , mh.


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(iii) M is called finitely generated (f.g.) if you can find m1, m2, . . . , mm Msuch thatM = Rm1 + Rm2 + . . . R mh. We may also sayM is m-generated.

(iv) LetA be a (possibly infinite) subset of M. Consider the subset of M describedas {nk=1 rkak|n N, rk R, ak A}. This is a submodule of M, called thesubmodule generated by the subset A.

Proposition 3.5. LetN1, N2 be submodules of an R-module M. Then

(i) N1 + N2, defined as {n1 + n2|n1 N1, n2 N2}, is a submodule of Mcontaining both N1 and N2.

(ii) N1 N2 is a submodule of M contained in both N1 and N2.Proof. Michael Young.

Definition 3.6. Let M, N be R-modules. A homomorphism of R-modules isa map : M N with the properties (i) (m1 + m2) = (m1) + (m2) and (ii)(rm) = r(m).

It is easy to verify that (0M) = 0N, (m) = (m), (nk=1 rkmk) =nk=1 rk(mk).

In linear algebra, homomorphisms of vector spaces are called linear transfor-mations.

Here is a very important example of a homomorphism: let H be a submoduleof the module M. Then the map M M/H, given by m m is a surjectivehomomorphism. It is called the canonical homomorphism from M to M/H.

We can now define monomorphisms (injective homomorphisms), epimorphisms(surjective homomorphisms) and isomorphisms of modules.

Proposition 3.7. LetM, N be R-modules, : M N a homomorphism.(i) image is a submodule of N.

(ii) ker =

{m

M

|(m) = 0N

}is a submodule of M.

(iii) is a monomorphism iff ker = {0M}.Proof. Jesse Bassett Proof of i: Two elements in image have the form

(a), (b), for some elements a, b M. We need to show that the image is anabelian group and closed under scalar multiplication.(a b) M (a) (b) = (a b) image .AndGiven a M and r R then ra M since M is an R-module. Which implies:r(a) = (ra) image .image is an abelian group and closed under scalar multiplication.Thus, making the image a submodule of N.Proof of ii: The kernel of is made up of all elements in M that are sent to zeroin N. Let a, b

ker

(a), (b) = 0N. Then (a

b) = (a)

(b) = 0. Then

(a b) ker . Similarly, given r R and a ker . So, (ra) = r(a) = r(0) =0. ra ker .Therefore, ker is a submodule of M.Proof of iii:() Suppose is a monomorphism, that is one-to-one. Let a ker . We needto show that a = 0M. Since a ker , we know that (a) = 0. So, (a) = (0).Therfore, a = 0M() Suppose ker = {0M}. We need to show that is a monomorphism. Suppose


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3. MODULES OVER A RING 9

a, b M and (a) = (b). Working with (a) = (b) one can move the right handside over and see that (a) (b) = 0. So, (a b) = 0 a b ker . Thus,a

b = 0M

. Thus a = b when the images are equal. Thus, is a monomorphism.

Theorem 3.8. First Isomorphism Theorem Let : M N be a homomor-phism ofR-modules. Then the map : M/ ker image given by(m) = (m)is well-defined and is an isomorphism.

Proof. Steve Burkett Define by (m) = (m).

Then is well defined, for if m = m then m m ker , so 0 = (m m) =(m) (m) = (m) (m), giving m = m (m) = (m).

is a homomorphism:(1) (m + m) = (m + m) = (m + m) = (m) + (m) = (m) + (m).(2) (rm) = (rm) = (rm) = r (m) = r (m)

is onto: n im m M s.t. n = (m) = (m)

is one to one: (m) = (m) (m) = (m) (m) (m) = 0 (m m) = 0 m m ker m = m.

Since is a homomorphism that is both one to one and onto, it is an isomor-phism.

In view of the First Isomorphism Theorem, if : M N is an epimorphism,we can identify N with M/ ker , with a commuting diagram:

M

can$$ $$H

HHHH

HHHH

// // N

M/ ker

=::vvvvvvvvv

Theorem 3.9. Quotient of quotients Let M be an R-module, H a submoduleof M and K a submodule of H. (K H M.) Then

M/H = (M/K)/(H/K).Proof. Charles Macaulay. Define : M/K M/H by a + K a + H. It is

easy to see that this is a well-defined function. For any a, b M, ifa + K = b + K,then a

b

K. Since K

H, then a

b

H. This gives us a + H = b + H.

Therefore, (a + K) = (b + K). Hence, is well-defined.We can easily see that is a homomorphism since for any a, b M, we have

((a + K) + (b + K)) = ((a + b) + K)

= (a + b) + H

= (a + H) + (b + H)

= (a + K) + (b + K).


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Likewise, for scalar multiplication. It is is also easy to see that is surjectivesince for any a M, a + H M/H (a + K) = a + H by definition of . Thus,image = M/H.

Now, in order to apply the First Isomorphism Theorem, we have to show thatker = H/K. Since 0 + H = H is the zero element of M/H and a + K H iffa H, then ker is simply the quotient module H/K. Hence, we can apply theFirst Isomorphism Theorem, and we get

M/H = (M/K)/(H/K).

Theorem 3.10. LetH and K be two submodules of an R-module M. Then

H/(H K) = (H+ K)/K.Proof. Ashley Neal

Consider the mapping f : H

(H + K)/K, where f(h) = h + K. This is ahomomorphism since f(h1 + h2) = (h1 + h2) + K = (h1 + K) + (h2 + K) = f(h1) +f(h2) and f(rh) = rh + K = r(h + K) by proposition 3.3 = r f(h). f is also onto.If (h + k) + K is an arbitrary element of (H+ K)/K, with h H and k K, thenf(h) = h+K = h+(k+K) = (h+k)+ K. So image f = (H+K)/K. Consequently,the first isomorphism theorem says H/ker(f) = (H+ K)/K. So to finish the proofwe need to show kerf = H K. Let x ker(f), so f(x) = 0(H+K)/K = K. ThenK = f(x) = x + K, so x K. Since ker(f) H, x H. Hence, x H K, soker(f) H K. Let x H K. Then x K, so f(x) = x + K = K. Thereforex ker f, so H K ker f. Hence ker f = H K, so the proof is complete

Theorem 3.11. LetR be a ring, andI an ideal ofR. LetM be anR-module. Ifa m = 0M for anya I and m M, then there is a natural R/I-module structureon M. ie. scalar multiplication by elements inR/I can be defined unambiguouslyon M.

Proof. Jamie Perkin

Nice terminology for this result is to say that if M is annihilated by I, thenM is actually an R/I-module.

Corollary 3.12. LetN be an R-module, I an ideal of R. Denote by IN thesubmodule of N generated by the the subset {an|a I, n N} (see 3.4,iv). ThenN/IN, which is the quotient R-module, is actually an R/I-module.

Proof. Ryan Seman

4. Free modules

Let M1 and M2 be two R-modules. The direct sum of M1 and M2 is definedas

M1 M2 = {(m1, m2)|mi Mi, i = 1, 2}.

It is an easy check that M1 M2 is also an R-module, with zero element(0M1 , 0M2). Recall that the strength of ordered pairs is that ( m1, m2) = (m3, m4) m1 = m3, m2 = m4.


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Corollary 4.6. Let H be any n-generated R-module. Then there is an epi-morphism Rn H. In particular, if K is the kernel of this epimorphism, H =Rn/K.

Proof. Steve Burkett Let M = Rn with the standard basis and let {h1, h2,...,hn}be generators of H. Then the homomorphism guaranteed by Proposition 4.5satisfies hi = (ei) for all 1 i n, and is onto because if h H, thenh =

rihi =

ri(ei) = (

riei).

Since im() = H, the map : Rn/K H by (m) = (m) is an isomorphismby Theorem 3.8

Theorem 4.7. (Universal property of a free module) Let M be a f.g. free

R-module. Suppose the following information is given.

H

// K // 0

M

OO

(This diagram is another way of indicating that is an epimorphism ofR-modules.)Then there exists a lifting of the homomorphism to give the commuting diagram(ie. = ):

H

// K //

0

M

``BBBBBBBB

OO

Proof. Charles Macaulay Let {m1, m2, . . . , mn} be a basis for M, and letm M. Then we have

m =ni=1

rimi

for some ri R. Then applying to m, we get

(m) =ni=1

ri(mi).

Since is an epimorphism, for each (mi) K, there exists an hi H such that(hi) = (mi). Then by Proposition 4.5, given h1, h2, . . . , hn in H, there exists aunique homomorphism : M Hsuch that for each mi {m1, m2, . . . , mn}, (mi) =


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4. FREE MODULES 13

hi. Hence ((mi)) = (hi) = (mi) for all i. Therefore,

(m) =

ni=1 r

i(mi)

=ni=1

ri((mi))

=ni=1

ri()(mi)

= ()(ni=1

rimi)

= ()(m)

Since the domains of and are the same, we conclude that = .

Proposition 4.8. Let I be an ideal of R. Suppose M is a f.g. free R modulewith basis {m1, m2, . . . mn}. ThenM/IM is a f.g. free R/I module (see 3.12) withbasis {m1, m2, . . . mn}.

Proof. Ashley Neal

Theorem 4.9. Let V be a finite dimensional vector space over a field F. Let{v1, v2, . . . , vn} and {w1, w2, . . . , wm} be two bases for V over F. Then n = m. ie.the dimension of a finite dimensional vector space is well defined.

Proof. Steve Burkett To begin, we will assume that m > n and then showthat if{v1,...,vn} is a basis, then {w1,...,wm} must be a linearly dependent set.

Since {v1,...,vn} spans V, we can write each wi as a linear combination of thevis, that is, wi = ai,1v1 + ai,2v2 + ... + ai,nvn, i, or in matrix notation:

a1,1 . . . a1,n

.... . .

...

am,1 . . . am,n

v1...

vn

=

w1

...

wm

We recall from elementary linear algebra that since the first matrix has more

rows than columns, the rows must be linearly dependent. It is then clear that thewis are linearly dependent as well.

The result used above is not trivial, so we now give a slightly technical alternateproof that does not rely on matrices. We will show that if m > n, then {w1,...,wn}must span V. This will mean that there are some ais for which wn+1 =

ni=1 aiwi,

so that the wis are linearly dependent.


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The proof is by induction. We show that whenever {w1, . . . , wk, vk+1, . . . , vn}is a set that spans V, {w1, . . . , wk+1, vk+2, . . . , vn} must span V as well. To see thatthis is true for k = 0, use the fact that the v

is form a basis to write w

1= n

i=1aivi.

There must be some ai = 0, and by reindexing the aivis so that a1 is our nonzerocoefficient, we have:

w1 = a1v1 +ni=2

aivi v1 = w1a1

1a1

ni=2

aivi

Since v1 is a linear combination of{w1, v2, . . . , vn}, this set spans V. For the generalcase, we write wk+1 =

ki=1 aiwi +

ni=k+1 bivi. There must be some bi = 0, or

else the set {w1, . . . , wk+1} would be linearly dependent, a contradiction. Againreindexing {vk+1, . . . , vn} as needed:

wk+1bk+1

=1

bk+1

k

i=1aiwi + vk+1 +

1

bk+1

n

i=k+2bivi,

so vk+1 is a linear combination of {w1, . . . , wk+1, vk+2, . . . , vn}, as required. Byinduction, {w1, . . . , wn} generates V, and so the wis are linearly dependent when-ever m > n.

We conclude that if the wis are to constitute a basis, we must have m n.But by a symmetric argument we must also have n m, and therefore m = n.

Theorem 4.10. LetM be a f.g. free module over a ring R. Let{v1, v2, . . . , vn}and {w1, w2, . . . , wm} be two bases for M over R. Then n = m. ie. the rank of a

f.g. free module is well defined.

Proof. Rao. You will need the following fact which I will give you. Let R be

our ring. Then you can always find an ideal I in R which is a maximal ideal. Thisseems obvious, just keep enlarging an ideal till it becomes maximal. For the ringswe will study in this course (noetherian commutative rings) that will be true, and Iwill indicate it when we define noetherian. In general we must use Zorns lemmawhich is a deep assumption about mathematics.

Granted that you can find a maximal ideal I in R, you should be able to seethe rest of the proof from the previous couple of theorems.

5. Noetherian Rings

Definition 5.1. A ring R is called noetherian if every ideal of R is finitelygenerated.

It is a convenient definition since most of the rings that we come across will

have this property. For example, a field is noetherian since it has only two ideals(0) and (1), both principal. Z is noetherian since every ideal in Z is principal.The theorems we prove below will show us that when we build new rings fromnoetherian rings, they tend to be noetherian also. Here is an easy example of anon-noetherian ring: for any ring R consider the polynomial ring in infinitely manyvariables X1, X2, X3, . . . with coefficients in R. It must seem quite believable toyou that the ideal generated by X1, X2, X3, . . . is not finitely generated.

For convenience, we make a similar definition for modules:


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5. NOETHERIAN RINGS 15

Definition 5.2. An R-module M is called noetherian if every submodule ofM is finitely generated.

Note that a noetherian module is itself finitely generated, since it is a sub-module of itself. Also note that if R is a noetherian ring, as an R-module, it is anoetherian R-module since any submodule of R is just an ideal of R.

Let us also introduce the notation of a short exact sequence. Let M, M andM be three R-modules. Then

0 M f M g M 0is called a short exact sequence of homomorphisms if the homomorphisms f and gsatisfy

f is a monomorphism (exact on the left) g is an epimorphism (exact on the right) image f = ker g (exact in the middle).

Here are the most common ways in which an exact sequence may come up:

Suppose M g M is an epimorphism. Then we have a short exact se-quence

0 ker g M g M 0. Suppose M is a submodule of M. Then the canonical homomorphism

M M/M is an epimorphism with kernel M. So we get0 M M M/M 0.

Theorem 5.3. Let 0 M f M g M 0 be an exact sequence of R-modules.

(1) IfM is noetherian, so are M and M.(2) IfM and M are noetherian, so is M.

Proof. Since f is a monomorphism, we may certainly identify M with itsimage in M. Hence assume that M is a sub-module of M, with f equal to theinclusion. This lends more simplicity to the language below. The proof of the firstpart, I leave to Jaime Perkin.

For the second part, suppose N is a submodule of M, given that M and M

are noetherian. Then g(N) is a submodule of M (which is noetherian), hence ithas a finite number of generators which can be described as g(n1), g(n2), . . . , g(nr)for some n1, n2, . . . , nr N. In addition, N M is a submodule of M (which isnoetherian) , hence it has finite number of generators nr+1, . . . , nr+s. Naturally,we will try to show that n1, n2, . . . , nr, nr+1, . . . , nr+s in N generate N. So letn N. Then g(n) = a1g(n1) + a2g(n2) + arg(nr), since g(n) g(N). But theng[n a1n1 a2n2 . . . arnr] = 0. Hence n a1n1 a2n2 . . . arnr is in ker g. Byexactness, ker g equals image f = M. Hence n

a1n1

a2n2

. . . arnr

M. It

is also in N. Hence n a1n1 a2n2 . . . arnr is a combination of nr+1, . . . , nr+s,and we are done, since now n is a combination ofn1, n2, . . . , nr, nr+1, . . . , nr+s.

Corollary 5.4. IfM1, M2, . . . , M n are noetherianR-modules, thenM1M2 Mn is also noetherian.

Proof. Sketch for Ryan Seman. First prove for n = 2. How does M1 M2project to M2 and does that give a nice exact sequence? The induction process forgeneral n is easy.


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Theorem 5.5. Suppose R is a noetherian ring. Then any finitely generatedR-module is a noetherian module.

Proof. Since R is a noetherian ring, R is a noetherian R-module (whosesubmodules are the ideals of R). By the corollary above, Rn is also a noetherianR-module. Now if M is any finitely generated R-module, with say n generators,by Corollary 4.6, we get a short exact sequence 0 K Rn M 0 (where Kis the kernel of the surjection Rn M. Now use part 1 of the theorem above toconclude that M is also noetherian.

In many situations, we may have a ring homomorphism : R S. If M is anS-module, then we can use to make M into an R-module as well, in an obviousway; let r m be defined as (r)m. It may be that M is finitely generated as anS-module (we say finitely generated over S), but not finitely generated as anR-module. For example, R may be tiny and not provide enough scalars, while Smay have lots more scalars.

Proposition 5.6. Suppose R is a noetherian ring and there is a ring homo-morphism : R S. This makes S into an R-module and suppose also that S isthen a noetherianR-module. Then S is a noetherian ring.

Proof. Let I be an ideal of S. Then I is an R-submodule of the noetherianR-module S (check this in your head). Hence I is finitely generated over R, bythe theorem above. What this means is that you can find a set of generatorss1, s2, . . . , sk in I such that everything in I can be expressed as a linear combination

ri si =

(ri)si. But since each (ai) is a scalar in S, we see that s1, s2, . . . , skgenerate I as an S-module. Hence the ideal I of S is a finitely generated S-module.

Corollary 5.7. Let R be a noetherian ring, I an ideal. The R/I is a noe-therian ring.

Proof. Michael Young

Proposition 5.8. A noetherian module M satisfies the following additionalequivalent properties. So let M be any R-module. The following are equivalent:

(1) Any submodule of M is finitely generated.(ie.M is noetherian.(2) M satisfies the Ascending Chain Condition (ACC) for submodules which

states that you cannot find a sequence of larger and larger submodules ofM without coming to a halt after finitely many steps. Formally, we write ifM1 M2 M3 . . . is a chain of increasing submodules of M, thenthere is an index N such that MN = MN+1 = MN+2 = . . . .

(3) Any set of submodules ofM contains a maximal submodule.

Proof. (1)

(2)Jesse Bassett,.

(2)(3)Steve Burkett. Suppose M is a set of submodules of M that does notcontain a maximal submodule. (Here, maximal means M-maximal in the sense ofset theoretic inclusion). Then for any M1 M, there is some M2 M such thatM1 M2. But then M2 cannot be a maximal submodule since it is in M, andcontinuing by induction we can build an infinite chain M1 M2 M3 ... ofsubmodules of M, contradicting (2).

(3) (1)Charles Macaulay. Let N be any submodule of M, and let {N}be the collection of finitely generated submodules of N. Since N N M


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5. NOETHERIAN RINGS 17

for all , {N} is also a collection of submodules of M. By assumption, thereexists a maximal element, Nk, in the collection. Nk is a f.g. submodule of N, sayNk

= Rn1

+Rn2

+. . .+Rnk

. IfN= N

k, then there exists an element n

k+1 N

\Nksuch that Nk = Rn1 + Rn2 + . . . + Rnk Rn1 + Rn2 + . . . + Rnk + Rnk+1 N.

Since Nk is maximal, this is a contradiction. Thus, N = Nk.Therefore, any submodule N of M is finitely generated if any collection of

submodules of M has a maximal element.

A famous theorem in this subject is the Hilbert Basis Theorem, proved byHilbert in 1888. Noetherian rings were defined as such after Emmy Noether, onlymuch later, so Hilbert did not use the language of noetherian rings. This was one ofthe first non-constructive proofs and algebra was dominated then by a topic calledinvariant theory, where people tried to construct invariants. Many researchers didnot appreciate Hilberts proof. Paul Gordan advised against publication of the

proof saying Das ist keine Mathematik. (see Roger Cooke, Notices of the AMS,Vol 57, No 4, 2010)

Theorem 5.9. Hilbert Basis Theorem. Suppose R is a noetherian ring. ThenR[X] is also noetherian.

Proof. Let I be an ideal in R[X]. Elements in I are polynomials, of varyingdegrees. We want to find a set of possible generators for I.

Define In to be the set of all possible elements a R, where a = 0 or a is theleading coefficient of some polynomial f I, where f has degree n. Do this forn 0, giving I0, I1, I2, . . .

Then

In In+1 for each n 0.

IN is an ideal of R.

Proof of first Ashley Neal In In+1 for each n 0:Assume a1 In. We want to show that a1 In+1. If a1 = 0, we have

0 In for each n 0 by definition, so we are done. If a1 = 0, then a1 is theleading coefficient of some polynomial f I of degree n. Since X R[X], X f =X(a1X

n + . . . + an+1) = a1Xn+1 + . . . + an+1X I. This shows that a1 is the

leading coefficient of some polynomial f I of degree n + 1, therefore a1 In+1.Proof of second Ashley Neal In is an ideal of R:

0 In, so it is nonempty. Next, we need to show that In is an additivesubgroup of R. Let a1, b1 In, so a1Xn + . . . + an+1, b1Xn + . . . + bn+1 I forsome polynomials in I. This implies that (a1X

n+. . .+an+1)(b1Xn+. . .+bn+1) =((a1 b1)Xn + . . . + (an+1 bn+1) I, since I is an ideal. Hence, a1 b1 In.Now we must show that given any r

R and a1

In, that r

a1

In. Since

a1 In, we have a1Xn + + an+1 I for some polynomial. Also, r R R[X].So r (a1Xn + + an+1) = ra1Xn + + ran+1 I, since I is an ideal of R[X].Therefore ra In.

Let me continue the story. Since R is noetherian, by the ACC condition,there is an index N such that IN = IN+1 = IN+2 = . . . Lets choose a ratherlarge collection of polynomials as follows. Each In is finitely generated, since R isnoetherian. Choose in In a set of generators an,1, an,2, . . . , an,tn for In. Each an,iis the leading coefficient of some degree n polynomial fn,i I. Choose an fn,i for


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each an,i. Do this for n = 0, 1, 2 . . . , N . We get a whole bunch of polynomials in Igiving a finite set S.

Claim: this finite setS

generates I, which would complete the proof.We prove this by induction. Let f I be a polynomial of degree k. By

induction on k, we will show that f is a linear combination of polynomials in S.If k = 0, then f is a constant polynomial. It is in I0, hence is a linear combi-

nation of those elements in Sthat came from I0. Hence the statement is true fork = 0.

Suppose the statement is true for k < n. To check it for k = n. So suppose fhas degree n and f is in I.

Case(1) n N. Let a be the leading coefficient of f. Then a In. Hencea =

tni=1 rian,i for some scalars ri R. Therfore f

tni=1 rifn,i has its degree n

term equal to zero. This means that ftni=1 rifn,i, (also in I) has degree < n. Bythe inductive assumption, f

tni=1 rifn,i is a linear combination of polynomials in

S. Hence f is one as well.

Case (2) n > N. Jaime Perkin

Corollary 5.10. Suppose R is noetherian. Then the polynomial ring in nvariables R[X1, X2, . . . , X n] is noetherian, and any of its quotient rings R[X1, X2,. . . , X n]/I is noetherian.

Proof. Apply the Hilbert Basis Theorem repeatedly to go from R to R[X1]to R[X1, X2] (which equals R[X1][X2]!) all the way to R[X1, X2, . . . , X n]. Thenapply Corollary 5.7.


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CHAPTER 2

Types of domains

1. Factorization in domains

An element u in a ring R which is invertible in R will be called a unit in R. 1Ris a unit, called the unit element or the identity element. The set R of all unitsin R is a multiplicative group, (for example, the product of invertible elements isan invertible element, etc).

Z = {1}. In a field F, F is the set of all non-zero elements. For any domain D, D and D[X] are the same set. This may not

be the case in the presence of divisors of zero. For example if R is aring and a non-zero element a is found in R such that a2 = 0, then(1 + aX)(1 aX) = 1.

In a ring like Z/(10), the set of units is {1, 3, 7, 9}, given by the numbersrelatively prime to 10.

Two elements a and b of R are called associates (or equivalent upto a unit) ifyou can find a unit u such that a = ub. This gives an equivalence relation on R.

For a, b R, we use the notation a|b (to be read as a divides b or b is amultiple ofa) if you can find an element c R such that b = ca. We say that a isa proper divisor of b if neither a nor c is a unit.

It is always true that a|0, since 0 = 0 a. a|1 if and only if a is a unit. If x|a and x|b, then x divides any linear combination of a and b. We can define the notion of a greatest common divisor as follows: let

a, b R. d R is called a gcd for a and b if d = 0, d|a, d|b and d satisfiesthe condition: (d|a d|b) d|d

Lemma 1.1. (1) In a ring R, a|b b (a) (b) (a).(2) In an integral domainD, a is a proper divisor of b (b) (a) D.

Proof. Ryan Seman Note: In part 2, the inclusions are proper inclusions.part 1:

(i) (ii): (a) = {ca : c R}. Let a|b, then ca = b for some c R, soca = b (a).

(ii) (iii): (b) = {db : d R}. Let b (a), then ca = b for some c R. Letd R, then b = ca = db = dca, and dc R since d, c R, so db = dca (a).Therefore (b) (a).

(iii) (i): Let (b) (a). 1R b = b (b). Since b (b), b (a), so b = ca forsome c R, therefore a|b.

part 2:

19


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20 2. TYPES OF DOMAINS

: In an integral domain D, let a be a proper divisor of b. Clearly (b) (a) D, but we need to show these are proper inclusions. b = ca, where neithera nor c is a unit. If (a) = D, then a is a unit, so (a)

= D, therefore (a)

D. If

(b) = (a), then b = ca and a = c1b, which gives a = c1ca, and so by the cancellationlaw, 1 = c1c, therefore c is a unit, but c cannot be a unit, so (b) = (a). Therefore(b) (a). This gives us (b) (a) D.

: in an integral domain D, let (b) (a) D. Since (a) is a proper subset ofD, a is not a unit. Since (b) (a), a|b. So b = ca. If c is a unit, then there existsc1 R such that c1c = 1. But if that is true then we have c1b = c1ca c1b = a.But this gives b|a (a) (b), which is not allowed, since it is given that (b) (a),therefore c cannot be a unit. So b = ca where neither a nor c is a unit. Thereforea is a proper divisor of b.

A non-zero non-unit element a R is called an irreducible element if a has noproper divisors, ie. whenever a is expressed as a = bc, where b, c

R, then either

b or c is a unit.A nonzero non-unit element a R is called a prime element (or a prime) if

whenever a|bc, then either a|b or a|c.Lemma 1.2. In an integral domain D, prime elements are irreducible.

Proof. Michael Young Let x D be a prime, and suppose that x = mn whereneither m nor n is a unit. Since x|mn and x is prime, either x|m or x|n. Say thatx|m. Then c D such that cx = m. Since x = mn, then cmn = m and thusby the Cancellation Law in a domain, cn = 1D. Therefore n is an unit. It caneasily be proven that if x|n, then m is an unit. This is a contradiction and thus xis irreducible.

Theorem 1.3. In a noetherian domain, any non-zero non-unit can be factoredinto a finite number of irreducible elements.

Proof. Jesse Bassett Suppose a is a non-zero non-unit which cannot be fac-tored into a finite number of irreducible elements. Since a1 itself cant be irreducible,a1 = bc, some b, c, neither being zero or a unit. One of the two also cannot be fac-tored into a finite number of irreducible elements, otherwise a1 itself would be.Suppose b cant. Rename b as a2, giving (a1) (a2). Continue and finish theargument..

Definition 1.4. A domain D is called a unique factorization domain (UFD)if the following hold:

(1) Any non-zero non-unit a can be factored into a finite number of irreducibleelements: a = p

1p

2. . . p

n.

(2) Such an irreducible factorization of a is unique up to units. So the nfactors can at worst be replaced by associates.

Example 1.5. The classic way to get a ring which does things you want it todo is to quotient out by an ideal appropriately. Here is an example of a noetheriandomain which is not a UFD. We would like to have an element a which factorsas a = xy and a = wz, x, y, w, x all distinct irreducibles. So do this: let D =R[X, Y, Z, W]/(XYZW), the quotient of the polynomial ring by the ideal (XY


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1. FACTORIZATION IN DOMAINS 21

W Z). Let x = X D etc. Then xy = wz since XY W Z (XY W Z). Younow have to do work to verify that D is a domain, x, y, w, z are irreducible, and xis not an associate of w or z, etc. These are details I wont try to write down.

A field F is a UFD by default, since it has no non-zero non-units. We will seethat Z is a UFD as we proceed. A theorem I will not prove here is the followingtheorem originally done by Gauss for Z.

Theorem 1.6. Gauss. If D is a UFD, then so is D[X].

Proof. It will take too long to do it here. Most undergraduate textbooksin algebra prove a result called Gauss lemma, which compares factorization of apolynomial in Z[X] with its factorization over Q[Z]. Copy this proof using D and[D] in place ofZ,Q. Gauss Lemma is the heart of the proof of the theorem thatD[X] is also a UFD.

Definition 1.7. A domain D is called a principal ideal domain (PID) if every

ideal in D is a principal ideal.

Since principal ideals are finitely generated, a PID is a noetherian domain.Hence the factorization property of Theorem 1.3 applies.

A field is a PID since it has only two ideals, both principal. Z is a PID, as weshall soon see. Z[X] is not a PID since you would be hard put to show that theideal generated by 5 and X, (5, X), is principal.

Lemma 1.8. In a PID D, irreducible elements coincide with prime elements.

Proof. We already know that prime implies irreducible. Conversely, let a beirreducible. Suppose a|bc and suppose a | b.

Consider the ideal (a, b). It is principal, since D is a PID. So (a, b) = (d) forsome d. So d

|a and d

|b. Since a is irreducible, d is either a unit or an associate of

a. d cannot be an associate of a because if d = au for a unit u, then b = de forsome e, hence b = aue, or a|b. So d is a unit.Hence (d) = D. Thus (a, b) = R.

We conclude that 1 (a, b), So 1 = ar + bs, giving c = arc + bcs. Since a|bc,we must have bc = af for some f. Therefore, c = a(rc + f s) or a|c, which was whatwe wanted to prove.

Theorem 1.9. A PID is a UFD. More generally, a noetherian domain whereirreducible elements coincide with prime elements is a UFD.

Proof. We already saw that any element in a PID or a noetherian domaincan be factored into a finite product of ireducible elements. Just need to show thefactorization is unique. This is easy but boring, so I just show a simple example.Suppose p1p2 = q1q2, all four factors being irreducible in D. By the lemma above

or by the extra hypothesis , p1 is a prime element. p1|q1q2, so p1|q1 (say). Since q1has no proper factors and q1 = p1f, one factor, necessarily f is a unit, so p1 and q1are associates. ie. p1 and q1 are pretty much the same. Do this inductively forlarger products.

The next is a later addition of a result that I inserted Oct 8 since I needed it.

Theorem 1.10. Let D be a UFD. Then any irreducible element is a primeelement.


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Proof. Let p be an irreducible element and suppose that p|ab. So ab = pcfor some c. Breaking up into irreducible factorizations, we get a = p1p2 . . . pr,b = q

1q

2. . . q

s, c = u

1u

2. . . u

t, hence

p1p2 . . . prq1q2 . . . qs = pu1u2 . . . ut.

By unique factorization in D, the irreducible factor p must appear on the leaft aswell and be either a pi or a qj . Hence p|a or p|b.

2. Euclidean domains

Lets start with a ubiquitous piece of notation.

Notation. Let R be a subring of the ring S. Let x S. R[x] is defined as theset {ni=0 rixi|ri R}. R[x] is a subring of S which contains R.

Make a strong distinction between this R[x], where x S and R[X], where Xis an indeterminate. It may be that a polynomial expression in x, say

ni=0 rix

i

equals 0 in S, hence equals 0 in R[x], even though the coefficients ri are not allzero. But for the indeterminate X,

ni=0 riX

i cannot equal 0 if the coefficients are

not all 0. For example, consider Z Z[2] R. The quadratic expression in 2given by (

2)2 2 equals 0.

It may be that x S is actually in R itself, in which case R[x] = R, since everyexpression

ni=0 rix

i belongs to R.Lastly, note that there is a homomorphism R[X] R[x] obtained by sending

X to x, and elements of R to themselves. This is called an evaluation map, wherethe indeterminate X is sent to the value x S.

This notation (and this notion) can be used to consider a number of interestingdomains containing Z.

Example 2.1. Gaussian integers. Consider Z[i], where i = 1. This is calledthe ring of Gaussian integers, where Z Z[i] C. Since i2 = 1, we do not needto consider polynomial expressions in i of degree two or more, and as we learn inelementary algebra, an element a in Z[i] can be more simply written as a = a1 + a2iwhere a1, a2 Z.

Let N(a) = N(a1 + a2) = a21 + a

22 N, the product of a and its conjugate a

(or the square of the absolute value function for complex numbers). N is called thenorm function. We can use this function to discover the units and irreducubleelements in Z[i], since N(ab) = N(a)N(b). For example if u = u1 + u2i is a unitin Z[i], then v Z[i] such that uv = 1, hence N(u)N(v) = N(1) = 1 N. HenceN(u) = 1. So u21 + u

22 = 1. Thus the units in Z[i] are {1, i}.

Steve Burkett. It is also true that u is a unit in Z[i] whenever N(u) = 1.In this case, if u = u1 + u2i, then N(u) = u

21 + u

22 = 1, so we must have either

u1 = 1, u2 = 0 or u1 = 0, u2 = 1, and therefore u {1, i}.Claim: 1i is irreducible: Now suppose there are a, b Z[i] such that ab = 1+i.

Then N(a)N(b) = N(ab) = N(1 + i) = (1 + i)(1 i) = 2. So either N(a) = 1,N(b) = 2 or else N(a) = 2, N(b) = 1, because 2 factors only as 12 in N. Thereforeeither a is a unit or b is a unit, and 1 + i is irreducible. The argument for 1 i isidentical.


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2. EUCLIDEAN DOMAINS 23

Claim: 3 is irreducible: Suppose ab = 3, so N(a)N(b) = N(3) = 9. Now ninecan be factored as either 3 3 or 1 9. But the first case is impossible, for ifN(a) = 3 then there would be a

1, a

2 Z with a2

1+ a2

2= 3, which is not true. Thus

either N(a) = 1 or N(b) = 1, so that either a or b is a unit, and 3 is irreducible.Observe that in Z[i], the element 2 is not irreducible (or prime) since 2 = (1+ i)(1i).

Example 2.2. Z[5] is not a UFD.

Once again, since the square of5 is in Z, elements in Z[5] can be uniquely

written as a = a1 + b5. Again, let N(a) = a21 + b2 5, the norm function or the

square of the absolute value. Charles Macaulay:

Observe that if u is a unit, then uv = 1 for some v and since N(u)N(v) =N(uv) = N(1) = 1, and N(u) > 0 in N, we get N(u) = 1.

(1) The units of this ring consist of just 1.

Let u Z[

5]. Ifu is a unit ofZ[

5], then N(u) = 1. Since u =a1 + a25, where a1, a2 Z, thenN(u) = N(a1 + a2

5) = a21 + 5a22 = 1.Since only values ofa1, a2 that satisfy N(u) are a1 = 1 and a2 = 0, thenu = 1. Therefore, Z[5] = {1}.

(2) N(a) can never equal 2 or 3.Suppose N(a) = 2. Then N(a) = a21 + 5a

22 = 2. We must have a2 = 0

because if a2 = 0, then N(a) > 5. Then this implies a1 =

2, which is acontradiction since a1 Z. Hence, N(a) = 2. Similarly, N(a) = 3.

(3) 1 5 are irreducible elements ofZ[5].Suppose that ab = 1 +

5 for some a, b Z[5]. ThenN(a)N(b) = N(1 +

5) = 1 + 5 = 6.

Now, 6 factors as as 2 3 or 1 6 in N. We know that N(a) = 2 andN(a) = 3 by part (2). Now consider N(a) = 6. Since N(a) = 6, thenN(b) = 1. Hence, b is a unit, which implies that 1 +

5 is irreducible.Using a similar argument, it is seen that 1 5 is irreducible as well.

(4) 2 and 3 are irreducible elements ofZ[5].

Suppose ab = 2 for some a, b Z[5]. Then N(a)N(b) = N(2) = 4.Now, 4 factors as 14 and 22 in N. Suppose N(a) = 4. Then N(b) = 1,which implies that b is a unit. Hence, 2 is irreducible.

Similarly, suppose ab = 3. Then N(a)N(b) = N(3) = 9. 9 factorsas 3 3, 1 9 in N, and we know that N(a) = 3. If we let N(a) = 9,this implies that N(b) = 1. Therefore, b is a unit, which implies that 3 isirreducible.

(5) Thus Z[5] is not a UFD, because 2 3 = (1 + 5)(1 5).Since

2 3 = 6 = 1 + 5 = (1 + 5)(1 5),6 can be expressed as two distinct products of irreducibles. Hence, Z[

5]is not a UFD.

Example 2.3. Z[

2] and its units.Z[

2] = {a1 + a2

2 |ai Z}. Think of the conjugate ofa1 + a2

2 as a1 a2

2,


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and let N(a) be the product of a and its conjugate. So N(a) = N(a1 + a2

2) =a21 a22 2. N is multiplicative, but can have both positive and negative integervalues.

Ashley Neal(1) Ifu is a unit in Z[

2], then N(u) = 1.

If u is a unit, u u1 = 1 and u1 Z[2]. Since N(u u1) = N(1) = 1and N is multiplicative, N(u)N(u1) = 1 Z. Hence N(u) = 1 Z.

(2) IfN(u) = 1, then u is a unit in Z[2].If N(u) = 1 and u = a1 + a2

2, then u1 = a1a2

2

N(u) = a1 a2

2.

Since a1, a2 Z, a1, a2 Z, hence u1 Z[

2].(3) 1 +

2 is a unit.

N(1 +

2) = 12 12 2 = 1 2 = 1. Then by (2), 1 + 2 is a unit.(4) So are all of its positive powers, which are all distinct.

Since N is multiplicative, N(an) = N(a)n for all n N. So N((1 +2)n) = N(1 +

2)n = (

1)n =

1 and is hence a unit by (2), for all

n N. Z[2] is a subring ofR, a field. So Z[2] is an integral domain.Assume that (1 +

2)n = (1 +

2)m for some n = m. WLOG, assume

n m. So (1 +

2)n = (1 +

2)nm(1 +

2)m = (1 +

2)m. Thenby cancellation, (1 +

2)nm = 1. We can view elements in Z[

2] as

elements ofR. In R, (1 +

2) 1, so for n m 1, (1 + 2)nm 1,which is a contradiction. Hence, all of the positive powers of 1 +

2 are

distinct.(5) Thus Z[

2] has infinitely many units.

Since every positive power of (1 +

2) is a distinct unit in Z[

2], andthere are infinitely many positive integers, there are infinitely many unitsin Z[

2].

Definition 2.4. A domain D is called a Euclidean domain if there is a sizefunction : D \ {0} N = {0, 1, 2, . . . , } with the following division algorithmproperty:Let a, b D, with a = 0. Then there are elements q, r D such that b = aq + r,and either r = 0 or (r) < (a).

Example 2.5. Easy examples are:

Z for which the size function is (a) = |a|. F[X], where F is a field. Because we can divide polynomials in F[X],

getting a quotient and a remainder, as we learn in high school. A sizefunction on F[X] is given by (f(X)) = degree of f(X), since the degreeof the remainder polynomial (if non-zero) is always smaller than that ofthe divisor polynomial.

Theorem 2.6. A Euclidean domain is a PID.Proof. Jaime Perkin In an ideal, look for the element with smallest size.

Thus Euclidean PID UFD.Example 2.7. Z[i] is a Euclidean domain.

The method, fairly simple, is to use for the function the norm function N(a) =a21 +a

22. I will talk in greater detail in class on this, but here is the idea: Let a = 0 in

Z[i]. Give b Z[i], look at the complex number ba = z1 + z2i C. (the coefficients


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2. EUCLIDEAN DOMAINS 25

are rational numbers.) Choose q1, q2 integers within 0.5 distance from z1, z2. Thisgives q = q1 + q2i Z[i] as our proposed quotient. Show this q works for us.

Most examples of Euclidean domains among rings of the form Z[] seem touse properties of the norm function to obtain a size function . Here is anotherexample.

Example 2.8. Z[] is a Euclidean domain, where 3 = 1.

Let = e2i/3 = 12 (1 + i

3), the usual cube root of unity. Then 1, , 2 are thethree solutions of X3 1 = 0 in C, which factors as (X 1)(X2 + X + 1) = 0.Hence 2 = 1. Therefore

Z[] = {a1 + a2 | a1, a2 Z}.Let N(a) = |a|2 = a a, where a is the complex conjugate of a.So N(a) = N(a1 + a2) = (a1 + a2)(a1 + a2) = a

21 + a1a2 + a

22.

Claim 1. N : Z[] N is a size function.In fact, let a = 0 in Z[], and let b Z[]. Consider the complex number

ba . Since{1, } span the R-vector space C, we can write ba = z1 + z2. Choose q1, q2 integers

within 0.5 distance from z1, z2, Let q = q1 + q2 Z[]. ThenN( ba q) = N((z1 q1) + (z2 q2)) = (z1 q1)2 + (z1 q1)(z2 q2) + (z2 q2)2 14 +

14 +

14 .

Hence N(b qa) = N(a)N( ba q) 34 N(a). Thus this q is the required quotient,and b qa the required remainder in the definiton of a Euclidean domain.

If you look carefuly at this last example, it also shows some of the limitationsof this tactic to find a size function, since if the norm function were not as nice asa21 + a1a2 + a

22 above, we would be hard put to get a bound like

34 later. Indeed the

Euclidean domains to be found among quadratic extensions rings of the formZ[] are just a handful. I will try to list them later.

Another thing to note is that Z[3] Z[]. You might ask why we didnot treat the simpler domain Z[

3]. It turns out that it is not so simple; forexample (1 +

3)(1 3) = 4 = 2 2, and you can show Z[3] is not a UFD,let alone a Euclidean domain. In fact, even unique factorization of ideals seems tofail since I cant explain the above non-unique factorization in terms of ideals, as wedid for Z[

5]. It turns out that Z[] is the better ring to study here, becauseit is integrally closed, something we will define soon.

Let us finish with some lemmas.

Lemma 2.9. Let a = 0 belong to the domain D. a is a prime element iff theprincipal ideal (a) is a prime ideal.

Proof. Ryan Seman

Lemma2.10

.In a PID D, every non-zero prime ideal is a maximal ideal.

Proof. Ryan Seman Let I D be a non-zero prime ideal. Since D is a PID,I is also a principal ideal. So I = (a) for some a D. Let J in D contain I. J is aprincipal ideal, so let J = (c). a J, therefore a = bc for some b D. Because I isa prime ideal, a is a prime element, by Lemma 2.9. Since a is a prime element, byLemma 1.8, a is an irreducible element. This means that either b is a unit, or c isa unit. Ifb is a unit, then a and c are associates, and so (a) = (c), therefore I = J.If c is a unit, then (c) = J = D. Therefore I is a maximal ideal.


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Example 2.11. In general, in a domain, an irreducible element need not gen-erate a prime ideal. See for example Z[

5]. Jesse Bassett finish this remark.

3. Integral extensions of rings.

Let R be a subring of a ring S. Let x S. We say that x is integral over R ifthere is some monic polynomial expression xn + an1xn1 + + a1x + a0 whichis zero in S (hence in R[x]).

Clearly, any x in R itself is integral over R (by considering a monic linearexpression).

Lemma 3.1. If x is integral over R, then the ring R[x] is a finitely generatedR-module.

Proof. Steve Burkett We have already seen that R[x] is a ring containing Rand thus an R-module. We must show that it is finitely generated over R. Since xis integral over R, there is some monic expression xn + an1xn1 + . . . + a1x + a0

equal to zero, i.e,

(3.1) xn = n1i=0

aixi

Any element y of R[x] can be writtenm

i=0 rixi, of degree m. We claim

that any such element is a finite linear combination of {1, x , x2, . . . , xn1}. This isobvious whenever 0 m n 1, so for a proof by induction, assume that it is alsotrue for expressions of arbitrary degree k. Then for y R[x] with degree k + 1:

y =k+1i=0

sixi =

ki=0

sixi + x(sk+1x

k) =ki=0

sixi + x

ki=0

tixi,

where ti = 0 for i < k, and tk = sk+1. Applying the inductive hypothesis twice

yields:

y =

n1i=0

bixi + x

n1i=0

cixi =

n1i=0

bixi +

n1i=1

ci1xi + cn1xn,

and by replacing xn with the right hand side of expression (1), we see that

y =

n1i=1

(bi + ci1 cn1ai)xi + b0 cn1a0

which is a linear combination of {1, x , x2, . . . , xn1} with coefficients in R.

Lemma 3.2. Let R, S, and T be rings such that R S T. Suppose S is afinitely generated R-module and T is a finitely generated S-module. Then T is a

finitely generated R-module.

Proof. Charles Macaulay Let S be finitely generated by U = {ui}ni=1 as aR-module, and let T be finitely generated by V = {vj}mj=1 as a S-module. Weclaim that T is finitely generated by W = {uivj |ui U, vj V} as a R-module.

Suppose x T. Thenx =

mj=1

sjvj ,


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3. INTEGRAL EXTENSIONS OF RINGS. 27

where sj S. Each sj can be expressed as

sj =

n

i=1 rj,iui,

where rj,i R. Hence,

x =mj=1

sjvj =mj=1

ni=1

rj,iuivi.

Therefore, x is expressed as a R-linear combination of the elements of W. Since|W| = mn, T is a finitely generated R-module.

Lemma 3.3. LetR be a subring ofS. Suppose {xi}ni=1 is a collection of elementsin S which are integral over R. Then R[x1, x2, . . . , xn] is a finitely generated R-module.

Proof. Charles Macaulay We will prove this lemma by induction.

Suppose n = 1. We apply Lemma 3.1 to see that R[x1] is a finitely generatedR-module.

Suppose that R[x1, x2, . . . , xk1] is a finitely generated R-module. Let xk beintergal over R. Then there exists a monic polynomial expression

g(xk) = xmk + bm1x

m1k + + b1xk + b0 = 0S ,

where bj R. Since R R[x1, . . . , xk1], xk is also integral over R[x1, x2, . . . , xk1].From Lemma 3.1, we see that R[x1, x2, . . . , xk1, xk] is a finitely generated R[x1, x2, . . . ,[0]xk1]-module. By Lemma 3.2, R[x1, x2, . . . , xk1, xk] is a finitely generated R-module.

By induction, our Lemma holds for all n.

Recall from linear algbraCramers Rule.

Square matrices and determinants can be defined over a ring R just as we do ina linear algebra course. If A is a square matrix, we define the adjoint matrix A

to be the transpose of the matrix whose (ij)-th entry is (1)i+j times the (ij)-thminor of A. Then AA = (det A) I.Therefore, if AX = b is a system of n equations in x1, x2, . . . xn, with coefficientmatrix A and right hand side equal to a vector b1, b2, . . . bn, when we multiply byA, we get det A xi = bi, where b = A b. If in particular b is the zero vector,each det A xi = 0, i = 1, 2, . . . , n.

Lemma 3.4. Suppose R is a subring of R1 and R1 is a finitely generated R-module. Then every element s R1 is integral over R.

Proof. Let u1, u2, . . . , un be a set of generators for the R-module R1 over R.

The for each i, sui, which is in R1, is some linear combination of u1, u2, . . . , un:

sui = ai1u1 + ai2u2 + + ainun, aij R.This can be read as a matrix equation s I u = A u, or (A sI) u = 0. ByCramers Rule, det(A sI)ui = 0 for each i. Since 1 R1 is a linear combinationof u1, u2, . . . , un, it follows that det(A sI) 1 = 0. Thus det(A sI) = 0 in R1.But it is obvious that det(A sI) is a monic polynomial expression of degree n inR1, with coefficients in R. This finishes the argument.


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Theorem 3.5. The integral closure of a ring is a ring.LetR be a subring on a ring S. Let R = {x S|xis integral overR}.

R R S.R is called the integral closure of R in S, and R is a subring of S.

Proof. The inclusions are obvious. Let x1, x2 R. By Lemma ??, R[x1, x2]is a f.g. R-module. By Lemma 3.4, every element in R[x1, x2] is integral over R.Now x1 + x2, x1 x2, and x1x2 are all elements of R[x1, x2]. Hence they are allintegral over R.

Thus, ifx1, x2 R, then x1 + x2, x1 x2, x1x2 R as well, making R a subringof S.

Definition 3.6. Let R be a subring of S. R is said to be integrally closed inS if the integral closure R of R in S equals R; R = R.

Proposition 3.7. LetR be a subring of S. Then R is integrally closed in S.

Proof. Charles Macaulay

Proof. We know that R R S by Theorem 3.5. We need to show R = R.R R follows from Theorem 3.5. Therefore, assume x R. Then there exists

a polynomial expression

f(x) = xn + an1xn1 + + a1x + a0 = 0S ,where ai R. By Lemma 3.3, R[a0, . . . , an1] is a finitely generated R-module.Since ai R[a0, . . . , an1], x R[a0, . . . , an1]. By Lemma 3.1, R[a0, . . . , an1, x]is a finitely generated R[a0, . . . , an1]-module. By Lemma 3.2, R[a0, . . . , an1, x] isa finitely generated R-module. Since x R[a0, . . . , an1, x] and R R[a0, . . . , an1, x],by Lemma 3.4, x R. Hence, R R.

Therefore, R = R, which proves that R is integrally closed in S.

Proposition 3.8. LetD be a PID. Then D, which is a subring of the field offractions [D], is integrally closed in [D].

Proof. Ashley NealWe want to show that D = D = {x [D] such that x is integral over D}. Recallfrom theorem 3.5, D D [D]. So we only need to show D D. By definitionof [D], p1q1 =

p2q2

if and only if p1q2 = p2q1. We can assume that the fractions are in

reduced form. A PID is a UFD, so any non-zero non-unit can be factored into afinite number of irreducible elements. Let p = p1 . . . pr and q = q1 . . . qs representthese factorizations. Reduced form for pq [D] implies qi pj for all i, j. For any

x D, x =p

q [D], and there is a monic polynomial (p

q )n

+ an1(p

q )n+1

+ . . . +a1(

pq ) + a0 = 0[D]. Suppose that x D and x / D, so q is not a unit, or s 1. We

can multiply everything by qn to cancel the denominators using the definition ofequal in [D], giving pn+an1pn1q +. . .+a1pqn1 +a0qn = pn+q(an1pn1 +. . .+a1pq

n2 + a0qn1) = 0. Therefore, pn = q(an1pn1 + . . . + a1pqn2 + a0qn1).So q | pn. This means q1q2 . . . qs | (p1p2 . . . pr)n in terms of irreducible elements.Since irreducible elements are prime by Theorem 1.10, qi | pnj for some i, j. Thisimplies qi | pj, which is a contradiction. Therefore q is a unit, so x = p D.


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4. A QUICK LOOK AT FIELD EXTENSIONS 29

Proposition 3.9. LetD be a UFD. Then D, which is a subring of the field of

fractions [D], is integrally closed in [D].

Proof. Jaime Perkin The proof should be mostly identical to the last proof.

4. A quick look at field extensions

Much of the last section can be applied to fields, though the terminology isslightly different.

Let F be a subfield of a field L, and let x L. x is said to be algebraic over Fif there is a non-zero polynomial f(X) F[X] such that its evaluation at x equalszero; ie. f(x) = 0 in L.

Notice that if such an f(X), can be found, since F is a field, we can divide by

the leading coefficient and assume that f(X) is a monic polynomial. Hence, we aretalking about exactly the same notion as being integral.

Proposition 4.1. If x L is algebraic over F, there is a unique monic poly-nomial f(X) F[X] of lowest degree which vanishes when evaluated at x. f(X)is called the minimal polynomial of x over F. For any other g(X) F[X],g(x) = 0 f(X)divides g(X).

Proof. Ryan Seman

If x is algebraic over F, the subring F[x] of L is a finite dimensional vectorspace over F. It is also an integral domain since any subring of a field is a domain.From Theorem 3.3, we get,

Proposition 4.2. If F is a subfield of a field L, the elements in L which arealgebraic over F constitute a subring F of L. F is called the algebraic closure ofF in L.

More is true. It turns out that F is a field, lying between F and L.

Theorem 4.3. Suppose F is a subfield of a field L. Suppose x L is algebraicover F. Then the ring F[x] L, which is a finite dimensional vector space overF, is a field.

Proof. Let f(X) be the minimal polynomial ofx over F, found in Proposition4.1. By the same proposition, I = {g(X) F[X]|g(x) = 0 L} is the non-zeroprincipal ideal in F[X] generated by f(X), and is the kernel of the evaluationhomomorphism F[X] F[x] 0.

According to the First Isomorphism Theorem for rings, F[X]/I= F[x] and

F[x] is a domain. Hence F[X]/I is a domain. I is a prime ideal by Proposition1.7. In F[X], which is a PID, any nonzero prime ideal is maximal (Lemma 2.10).Hence F[X]/I is a field. Thus F[x] is a field.

Corollary 4.4. Given F F L, where F is the algebraic closure of F inL, F is a field.

Proof. Michael Young


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Notation. I want to include one more piece of notation. The notation F[x]refers to all polynomial expressions in x, where x is an element in a bigger field L.We define F(x) as the set of all rational expressions in x. So F(x) is the smallestsubfield of L which contains F and x. When x is not algebraic over F, F[x] is aproper subset ofF(x). When x is algebraic over F, F[x] = F(x). This was provedabove.For better or for worse, ring theory and field theory use different notation for thesame concept. Ifx is integral over a ring R, R[x] refers to the ring which is obtainedby adjoining x to R. If x is algebraic over a field F, F(x) refers to the ring F[x],to emphasize that it is more than a ring, it is a field.

Splitting field of a polynonomialSuppose F is a field, f(X) F[X] a monic polynomial, and L an extension field

ofF containing a root off(X). Hence in L, f() = 0. By the division algorithmin L[X], f(X) = (X

)q(X) + r, where r

L is the remainder. Plugging in

for X in this equation, we get r = 0. Thus f(X) factors in L[X] as a product(X)q(X). This is pretty much the Remainder Theorem from high school algebra.

Is there a field in which f(X) factors into a product of exclusively linear polyno-mials? If there were, we would get a factorization f(X) = (X1)(X2) . . . (Xn), over this field, where n = degree f(X) and 1, 2, . . . n would be all the rootsof f(X) (possibly with repetitions). (Note that unique factorization holds in thisEuclidean domain, hence the roots are determined uniquely up to order.)

A little algebra then shows that the coefficients an1, an2, . . . , a0 of f(X)(which belong to F), can be expressed in terms of the roots. The easiest expressionsamongst these are an1 = (1 + 2 + + n) and a0 = (1)n[1 2 n].

Finding such a field is not so hard if we do it abstractly. We first find a fieldcontaining F in which f(X) has one root 1. To do this, let g(X)be an irreducible

factor of f(X) in F[X]. Construct F[X]/(g(X)). Since g(X) is irreducible, theprincipal ideal (g(X)) is maximal, hence F[X]/(g(X)) is a field. Calling it L1, let1 denote the coset X. Elements in L1 are then polynomial expressions in 1 withF-coefficients (of degree less than the degree of g(X)). Hence F can be viewed asa subfield ofL1. Lastly, g(1) = 0 in L1 by construction. Hence by the RemainderTheorem, g(X) and therefore f(X) factors in L1[X]. Let f(X) = (X 1)q(X)be the factorization in L1[X]. Repeat the process for q(X) in L1[X], getting a stilllarger field L2 containing a root 2 ofq(X). By induction this process ends after afinite number of steps (in fact, the number of steps is the degree of f(X)).

When the process is over, we end up with a field L which contains roots1, 2, . . . , n of f(X), where f(X) splits into the product of the linear factors(X 1), (X 2), . . . , (X n). This field L is called a splitting field over F ofthe polynomial f(X).

Observe also that L1 = F(1), L2 = L1(2) = F(1, 2) etc., so the splittingfield over F of the polynomial f(X) is obtained by adjoining the roots of f(X) toF.

Proposition 4.5. Let F be a field, f(X) an irreducible polynomial in F[X].Letx1 and x2 be two roots of f(X) in an extension field L (which could even be asplitting field). Then given g(X) F[X], g(x1) = 0 g(x2) = 0.


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5. QUADRATIC NUMBER FIELDS 31

Proof. This follows directly from Proposition 4.1. Since f(X) is irreducibleand f(x1) = 0, f(X) is the minimal polynomial for x1. Hence g(x1) = 0 g(X) =f(X)h(X)

g(x

2) = f(x

2)h(x

2) = 0

h(x

2) = 0. Likewise, the other direction.

This underlies the well known tool we use in complex analysis: if a real poly-nomial vanishes on a1 + a2i, ai R, it also vanishes on a1 a2i. This is becauseboth a1 + a2i and a1 a2i are roots of the irreducible (when a2 = 0) real quadraticpolynomial (X a1)2 + a22. Similar remarks can be made about rational coefficientpolynomials and expressions like a1 + a2

3, ai Q.

5. Quadratic number fields

Definition 5.1. An algebraic number field is a field L containing Q and whichis a finite dimensional vector space over Q. The dimension n of this vector space isdenoted [L : Q], and L is called an algebraic extension ofQ of degree n.

Definition 5.2. Let L be an algebraic number field. The ring of integers in Lis defined as the integral closure ofZ in L, and is denoted OL.

The easiest examples are when n = 2, quadratic extensions ofQ.Start with an irreducible quadratic polynomial f(X) (make it monic) in Q[X].

By Lemmas 2.9 and 2.10, the ideal (f(X)) is maximal, hence Q[X]/(f(X)) is a fieldL. Let x denote the coset X + (f(X)). Then in L, f(x) = 0, hence any elementin L can be written as a1 + a2x, a1, a2 Q. In fact, {1, x} gives a basis for L as aQ-vector space.

Such a field L is called a quadratic number field, and is written Q(x). We oftendescribe it by means of a convenient complex number.

For example, L = Q[X]/(X2 + X + 1)) = Q[] = Q(), where = e 2i3 andL = Q[X]/(X2 + 3))

= Q[

3] = Q(

3).

Surprisingly, as subfields ofC, Q() = Q(3) since = 12 (1 + 3) is aQ-linear combination of {1, 3} and vice versa. But Z(3) Z().

Example 5.3. The ring of integers in Q(3) is Z(). Jesse Bassett

Proposition 5.4. Any quadratic number field is isomorphic to a subfield ofCof the formQ(

d), where d is a square-free integer inZ. If d > 0, the quadratic

number field is called a real quadratic field. If d < 0, it is called an imaginaryquadratic field.

Proof. Any irreducible polynomial f(X) in Q[X] has a root in C, as we learnin complex analysis. If C is this root, is algebraic over Q, and the mapQ[X]/(f(X)) Q[] = Q() is an isomorphism.

When f(X) is quadratic, Q() has a Q-vector space basis of {1, }. Cleardenominators in f(X) to get f(X) = aX2 + bX + c, with integer coefficents. Letd = b2 4ac. Q(d) has a Q-vector space basis of{1, d}.

We know from the quadratic formula that can be expressed as a rationallinear combination of 1 and

d. Conversely,

d is a rational linear combination of

1 and . Hence Q() = Q(

d).

If d has a square factor, d = p2d1, then

d = p

d1, and we easily see thatQ(

d) = Q(

d1).


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Proposition 5.5. Letd be a square free integer (positive or negative) and let

L = Q(

d) be the quadratic number field defined by

d. Then the ring of integers

in L is given byZ[] where

=

d, if d 1 mod 4,

12 (1 +

d), if d 1 mod 4.

Proof. Steve Burkett One form of Gauss lemma, which was discussed brieflyin the proof of Theorem 1.6 states:Gauss: If D is a UFD, and f D[X] factors over [D], then f factors over D.

Lemma: If D is a UFD, and x is integral over D, then the minimal polynomial ofx over [D] lies in D[X].

Proof. Let g be the minimal polynomial of x over [D]. Note that g is irre-ducible, lest we have g = g1g2 gm and some gi(x) = 0 with degree strictly lessthan g.

Because x is integral over D, there is some monic f D[X] such that f(x) = 0.By Proposition 4.1, h [D][X] with f = hg. Since f factors over [D], it factorsover D by Gauss, that is, f1f2 fn = f = hg, with fi D[X], i. But [D] is aUFD because it is a field, so [D][X] is a UFD by Theorem 1.6. Thus fi = ag for somei, where a [D] is a unit. Since f and g are monic, a = 1, so g = fi D[X].

Proof of Prop 5.5:

Proof. We first show that OL Z[]. Let x = a + b

d Q(d) be integralover Z. We need to show that x Z[]. Ifx Z this is obvious, and if x Q \ Z,x cannot be integral over Z by Proposition 3.8, so assume that x / Q. Now x is aroot of the monic polynomial

f = (X x)(X x) = (X (a + bd)(X (a bd) = X2

2aX+ a2

b2

d.Since x / Q, f cannot have a linear factor in Q[X], so f is irreducible in Q[X]. Infact, f is the minimal polynomial of x over Q. To see this, let g be the minimalpolynomial for x. By proposition 4.1, we must have g|f. But f is irreducible, sothe degrees of f and g are the same, that is, f = bg for some b in Q. Since f andg are monic, b = 1 and f = g.

Since f is the minimal polynomial of x over Q = [Z], f Z[X] by the lemma,that is, f has integer coefficients. We conclude that 2a Z and a2 b2d Z. Given2a Z, there are two cases to consider:

(1) If a Z then b2d must also be an integer since a2 b2d Z. But thismeans that b must be an integer as well, for d is squarefree by assumption,and so cannot clear the squared factors in the denominator of b. Since

a, b Z, x Z[] as required, assuming d 1 mod 4.(2) If a / Z, then a = m + 12 a2 = m2 + m + 14 , for some m Z. Thusm2 + m + 14 b2d Z, so b / Z. We also have that 4b2d = (2b)2d Z,so 2b must be an integer. Since b / Z, b = l + 12 , for some l Z, and(m + 12 ) + (l +

12 )

d = m l + (2l + 1) 12 (1 +

d) Z[].

We still need to show that we have case 2 above iff d 1 mod 4. We havem2 + m +

1

4 b2d Z b2d 1

4 Z (l + 1

2)2d 1

4 Z l2d + ld + 1

4d 1

4 Z


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6. LINEAR ALGEBRA 33

Which is true iff there is h Z such that d 1 = 4h d 1 mod 4. We haveshown that OL Z[].

The reverse inclusion is easily checked in the d

1 mod 4 by plugging a + bZ[] into the monic polynomial X22aX+a2b2d, and verifying that the expression

reduces to 0. In the d 1 mod 4 case, the polynomial X2 2aX bX+ a2 + ba b2

4 (d 1) has integer coefficients, and reduces to zero as required. End of Prop 5.5.

6. Linear Algebra

We will now introduce the notions of dual vector spaces, dual bases, bilinearforms. Since some of concepts work equally well over a ring, our definitions can bebroader. It is not efficient to make definitions for vector spaces where they can alsobe made for modules.

Definition 6.1. Let M, N be R-modules. Then

Hom(M, N) = {| is an R-module homomorphism from M to N }.In the special case where N = R, M is defined as M = Hom(M, R).

Proposition 6.2. There is a natural and obvious way in which Hom(M, N)can be viewed as an R-module.

Proof. Charles Macaulay

Example 6.3. Ashley Neal

(1) Let R be viewed as an R-module. then R = R.Construct f : Hom(R, R) R as follows: f(g) = g(1R) where g Hom(R, R). f is a homomorphism since f(g1 + g2) = (g1 + g2)(1R) =g1(1R) + g2(1R) = f(g1) + f(g2) and f(rg) = (rg)(1R) = r g(1R) =r

f(g). We need to show that f is 1-1 and onto. Suppose g

Kerf. So

f(g) = g(1R) = 0. Therefore g(a) = g(a 1R) = a g(1R) = a 0R = 0R forany a R. So g = 0 as a homomorphism from R to R. Thus f is 1-1. Letr R. We want to show that there exists g such that f(g) = g(1R) = r.Since R is free with a basis {1R)}, you can always build a homomorphismg : R R which sends 1R r. Therefore f(g) = r for this choice of g Hom(R, R), so f is onto.

(2) Let Z/(5) be viewed as a Z-module. Then [Z/(5)] = {0}.[Z/(5)] = Hom(Z/(5),Z) = {f : f is a Z-module homomorphism fromZ/(5) to Z}. Assume f [Z/(5)], then f(y) = yf(1) for all y Z.So f(1) = x for some x Z. Therefore, f(2) = 2x, f(3) = 3x, f(4) =4x, f(5)) = 5x, but f(5) = f(0) = 0 if f is a homomorphism. So 5x = 0in Z, and thus x = 0. Thus f(y) = 0 or f is the zero homomorphism.

Hence [Z/(5)] = 0.Theorem 6.4. The dual of a free module and dual bases.

Let M be a free R module of rank n. Let {m1, m2, . . . , mn} be a basis for M.Then M is a free module of rank n and has a basis {m1 , m2 , . . . , mn}, wheremi : M R is the homomorphism which takes mi to 1 and the remaining basiselements to 0. {m1 , m2 , . . . , mn} is called the dual basis for M and is succinctlydescribed by mi (mj) = ij , 1 i, j n, where the Kronecker symbol ij equals 1if i = j and 0 otherwise.


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Proof. Jaime Perkin Note that the construction of mi is obtained fromProposition 4.5. It remains for you to show that the dual basis is linearly inde-pendent and spans M

.

Definition 6.5. Let M be an R-module. A bilinear form , on M is amap M M R where the image of (m, m) is denoted by m, m, with thefollowing properties:

(1) , is linear in the first coordinate; ie.a1m1 + a2m2, m = a1 m1, m + a2 m2, m .

(2) , is linear in the second coordinate; ie.m, a1m1 + a2m2 = a1 m, m1 + a2 m, m2 .

Proposition 6.6. Let, be a bilinear form on the R-module M. There isthen a well defined homomorphismT : M M given by the formula T(m)(m) =m, m.

Proof. Ryan Seman First show T(m) belongs to M. Next show that Tis a homomorphism of R-modules. An ambitious student can also show that theconverse of this proposition is also true (ie. if there is a T, you can build a bilinearfor

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