Algebra Qual

7/28/2019 Algebra Qual

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Algebra Qualifying Exam

John Dusel

To MBM: we suffer together.


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Thanks to Paul Oesser, Nathan Manning (Algebra Qualmasters for2010,2009), Adam Katz, and Mei-Chu Chang.


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Contents

2011 Qual 13Part A 132011 A1 132011 A2 132011 A3 142011 A5 142011 A6 152011 A7 152011 A8 15Part B 16Part C 16

2010 Qual 17Part A 172010 A1 172010 A2 18

2010 A3 192010 A4 21Part B 222010 B1 222010 B3 22

2009 Qual 25Part A 252009 A1 252009 A2 unnished 262009 A3 26

2009 A4 272009 A5 unnished 282009 A6 282009 A7 unnished 292009 A8 29Part B 302009 B1 30

3


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4 CONTENTS

2009 B2 322009 B3 33

2009 B4 342009 B5 34Part C 372009 C1 372009 C3 372009 C5 38

2008 Qual 41Part A 412008 A1 412008 A2 42

2008 A3 432008 A4 462008 A5 48Part B 502008 B1 502008 B2 502008 B3 522008 B4 542008 B5 54Part C 552008 C1 55

2008 C2 552008 C3 582008 C4 58

2007 Qual 59Part A 592007 A1 592007 A2 592007 A3 602007 A4 60Part B 60

2007 B1 60An example where Hom(, M ) is not exact 612007 B2 622007 B3 unnished 632007 B4 632007 B5 63Part C 65


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6 CONTENTS

2004 A1 unnished 912004 A2 91

2004 A3 91Part B 922004 B1 922004 B2 932004 B3 932004 B4 932004 B5 942004 B6 95Part C 962004 C1 962004 C2 96

2004 C4 972004 C6 98

2004 Qual, version 2 (October) 101Part A 1012004v2 A1 unnished 1012004v2 A2 1012004v2 A3 1022004v2 A4 1022004v2 A5 102Part B 103

Part C 1042004 C1 1042004v2 C3 1052004v2 C4 106

2003 Qual 109Part A 1092003 A1 1092003 A2 unnished 1092003 A3 1092003 A4 110

2003 A5 1102003 A6 111Part B 1112003 B1 unnished 1112003 B2 unnished 1112003 B3 1112003 B4 unnished 111


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CONTENTS 7

2003 B5 1122003 B6 112

Part C 1132003 C2 1132003 C3 114

2002 Qual 117Part A 1172002 A1 1172002 A2 1172002 A3 1172002 A4 1192002 A6 119

Part B 1192002 B4 1192002 B5 1202002 B6 121Part C 1212002 C1 1212002 C3 1222002 C5 123

2001 Qual 125Part A 1252001 A1 1252001 A2 1252001 A3 1252001 A4 1262001 A5 126Part B 1262001 B1 1262001 B2 1272001 B5 127Part C 1272001 C2 1272001 C3 1282001 C5 1292001 C6 129

2000 Qual 133Part A 1332000 A1 unnished 1332000 A2 unnished 133


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8 CONTENTS

2000 A3 unnished 1332000 A4 133

Part B 1332000 B1 1332000 B2 1332000 B3 1342000 B4 1342000 B5 134Why Q is not a free Z-module 1352000 B6 1352000 B8 136

1999 Qual 137

Parts A and B 1371999 A1 unnished 1371999 A2 unnished 1371999 A3 1371999 AB5 1371999 AB7 1371999 AB8 1381999 AB10 139Part C 1391999 C3 1391999 C4 140

1999 C6 1411998 Qual 143

Part A 1431998 A1 unnished 1431998 A2 1431998 A4 143

1997 Qual 145Part B 1451997 B2 1451997 B3 1461997 B4 146Part C 1461997 C1 1461997 C2 147

1996 Qual 149Part B 149


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CONTENTS 9

1996 B1 1491996 B2 149

1996 B3 150Part C 1511996 C1 1511996 C3 1511996 C4 1521996 C5 152

1995 Qual 155Part A 1551995 A1 1551995 A2 155

1995 A3 1551995 A4 1561995 A5 156Part B 1561995 B1 1561995 B4 158Part C 1581995 C3 (incomplete) 158

1994 Qual 161Part A 161

1994 A1 1611994 A2 improvement needed 1611994 A3 1611994 A4 unnished 162Part B 1621994 B2 1621994 B4 163

1994 Version 2 Qual 165Part A 1651994v2 A1 unnished 165

1994v2 A2 unnished 1651994v2 A3 unnished 1651994v2 A4 165Part B 1651994v2 B3 1651994v2 B4 1661994v2 B5 166


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10 CONTENTS

1993 Qual 169Part A 169

1993 A6 169Part B 1691993 B2 1691993 B3 1701993 B4 1711993 B7 171Part C 1721993 C1 1721993 C3 1731993 C5 173

Qual Seminar Homework 17506/12/10 175Classify all groups of order 45 175

Exam Questions from Academic Year 20092010 177Part B 177201B Midterm 177201B Midterm Question 1 177201B Midterm Question 2 177201B Midterm Question 3 178201B Midterm Question 4 179

201B Midterm Question 5 180201B Final Exam 180Final Exam Question 1 180Final Exam Question 2 182Final Exam Question 3 183Final Exam Question 4 184Final Exam Question 5 185Final Exam Question 6 186Final Exam Question 7 187Part C 188Assignment 1 188

III.3 #3 188III.3 #4 190III.3 #6 190III.3 #7 192III.6 #1 192III.6 #2 193III.6 #3 193


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CONTENTS 11

III.6 #10 194Additional Exercise 1 195

Exercise 2 195Exercise 3 196Assignment 2 196VI.1 #1 196V.1 #3 197V.1 #7 197V.1 #14a 199V.1 #17 199V.1 #19 200Additional Exercise 1 200Exercise 2 201

Exercise 3 202Exercise 4 203Exercise 5 203Assignment 3 203V.2 #3 203V.2 #4 204Additional Exercise 1 205Additional Exercise 2 205Additional Exercise 3 207Additional Exercise 4 207Final Exam 208

201 C Final Exam Problem 1 208201 C Final Exam Problem 2 209201 C Final Exam Problem 3 210201 C Final Exam Problem 4 210201 C Final Exam Problem 5 211201 C Final Exam Problem 6 212201 C Final Exam Problem 7 212201 C Final Exam Problem 8 215

Bibliography 217


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12 CONTENTS

Any mistakes in this document are due to JMD. Comments andcorrections should be sent to the e-mail address at the end of this

document.


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14 2011 QUAL

Proof. From the denition I.8.8 of internal direct product we have(1) H, K G; (2) G = KH = {kh : k K, h H }; (3) K H = {e}.Dene a function K G/H by k kH . (1) implies is a map of groups, (2) inplies is surjective, and (3) implies is injective.

(Sketchy) Suppose H S 5 and S 5 = H A5. The fact impliesH S 5 and |H | = [S 5 : A5] = 2. This means H = (ij ) for somedistinct i, j {1, . . . , 5}. It is well-known that (in particular) for all S 5 we have (ij )1 = ( (i)( j )), and for all such choices of i, jthere exists a with ((i)( j )) = ( ij ). There is no way H can benormal, so there can be no such H .

2011 A3. Prove that a free abelian group is a free group if andonly if it is cyclic. Give an example to show that a cyclic group maynot be free.

Proof. See 2003 A4 and p.65 of Hungerford.Let A be a free abelian group.If A is cyclic then A = x for some element x A. Each element

of A has the form xn for some x Z . Dene A Z by xn Z andverify that it is an isomorphism of groups ( Z is the free group on onegenerator).

For the converse we contradict the contrapositive: suppose that Ais not cyclic and A is a free group. Then A is free on a set {x ,y, . . . }and A contains the reduced word xyx1y1 = 1. This implies A is notabelian, contradiction.

Example. Consider Z2, which is cyclic on 1. A fancy reason whyZ2 is not free is that Z has no nite subgroups: We have {1}Z2and {1} Z but no group homomorphism from Z2 into Z .

2011 A5. Suppose that (the group) H acts on a set S . Givens S dene the H -orbit of s. Prove that if s S is another elementthen either s belongs to the orbit of s or that the orbit sets of s ands are disjoint.

Definition. The H -orbit of s is Hs = {h.s : h H }.Proof. Dene a relation on S by

s s if and only if s Hs.It is enough to prove that is an equivalence relation on S .

Certainly is reexive: s = eH .s.Suppose s s , meaning s Hs : there exists h H such that

s = h.s . The denition of group-action givesh1.s = h1.(h.s ) = ( h1h).s = e.s = s,


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PART A 15

which means s Hs or equivalently s s.Suppose s s : s = h.s and s s : s = g.s . Then the denition

of group-action again shows s s easily.2011 A6. Prove that a group of order 200 cannot be simple.

Proof. Let |G| = 200 = 2 352. Sylow 3 indicates that n5(G) = 1,and Sylow 2 implies that this unique Sylow 5-subgroup is normal inG.

2011 A7. Let t be an indeterminate, and consider the ring of poly-nomials R[t]. Using the fact that R[t] is a principal ideal domain provethat the ideal generated by t2 + 2 is maximal. Suppose now that wework with the ring C[t]. Find a polynomial f such that the ideal gener-ated by f is proper and strictly contains the ideal generated by t2 + 2.

Proof. The ideal ( t2 + 2) R[t] comprises all f (t2 + 2) withf R[t]. Suppose I R[t] is an ideal and (t2 + 2) I R[t]. ClaimI = ( t2 + 2) or I = R[t]. Since R[t] is a principal ideal domain thereexists a (monic WLOG) polynomial p R[t] such that I = ( p). Thecontainment ( t2 + 2) ( p) implies that there exists g R[t] such thatt2 + 2 = pg. Considering degrees, there are three cases:

If deg p = 2 , deg f = 0 then p is just a multiple of t2 + 2 andso ( p) (t2 + 2). So I = ( t2 + 2). If deg p = 0 , deg f = 2 then p is a unit and so ( p) = R[t].

If deg p = 1 = deg f then we have a contradiction: t2 + 2 isirreducible over the reals. (Write down a factorization intolinear polynomials, FOIL, and go forward until an equation isfound which does not hold in the reals.)

Over the complex numbers, t2 +2 = ( t i 2)(t + i 2) is reducible.This shows that ( t2 + 2) (t + i 2). Any element of the latter hasdegree at least 1, which precludes membership of any units. Thus wehave found f = t + i 2 as desired.

2011 A8. Suppose that R = Z6 and S = {2, 4}. Prove that S 1Ris a nite eld and identify the eld.Proof. Doing something like

RS :=Union[

Map[{#, 2} &, Range[0, 5]],Map[{#, 4} &, Range[0, 5]]]


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16 2011 QUAL

eqclass[x_List] :=Select[

RS,Or[2 (x[[1]] #[[2]] - x[[2]] #[[1]]) == 0,4 (x[[1]] #[[2]] - x[[2]] #[[1]]) == 0] &

]

Map[#[[1]] &, Union[Map[eqclass[#] &, RS]]]

will show that S 1R = {0/ 2, 2/ 2, 1/ 2, 1/ 4, 3/ 2, 3/ 4, 4/ 2, 5/ 2, 5/ 4}, or-der 9. Once we show that every element not equal to 0 / 2 has a multi-plicative inverse, it will follow that S 1F = F9. Tedious.

Part B

Part C


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2010 Qual

Part A

2010 A1. Let G be a group and let A be an abelian group. Let : G Aut( A) be a group homomorphism. Let A G be the setA G with the binary operations

(a, g)(a , g ) = ( a + (g)(a ), gg ).(1) Prove that A G is a group.(2) Find : Z2 Aut( Zm ) such that the dihedral group Dm isisomorphic to Zm Z2. Do not forget to prove the isomor-

phism!

1. Must prove(1) A G is closed under the binary operation(2) The binary operation is associative(3) A G has an identity

(4) all element of A G have an inverse.1 is clear. 2 is straightforward and tedious. 3. Let e = ( eA , eG ). 4.The inverse of (a, g) is ((g1)(a), g1).

2. The dihedral group of order 2 m is D2m = r, s : |r | = 1 , |s| =m,rs = s1r . Accordingly, there exist group isomorphisms : s =Zm , : r = Z2.

The plan is to nd : r Aut( s ) and show D2m = s r .Use the isomorphisms , to get : Z2 Aut Zm and D2m = Zm Z2.Dene : r Aut( s ) by (r ) = rsr (= rsr 1). The relations inD2m give (r )(s) = rsr = s1r 2 = s1 and g g1 is an automorphismfor any group G. All elements of D2m can be expressed as s

i

r j

wherei, j Z . Dene a map

D2m s ir j

(s i , r j ) s r .Verify that is an isomorphism.

The isomorphism induces an isomorphism : Aut s Aut Zmwhereby : f f = f 1. This yields an isomorphism : Z2 17


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18 2010 QUAL

Aut Zm , where = 1 as shown.

r / /

Aut s

Z2 / / _ _ _ Aut Zm .

Now let : s r Zm Z2 be (s i , r j ) = ( (s i), (r j )). InconclusionD2m

= s r

= Zm Z2.

2010 A2.

(1) Let G be a group of order np where p is a prime. Find asufficient condition for a subgroup of order p to be normal.

(2) List all isomorphism classes of abelian groups of order 120.(3) Is there a simple group of order 120?(4) What is the maximal possible number of elements of order 5

in a group or oder 120?

Answer (1). Pending.

Answer (2). Recall

Theorem (FTFAG) . Every nite abelian group is isomorphic to a group of the form

Z pn 11 Z pn kkwhere the pi are primes whose powers pn ii are uniquely determined by the group.

Given a prime p and a positive integer k there is in general oneabelian group of order pk for each partition of k. In other words if k = r 1 + + r t is a partition of k then

tr =1 Z r i is an abelian group

of order pk . The uniqueness portion of the Theorem indicates thatdistinct partitions of k yield distinct isoclasses.

Accordingly, if G is abelian of order 120 = 23 3 5 then G isisomorphic to one of the following:Z8 Z3 Z5, Z2 Z4 Z3 Z5, Z2 Z2 Z2 Z3 Z5.

Answer (3). We will need the following:

Lemma. If |G| = pn m with gcd(m, p) = 1 then n p(G)|m.


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PART A 19

This is proved below.Assume G is simple of order 120. Sylow 3 impliesn5 1 (mod 5)

and n5|24. Accordingly n5 {1, 6}, and simplicity forces n5 = 6because a unique Sylow 5-subgroup would be normal by Sylow 2.Dene an action of G on Syl5(G) = {P 1, . . . , P 6}by conjugation.This induces a homomorphism G S 6 whereby g(i) = j iff gP ig1 =P j . Obviously this is a nontrivial homomorphism, and since its kernel

is normal in G simplicity forces it to be a monomorphism G S 6.Identify G with the subgroup of S 6 to which it is isomorphic.If G < A 6 then G A6 is a proper subgroup of G. Then

|S 6| = |GA6| = |G||A6|

|G A6|=

120360|G A6|

and Lagranges theorem implies [G : G A6] = 2, meaing G A6 G.This cannot be true unless G A6 = G, so G < A 6.Now [A6 : G] = 3 by Lagranges theorem, so the action of A6 onG by conjugation yields another homomorphism A6 S 3 which isnontrivial because [A6 : G] = 3. But A6 is simple, so A6 S 3 must bea monomorphism. This is a contradiction since |A6| = 360 > 6 = |S 3|.In conclusion, G cannot be simple.

Proof of lemma. Let P be a Sylow p-subgroup of G. Sylow 2says that every other Sylow p-sobgroup is conjugate to P ; so there areas many Sylow p-subgroups of G as there are conjugates to P . Thusn p(G) = [G : N G (P )], and since

|G

|= [G : N G (P )]

|N G (P )

|we see

that nq(G) divides pn m. The conclusion of Sylow 3 is that n p(G) 1(mod p), which is to say that n p(G) is coprime to p Therefore, for therelation n p(G)| pn m to hold it must be true that n p(G)|m.

Answer (4). Since |G| = 120 < a subgroup P is a Sylow 5-subgroup if and only if |P | = 5. But in this case P is cyclic, P = afor some a G. On the other hand, a G has order 5 iff | a | = 5iff a Syl5(G). Now, if P and Q are distinct Sylow 5-subgroups of G, then a = P = Q = b implies P Q = {e}. (Any non-identityelements generates the whole cyclic group.) So when n5 = 6 (largestpossible by part 3) G contains 6

(5

1) = 24 (exclude the identity)

elements of order 25.2010 A3. Let G be a group acting on a set X . We say that the

action of G on X is transitive if & only if for all x, x X thereexists g G such that gx = x . Prove the following statements.

(1) For all x X , Gx = X . In particular, [ G : StabG (x)] = |X |and if |G| < then |X | divides |G|.


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20 2010 QUAL

(2) The subgroups StabG (x) are conjugate for all x X .(3) Suppose that |X | = k and let : G S k = Bij( X ) be the

group homomorphism given by the action. Then k dividesn = [G : ker ] and n divides k!.(4) Suppose that G is a nite simple group, |G| 3 and H is aproper subgroup of G with index k. Then G is isomorphic to

a subgroup of Ak . In particular, |G| divides k!2 .Hint. Consider the action of G on the set G H .

1. Fix x X . That Gx X is trivial. The reverse containmentGx X is immediate from transitivity: if x X then there existsg G with gx = x , so x Gx. In particular, by [Hungerford II-4.2]the index [G : StabG (x)] equals the cardinal number of xs orbit

[G : StabG (x)] = {gx : g G} = Gx = X .If G is a nite group, then by Lagranges theorem |G| = [G : Stab G (x)]|X |shows that |X | |G|.

2. Let x, y X ; use transitivity to nd g G such that y = gx. If h Stab G (x) then

(ghg1)x = ( gh)(g1x) = ( gh)x = g(hx) = gx = ywhence StabG (x) g StabG (y) g1. Rephrasing, under the innerautomorphism of G given by g g1 we have established thatStabG (x)

g StabG (y) g1. Taking the inverse of this automorphism

and applying it to Stab G (y) provides the reverse inclusion. Therefore,equality holds.

Remark: As there is an action of G on the set of its subgroups byconjugation, this result says that if G acts transitively on X then it alsoacts transitively on the set of stabilizers of elements of X .

3. The homomorphic image of a subgroup is a subgroup, so Im q , and suppose G is a groupof order pn q for some n > 0. Prove that G is not simple.

Lemma. Under the stated hypotheses, n p(G) divides q .

A more general statement is true: replace q with m such that( p, m) = 1.

Proof. Sylows second theorem says that any two Sylow p-subgroupsP,Q < G are conjugate, i.e. there exists g G satisfying gQg1 = P .It follows that n p equals the size of the conjugacy class of Q underconjugation

n p = {gQg1 : g G}.It is a theorem that the latter quantity is exactly [ G : N G (Q)]. ByLagranges theorem, pn m = n p|N G (Q)| and so

n p| pn m.Sylows third theorem says that n p 1 (mod p); in particular

p | n p = (n p, p) = 1since p is prime. Accordingly, it must be true that n p|m.

Proof of 2008 A2. From the Lemma it follows in particular that

n p(G) q < p.By Sylows third theorem, n p(G) 1 (mod p), i.e.

n p(G) = 1 + kp


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PART A 43

for some k N. These facts necessitate k = 0, otherwise

n p

(G)

1 + p > p,

contradictory to the Lemmas implication. Thus n p(G) = 1, whichmeans that G has a unique Sylow p-subgroup P G. I know thecontainment is proper since the order of P is by denition a power of p.

Sylows second theorem indicates that all Sylow p-subgroups areconjugate to one another. As P is unique, it follows that any conjugateof P is equal to P . But this is what it means for P G; hencethere exists a non-trivial normal subgroup of G. Accordingly G is notsimple.

2008 A3. Let G be a group with |G| = 120 = 2 3 3 5.Lemma. Let , S 6 and suppose has the (unique) disjoint cycle

decomposition = ( a1 . . . ak1 )(b1 . . . bk2 ) . . .

Then 1 has the disjoint cycle decomposition

1 = (a1) . . . (ak1 ) (b1) . . . (bk2 ) . . .

Proof. If (i) = j , then

1 (i) = (i) = ( j ).Accordingly, if the ordered pair i j appears in the disjoint cycle decom-position of , then the ordered pair (i) ( j ) appears in the disjointcycle decomposition of 1.

The lemma will be used frequently in the sequel without reference.(i) List all isomorphism classes of abelian groups of order 120.

Theorem (Fundamental Theorem of Finite Abelian Groups) . Ev-ery nite abelian group is the direct product of cyclic groups whose

orders have a certain property.Solution. The isoclasses are

G A Z120 ,Z2 Z4 Z3 Z5Z2 Z2 Z2 Z3 Z5 .


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44 2008 QUAL

(ii) Can G be simple?

Answer. Suppose G is simple. Playing around with Sylows the-orems shows that n5 = 6. Let G act by conjugation on the set

Syl5 = {F 1, F 2, F 3, F 4, F 5, F 6}of Sylow 5-sybgroups of G. This provides a group homomorphism : G S 6 wherebyverify homoM

(g)(i) = j if and only if gF ig1 = F j .The kernel of is a normal subgroup of G; certainly ker = G, so thesimplicity of G necessitates ker = {e}. Thus is a monomorphism;identify G and im , from now on regarding G < S 6.

Ask whether or not G < A 6. If not , let G be an odd permutationand observe that for any S 6,

GA6 ( 1) = S 6so that GA6 = S 6. Since every group involved is nite, the producttheorem ([2], I.4 Theorem 7) gives

|S 6| = |GA6| = |G||A6|

|G A6|= | G A6| = |

G||A6||S 6|

= 60.

By Lagranges theorem, |G| = [G : G A6]|G A6|, whence [G :G A6] = 2. But this means G A6 G which is forbidden. ConcludeG < A 6.Choose a Sylow 5-subgroup S of G. Since n5(G) = 6, the index

[G : N G (S )] = 6 as well. By Lagranges theorem,

120 = |G| = [G : N G (S )] |N G (S )| = 6 |N G (S )|which implies |N G (S )| = 20. Noting the obvious fact that S < G < A 6,The next step is to show that A6 contains no subgroup of order 20.

If K is a subgroup of A6 with order 20 = 2 2 5, then Sylowstheorems imply that n5(K ) 1 (mod 5) and n5(K )|4, in other wordsn5(K ) = 1. The unique Sylow 5-subgroup K 1 of K is thus normal.As a group of order 5, K 1 is cyclic with generator equal to some 5-cycle in A

6. Any 5-cycle xes one element of

{1, . . . , 6

}; it may be

assumed without loss of generality that 6 is xed by K 1s generator.For simplicity it may also be assumed, again without loss of generality,that

K 1 = (12345) .Since any conjugate of (1 2 3 4 5) is again a 5-cycle, for simplicity theambient group may be taken to be S 5 rather than S 6.


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PART A 45

Any permutation N S 5 (K ) satises

(12345) 1 = (1) (2) (3) (4) (5) = (1 23 4 5) k

for some k {1, 2, 3, 4}. Certainly there are exactly 5 4 = 20 such , which means that |N S 5 (K )| = 20. Since K 1 K , K is contained inthe normalizer of K 1 and hence K = N S 5 (K 1). But (1 2 4 3) N S 5 (K 1)since it is an odd permutation satisfying

(1243)(12345)(1243)1 = (2 4 1 35) = (1 23 4 5) 2.In conclusion, A6 contains no subgroup of order 20. At this point it isclear that G cannot be simple, so there are no simple groups of order 120 .

(iii) What is the maximal possible number of elements of order 5 in

G? Solution. A Sylow 5-subgroup of G has order 5, so it is cyclic.

Sylows second theorem states that if g G has order 5 then g F ifor some i. Furthermore, each F i is realized through conjugation: foreach i, Syl5 = {xF ix1 : x G}.There can be either 1 or 6 Sylow 5-subgroups (cf. part (ii)). DistinctSylow 5-subgroups have trivial intersection: If F i = F j then there existsx F i F j . Each such x generates F i , thus x, x 2, x3, x4 / F j , soF i F j = {e}.Summarily, each Sylow 5-subgroup contains exactly 4 elements of order 5, and each element of G having order 5 is contained in someSylow 5-subgroup. Since distinct Sylow 5-subgroups have distinct non-identity elements, there can be at most 6 4 = 24 elements of order 5in G.

(iv) How many conjugacy classes are there in S 6? The answer pro-vided here works for any n, but I am going to specialize to n = 6.

Definition. If S 6 is the product of disjoint cycles of lengthsn1, . . . , n r with n1 n r (including 1-cycles) then the list n1, . . . , n ris called the cycle type of .

Definition. A partition of 6 is a non-decreasing sequence of pos-

itive integers whose sum is 6.Proposition. , S 6 are conjugate if and only if they have the

same cycle type.

Proof. ( ) is immediate from the Lemma.( ) Suppose 1and 2 have the same cycle type. Including 1-cycles,

order their cycles in non-decreasing length somehow. If 2 cycles in


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46 2008 QUAL

1/ 2 have the same length this is not unique. Cover the parentheseswith your nger, this gives two lists of the numbers 1 through 6 where

in each number appears exactly once. Say1 = ( a11 . . . a1n 1 ) . . . (ar 1 . . . a rn r )

a11 , . . . , a 1n 1 , . . . , a r 1, . . . , a rn r2 = ( b11 . . . b1n 1 ) . . . (br 1 . . . brn r )

b11 , . . . , b1n 1 , . . . , br 1, . . . , brn r .Let S 6 denote the permutation dened by

(a ij ) = bij .

Since the parentheses appear in the same positions in 1 and 2s dis-

joint cycle decompositions it follows from the Lemma that 1 1

=2.

So there exists a bijection between conjugacy classes of S 6 and cycletypes. But cycle types are in one-to-one correspondence with partitionsof 6. There are 11 partitions of 6, so there are 11 conjugacy classes inS 6.

2008 A4. Let Z[i] = {x + yi : x, y Z}, i2 = 1. This is a unital ring and Z identies with a subring of Z[i].(i) Is the ideal of Z[i] generated by 5 prime? To say that (5) is

prime means for all x, y Z[i], xy (5) implies x (5) or y (5).This statement is false when x = 2 + i, y = 2 i:

(2 + i)(2 i) = 4 + 1 = 5 (5)but 2 i / (5). So (5) is not prime.(ii) Is Z[i] a domain? If so, describe its eld of fractions. Note thatZ[i] is a commutative unital ring, since Z is. Let a + ib,c + id Z[i]and suppose

0 = (a + ib)(c + id) = ( ac bd) + i(ad + bc).Accordingly,

ac bd = 0ad + bc = 0 .Multiplying the rst equation by c, the second by d and adding showsthat a(c2+ d2) = 0. As Z is a domain, it follows that a = 0 or c2+ d2 = 0.

If a = 0, then what we have is

ib(c + id) = 0 = ibc = bd


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PART A 47

a contradiction since ibc is purely imaginary while bd is an integer. Itmust be that c2 + d2 = 0, and since c2, d2 0 it follows that c2 = 0 = d2whence c = 0 = d. Consequently, c + id = 0, which shows that Z[i] isa domain.

Let F denote the eld of fractions of Z[i]. Formally, F consists of equivalence classes of ordered pairs (x, y) Z[i] (Z[i] {0}) underthe relation

(x, y) (x , y ) xy = yx .The equivalence class of (x, y) is denoted x/y .

The eld of fractions of Z is Q , so any eld containing Z and i mustcontain

Q[i] = {r + is : r, s Q}.In particular, F Q[i]. The multiplicative inverse of c + id/ 1 in F is

1c + id

= idc2 + d2

,

soa + ibc + id

= ( a + ib) idc2 + d2

=bd

c2 + d2+ i ad

c2 + d2Q[i]

whence F Q[i]. In conclusion, F Q[i].(iii) Choose a maximal ideal P in Z[i] and describe the localization

of Z[i] at P . An element x Z[i] is irreducible if and only if the ideal(x) is maximal in the set of all proper principal ideals of Z[i]. As

the Gaussian integers form a euclidean domain, they are a principalideal domain. To answer this question it suffices to nd an irreducibleelement of Z[i], take its principal ideal, and write down the localizationof Z[i] at said ideal.

The element 2 Z[i] is not irreducible since 2 = (1 + i)(1 i)and neither factor on the right belongs to Z[i] . However, 3 Z[i] is irreducible: it is irreducible over Z and any purported factorization

3 = ( a + bi)(c + di)

must satisfy ad + bc = 0 and ac bd = 3. Tedious considerationof the latter equations shows that such a factorization cannot occur.Consequently, the ideal (3) is maximal.The localization of Z[i] at (3) is dened as follows: Let S = Z[i](3), observe this set is multiplicative. Then the ring of fractions

S 1Z[i] = a + bic + di

: 3 | c + diis the desired localization.


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48 2008 QUAL

2008 A5.

Definition ([2], X 3.1). A morphism f : C D of a category Cisepic if and only if for all objectsE of Cand all morphisms g, h : D E the following implication is trueg f = h f g = h.

(i) Give an example of a category in which a morphism between twoobjects is epic if and only if it is surjective.

Example. The example is the category Set whose objects are setsand whose morphisms are functions. Here is a proof of the equivalenceof the two concepts.

( ) Assume f : C D and D = f (C ). Given an object E of Cchoose g, h : D E to satisfy g f = h f . Given y D, byhypothesis there exists x C such that y = f (x). But nowg(y) = g f (x) = h f (x) = h(y)

and since y was chosen arbitrarily it follows that g = h.( ) Assume f is epic and take E = D. The identity map 1 D : D D is available, and by virtue of the setup we also have the inclusion

map : f (C ) D. Note that Hom Set f (C ), D HomSet (D, D ) and,for all x C f : x f (x) f (x)

1D f : x f (x) f (x).Accordingly f = 1 D f ; by hypothesis = 1 F . In particular it istrue that

f (C ) = dom = dom 1 D = D

which means f is surjective.

(ii) Give an example of a category Cand of an epic morphism be-tween two objects in Cwhich is not surjective.Lemma. If R is a ring and f, g are ring homomorphisms Q Rwith f |Z = g|Z , then f = g.

Proof (A. Katz). Observe

f (1) = f (1 1) = f (1)f (1) = f (1)g(1).


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PART A 49

When n Z {0},

f (1) = f nn

= f nn

gnn

= f (n)f 1n

g(n)g1n

.

Here is a long chain of implications with the hypothesis invoked at justthe right time. Im still using n Z

{0

}.

f (n)f nn

= f (1) = f (n)f 1n

g(n)g1n

f 1n

f (n)f nn

= f 1n

f (n)f 1n

g(n)g1n

f nn

f 1n

= f nn

f 1n

g(n)g1n

f (1)f 1

n= f (1)f

1

ng(n)g

1

n

f 1 1n

= f 1 1n

g(n)g1n

f 1 1n

= f 1n

f (n)g1n

f 1n

= f 1n n g

1n

f 1n

= f (1)g1n

f 1n

= g(1)g 1n

f 1n

= g 1 1n

f 1n

= g1n

.


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50 2008 QUAL

Now for any mn Q ,

f mn = f (m)f

1n

= g(m)g1n

= gmn

whence f = g.

Example. Let : Z Q denote inclusion; it is evidently a ring ho-momorphism. This map is the candidate for an epic and non-surjectivemorphism (in the category Ring of course). The latter is obviously true,for example 12 Q im , so we need only verify that is epic. For anyring R choose ring homomorphisms f, g : Q R to satisfy f = g.But this just means f |Z = g|Z , whence f = g by the lemma. Thus satises the denition of epic.

Part B

All rings are assumed to be unital, and all modules are assumed tobe unitary unless otherwise stated.

2008 B1. Prove that

Q[x]/ (x5 4x + 2)is a eld. Show, on the other hand, that

Z[x]/ (x5 4x + 2)is not a eld.

Proof. Denote f (x) = x5 4x +2. Eisensteins criterion (Hunger-ford III.6.15) with p = 2 Z shows f to be irreducible in Q[x]. Butthen ( f ) is maximal and so Q[x]/ (f ) is a eld.

It is true that f is irreducible in Z[x], but false that ( f ) is maximal

in Z[x], because the latter is not a PID. (See Gallians ( ) proof of Theorem 17.5). In particular ( f ) (x, 2). Appeal to HungerfordIII.2.20(ii).

2008 B2. Let R be a ring, and let A,B,C be three R-modules such that B is a submodule of A, and C A/B . Prove that if C is a projective R-module, then A B C .


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PART B 51

Definition. An R-module C is projective if and only if givenany diagram of R-modules and R-module maps

C f

X g / /Y / /0

with the bottom row exact there exists a map of R-modules h : C X such that g h = f .Lemma (Short Five Lemma) . Let R be a ring and

0 / /X f

/ /

Y g

/ /

Z / /

0

0 / /X f / /Y g / /Z / /0

be a commutative diagram of R-modules and R-module maps with short exact rows. Then

(1) If , are monomorphisms, then is a monomorphism.(2) If , are epimorphisms, then is a epimorphism.(3) If , are isomorphisms, then is a isomorphism.

Proof that A B C . Let j : C A/B denote the isomor-phism given in the hypothesis. From the setup there is a short exactsequence

0 B

A

C 0,where , denote inclusion and projection, respectively. Consider thediagram

C 1C

A / /C / /0

(the bottom row is exact). Since C is projective, there exists a map of R-modules h : C A such that h = 1 C .Dene a map of R-modules : B C A by verify

(b, c) = (b) + h(c),


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52 2008 QUAL

and consider the following diagram:

0 / /B

/ /1B

B C

/ /

C / / j

0

0 / /B

/ /A

/ /A/B / /0

Since = 0 and h = 1 C , it follows that this diagram is commu-verify tative. Because j, 1B are isomorphisms, the Short Five Lemma impliesthat is also an isomorphism.

2008 B3. Let R be a commutative ring and I an ideal of R. Let A be an R-module and denote by IA the submodule of A generated by

all elements of the form ra with r I and a A. Prove that there is an isomorphism of R-modules

(R/I ) R A A/IA.

The rst proof is more elementary than the second.

Proof. Since R/I is an R-bimodule it is permissible to view the(R/I ) R A as a left R-module

r (s + I a) = ( rs ) + I a.

Note that I acts trivially on the latter. Running this process backwardsr + I a = r (1 + I a) = (R/I ) R A = R(1 + I A)

shows that {1 + I a : a A}is a set of generators for (R/I ) R Aover R.Dene : A (R/I ) R A by

(a) = 1 + I a.

Verify that is an R-module homomorphism, the last paragraph makesit clear that is in fact an epimorphism. Observe that if r ia i IA

(nite sum), then r ia i = (r ia i) = 1+ I r ia i = r i+ I a i = 0 a i = 0

so that IA ker .Passing to the quotient yields an R-module epimorphism

: A/IA (R/I ) R A, (a + IA) = 1 + I a.


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PART B 53

To show that is an isomorphism it suffices to exhibit a R-modulehomomorphism whose composite with the former is the identity on

A/IA . Dene a map(R/I ) A (r + I, a )

ra + IA A/IA.Verify that is well-dened and middle-linear. The universal propertyof the tensor product provides a unique ( R-module) homomorphism

: (R/I ) R A A/IA, (r + I a) = (r + I, a ).For all a + IA A/IA ,

(a + IA) = (1 + I a) = a + IA.Therefore is an inverse for , both of which are consequently R-module isomorphisms.

The second proof is more elegant than the rst.

Proof. The following sequence is exact

0 I

R

R/I 0where , denote the canonical injection, projection R-module homo-morphisms respectively. Proposition IV 5.4 says that R A is right-exact, so the resulting sequence

I R A 1A

R R A 1A

R/I R A 0is exact. In particular, 1A is an R-module epimorphism. By therst isomorphism theorem

(R R A) (ker 1A) R/I R A,

and by exactness ker 1A = im 1A . Observe that (all sums arenite)

im 1A = 1A n i(r i a i) : r i I, a i A

= n i 1A(r i a i) : r i I, a i A

= n i (r i) a i : r i I, a i A

I R A.Theorem IV 5.7 indicates that there are isomorphisms

R R A A, I R A IA.

Accordingly

A/IA (R R A) (im 1A) = ( R R A) (ker 1A) R/I R A.


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54 2008 QUAL

2008 B4. Let V and W be two vector spaces over a eld k, and f : V W be a linear map. Prove that f is surjective if and only if its dual map f is injective.

Definition. The dual of f is dened by

W f () V where f () = f . Its easy to verify that f is a linear map.

The dual map provides an arrow between arrows.

Proof. (f surjective f injective) Hypothesis: ker f = {0}.Let w W be given, by way of contradiction suppose w W im f .A non-zero element of W is given by w , where

w (x) =0 if x = w,1 if x = w.

For all v V ,

(f (w )) ( v) = w f (v) = w (f (v)) = 0by the way in which w was chosen. Hence w ker f , contradictoryto the hypothesis. In conclusion f is surjective.

(f surjective f injective) Hypothesis: im f = W . Assume

, W satisfy f () = f (). For all w W there exists (byhypothesis) v V such that w = f (v). Thus

(w) = f (v) = ( f ()) ( v) = ( f ()) ( v) = f (v) = (w),and since this is valid for all w it follows that = . In conclusion f is injective.

2008 B5. Find the Jordan normal form of the following matrix over the eld of complex numbers:

0 1 0 1

0 0 0 10 0 0 10 0 0 0

Solution. Call that matrix M . Make me a sandwich. Cofactorexpansion along row 4 of I 4 M gives chA() = 4, so = 0 is the


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PART C 55

only eigenvalue of M . Accordingly, 0 is the only entry appearing onthe diagonal of M s Jordan form:

J =

0 ? 0 00 0 ? 00 0 0 ?0 0 0 0

.

By denition the algebraic multiplicity of = 0 is 4. It follows that thesum of the sizes of the blocks of J equals 4. The eigenspace of = 0equals the kernel of M ; a quick computation shows that

geom. mult.(0) = nullity( M ) = 2 .

Consequently, J contains exactly 2 blocks. Finally, observe that M 2 =

0 but M 3 = 0; this means that M s minimal polynomial is x3, hencethe size of the largest block in J is 3. Putting all this together gives

J =

0 1 0 00 0 1 00 0 0 00 0 0 0

which is the Jordan normal form of M .

Part C

Do any 3 problems.2008 C1.

2008 C2. Let F be a splitting eld of f K [x] over K . Prove that if an irreducible polynomial g K [x] contains a root in F , then gsplits into linear factors over F .

Corollary ([2], V 1.9, p.236). Let E and F be extension elds of K and let u E and v F be algebraic over K . Then u and v are roots of the same irreducible polynomial f K [x] if and only if there is an isomorphism of elds K (u) K (v) which sends u onto v and is the identity on K .

Theorem ([2], V 3.8, p.260). Let : K L be an isomorphism of elds, S = {f i}a set of polynomials (of positive degree) in K [x],and S = {f i}the corresponding set of polynomials in L[x]. If F is a splitting eld of S over K and M is a splitting eld of S over L, then is extendible to an isomorphism F M .


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56 2008 QUAL

Proposition ([2], V 3 exercise 2, p.267). If F is a spliting eld of S over K and E is an intermediate eld, then F is a splitting eld of

S over E .Proof. First it will be shown that if K is an algebraic closure of

K containing F , then for any K -monomorphism : F K it is truethat im = F . The desired conclusion will then be deduced.Write

f (x) =n

i=1

kixi , ki K

and let a K -monomorphism : F K and a root u F of f begiven. Observe that 0 = f (u) implies

0 = f (u) = n

i=1

kiui =n

i=1

(kiui) =n

i=1

(ki)(ui) =n

i=1

ki(u)i

which is nothing but f (u) ; in other words (u) K is also a rootof f .

The hypothesis that F is a splitting eld of f over K can be usedto express

f (x) = kn

i=1

(x

ui), k K, u i F.

Since K is a eld, K [x] is a euclidean domain and hence a unique fac-torization domain. Thus the previous display is the unique factoredrepresentation of f as an element of F [x] K [x]. Since u is a root of f if and only if (x u) f , it follows that u1, . . . , u n is the exhaustive listof f s roots. Combining this with the observation made in the previousparagraph shows that (u i) {u1, . . . , u n}for all i = 1 , . . . , n . As isinjective, it follows that the action of on the set of f s roots is merelyto permute its elements.

The facts that F = K (u1, . . . , u n ),

|K = 1 K , and permutes

{u1, . . . , u n} clearly imply that (F ) = F . Now we can prove thedesired result.Let g K [x] be irreducible and suppose u F is a root of g. By

denition K contains all roots of g. If v K is a root of g, then bythe Corollary there exists a K -isomorphism of elds : K (u) K (v)such that (u) = v. Here is why extends to a K -automorphism of K : The algebraic closure K is the splitting eld for K [x] over K ; in


57/217

PART C 57

particular g splits in K [x]. The setup is

K

F K

K (u) / /K (v)

K K whereas the setup for the Theorem is

F M

K / /L

In the present situation K must be used instead of F , which is anotherreason F K is necessary; also S = {g}, S = {g}, and we shouldbe able to use K (u) for K and K (v) for L. For the latter is mustbe known that K is a splitting eld for g over K (u) and for g overK (v); this is guaranteed by the Proposition. Thus extends to aK -isomorphism : K K .Now

|F : F

K is a K -monomorphism. It follows from the

preliminary result of this proof that im = F , so v = (u) F . Asv was an arbitrary root of g, it is clear that all roots of g are membersof F . Therefore g splits over F .

Summary. Using Hungerfords method

Show that if F K and : F K is a K -monomorphism,then im = F . Observe that sends roots to roots. Use splitting eld to factor f , note uniqueness. This gives

the list of f s roots. Use monomorphism to conclude that just permutes the

roots. Use K - to get the result. Prove that if v K is a root of g, then v F . Pick u F (hypothesis), quote result giving a K -isomorphism

: K (u)toK (v) with (u) = v. Quote the Theorem and Exercise to extend : K K . Restrict |F : F K to get a K -monomorphism and usethe rst part. v F .


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58 2008 QUAL

2008 C3.

2008 C4. If Z p F is a eld extension of degree n then x

x p

is a Z p-automorphism of F of order exactly n whose xed eld is Z p.Proof. Denote (x) = x p. By hypothesis, there exists a basis

{ui : 1 i n}of F over Z pF =

n

i=1

a iui : a i Z p .

As there are p choices for each a i , it follows that |F | = pn .Consider F = F {0}, a nite abelian group under multiplication.By the Fundamental Theorem of Finite Abelian Groups,F

M

i=1Zm i , m1|m2| . . . |mM .

Call the group on the right G. Lagranges Theorem implies that, forall i, the order of each a Zm i divides m i and hence divides mM . ThusmM a = 0 for all a G, which means that um M = 1 for all u F . Soall pn 1 elements of F are roots of xm M 1. The latter has at mostmM distinct roots in F , it must be that G = Z pn 1, whence

F Z pn 1.By Lagranges Theorem |u| divides pn 1 for all u F , hence theequation u pn 1 = 1 F holds for all u F . Consequently, u pn = u for

all u F , which is to say that each u F is a root of x pn

x F [x].Trivially, 0 is also a root of this polynomial.Observe that i(u) = u pi for all u F and all i = 1 , . . . n 1. Fori = n, n (u) = u = 1 F (u); thus || divides n. If there exists i < n suchthat i = 1 F , then

u pi

= u u F or, equivalently

u pi

u = 0 u F whence all pn elements of F are roots of x pi x F [x]. But the latterhas at most pi(< p n ) roots in F , a contradiction.


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2007 Qual

Part A

Do four out of the ve problems.

2007 A1. This question is similar to 2010 A1 and 2008 A1.

2007 A2.Let R be a principal ideal domain and dene : R

{0} N0 by (a) = n if a = up1 p2 . . . pn for a unit u and primeelements pi R. Consider the condition(*) a1, a2 R d R : a1R + a2R = ( a1 + da2)R.

(1) Show that if R satises the condition (*), then R is a Euclideandomain with Euclidean function .

(2) Show that (*) implies that U (R) U (R/A ) is onto for eachideal A or R, where U (T ) denotes the group of units of thecommutative ring T .(3) Show that Z does not satisfy (*).

Another way to write (*) is ( a1) + ( a2) = ( a1 + da2).

1. Let a, b R. If b = 0 then certainly (a) (ab). We mustnd q, r R such that a = qb+ r with r = 0 or (r ) < (b). Thereexists d R such that ( a) + ( b) = ( a + db); write

a = ( d)b + ( a + db).Since (a + db)|b it is clear that (a + db) (b). If strict inequalityholds then use q = d, r = a + db. If equality holds then by thedenition of we have a + db = ub for some unit u. Then

a = ( u d)bso use q = u

d, r = 0.

2. Let A = ( a) and select a unit v + ( a) R/A . Then there existst R such that vt + ( a) = 1 + ( a), in other words vt 1 A. Writevt 1 = ma for some m R. Then 1 (vt) + ( ma ) = ( v) + ( a) andso R = ( v) + ( a) = ( v + da) for some d by (*). Thus 1 = s(v + da) forsome s R, in other words v+ da is a unit. Now v+( a) = v+ da +( a),and so the canonical map is surjective.

59


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60 2007 QUAL

3. If it were the case that Z satised (*), then (2) would be true.For the ideal (5), there are 4 units in Z / (5) but only two units in Z .

The canonical map is not surjective.2007 A3. Let S 5 operate on itself by conjugation. How many

orbits does S 5 have?

Proof. This proof references results from [2008 A3(iv)]. In thiscontext, orbit means conjugacy class. Permutations , S 5 areconjugate if and only if they have the same cycle type. The cycle typesof S 5 are in one-to-one correspondence with the partitions of 5. Thereare seven partitions of 5

5

4 + 13 + 23 + 1 + 12 + 2 + 12 + 1 + 1 + 11 + 1 + 1 + 1 + 1

so there are seven conjugacy classes in S 5.

2007 A4. Duplicates 2008 A2.

Part B

Solve four problems out of ve. All rings are assumed to be unitaland all modules are assumed to be unitary and left unless speciedotherwise.

2007 B1. Give an example of a ring R and an R-module M such that

(i) R M is not exact. Dummit and Foote [ 1] p.403 (4) says thatM is at if and only if R M is exact, so it would suffice to nd anon-at R-module. Item (5) says that the Z-module (abelian group)Q / Z is injective and not at, so this will be the example.

Example. Let R = Z and M = Q/ Z . The sequence

0 Z

Zwhere (n) = 2 n is exact, but the induced homomorphism

1Q / Z : Z Z (Q / Z) Z Z (Q / Z)


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PART B 61

is not injective. It is closer to the problem statement, but uglier, tosay that the induced sequence

0 Z Z (Q / Z) 1Q / Z

Z Z (Q/ Z)is not exact. At any rate, to see this just observe that

0 = 1 1/ 2 + Z 2 1/ 2 + Z = 1 0 + Z = 0i.e. ker 1Q / Z = 0.

(ii) Hom R (M, ) is not exact. [1] p.402 (1) says that Hom R (M, )is exact if and only if M is projective. So it would suffice to nd anon-projective R-module. I know Q is one of those, but Ive used thattoo many times. Lets go with Z2.

Example. Let R = Z and M = Z2, and write Hom for HomZ . Thesequence given by canonical projection

Z Z2 0is exact, but the induced sequence

Hom(Z2, Z)

Hom(Z2, Z2) 0is not exact. Why?

The integers Z have no nonzero nite subgroups (all subgroups areisomorphic to nZ for some n Z), and the homomorphic image of Z2in Z would be a nite subgroup, so it is clear that Hom( Z2, Z) 0. But

Hom(Z2, Z2) is comprised of the zero and the identity homomorphisms,so it must be that is the zero map and therefore is not surjective.

An example where Hom (, M ) is not exact. This is the equiv-alent to showing that Z2 is not an injective Z-module, which is truebecause it is not a divisible abelian group.

Apply the functor (with R = Z and M = Z2) to

0 Z2 Z4 Z2 0and obtain

0 Hom(Z2, Z2) Hom(Z4, Z2) Hom(Z2, Z2) 0.All three terms in the new sequence are isomorphic to Z2 for various(obvious) reasons, so it can be re-written

0 Z2 Z2 Z2 0.This sequence is not exact: the rst map is an injection between twosets of the same (nite) cardinality and hence is an isomorphism. If thesequence is assumed to be exact, the kernel of the second map equals Z2,


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62 2007 QUAL

and hence the third term in the sequence is actually 0. Contradiction:Z2 = 0.

2007 B2. Let R be a ring and M i , N i , i = 1 , 2, 3 be R-modules.Consider a diagram

0 / /M 1f 1

/ /M 2f 2

/ /

2

M 3 / /0

0 / /N 1g1

/ /N 2g2

/ /N 3 / /0

with exact rows (all maps are homomorphisms of R-modules).(i) Suppose that there exists 3 Hom R (M 3, N 3) such that g22 =

3f 2. Prove that there exists 1 Hom R (M 1, N 1) making the diagram

commute. Which conditions should 2 and/or 3 satisfy to ensure that 1 is injective?

Proof. To begin, check that im 2f 1 im g1. By exactness itsuffices to check that the former is a subset of ker g2. Given m1 M 1

g22f 1(m1) = 3f 2f 1(m1) commutativity= 3(0) exactness= 0 .

This ensures that the function1 := g11 2f 1

is well-dened. As a composite of R-module homomorphisms it is au-tomatic that 1 HomR (M 1, N 1). As for commutativity,

g11 = g1g11 2f 1 = 2f 1.To guarantee that 1 is injective it suffices to require that 2 be injec-tive; in this case 1 is a composite of three injective maps and mustitself be injective.

(ii) Suppose that there exists 1 Hom R (M 1, N 1) such that g11 =2f 1. Prove that there exists 1 Hom R (M 3, N 3) making the diagram commute. Which conditions should 2 and/or 3 satisfy to ensure that 3 is surjective?

Proof. To begin, verify that g22 is constant on ker f 2. If m2ker f 2, then

g22(m2) = g22f 1(m1) for some m1 M 1 by exactness= g2g11(m1) commutativity= 0 . exactness


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PART B 63

This ensures that the function

3 := g22f 12is well-dened. As a composite of R-module homomorphisms, it is clearthat 3 HomR (M 3, N 3). As for commutativity, if m2 M 2 then

3f 2(m2) = g22f 12 f 2(m2) = g22(m2).For 3 to be surjective it suffices to assume that 2 is surjective.

2007 B3 unnished.

2007 B4. Let R be a domain, A be (sic) a n n matrix over R.(i) Prove that if the system Ax = 0 has a non-trivial solution then det A = 0 .

Proof. Let v Rn {0}satisfyAv = 0 = 0 v.

Then 0 R is an eigenvalue of A with eigenvector v. One characteri-zation (from [2] VII 3.5(vii) and VII 5.5) of the determinant is

det A = ( 1)n 1 . . . nwhere i are the eigenvalues of A (with multiplicity). Since i = 0 forsome i, it follows that det A = 0.

(ii) Prove or provide a counterexample to the converse. The con-verse is: if detA = 0 then Ax = 0 has a non-trivial solution.

Proof. If 0 = det A = ( 1)n 1 . . . n , then since R is a domainit follows that i = 0 for some i. But then there is an eigenvectorv Rn {0}of A for 0, whence the system Ax = 0 has a non-trivialsolution.

(iii) Which, if any, of these statements remain true if we drop the assumption that R is a domain? Statement (i) remains true, since theformula used therein for the determinant is valid over any commutativering. That said, commutativity is essential.

Statement (ii) may fail in case R contains a zero-divisor, for in thissituation the equation ( 1)n 1 . . . n = 0 does not imply that i = 0for some i.

2007 B5. Let R, S be commutative rings, : R S be a ring homomorphism.


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PART C 65

(ii) Prove that the constant term of the characteristic polynomial of a matrix A Mat n (R) equals (1)n det( A).

Proof. Recall the denition of characteristic polynomial

A(x) = det( xI n A).Now, xI n A Mat n R[x] ; consider the evaluation homomorphisme0 : R[x] R dened by

e0(f ) = f (0).

By part (i), e0 extends to a ring homomorphism e0 : Matn R[x] Mat n (R) satisfyingdet e0(A) = e0 det( A)

for all A Mat n R[x] . Certainly, e0 det( xI n A) is the constantterm of As characteristic polynomial. Observedet e0(xI n A) = det(0 I n A) = det( A) = ( 1)n det( A).

Part C

Attempt any ve. All are worth 10 points.

2007 C2.Prove that Q( 2, 3) = Q( 2 + 3). I can do this with a, b R

rather than 2 , 3.

Proof. The inclusion Q( a + b) Q( a, b) is obvious.Since Q( a + b) is a eld,

Q( a + b) 1 a + b =1

a b( a b)

so ( a

b) Q( a + b). Then a = ( a + b) + ( a b)

2Q( a + b)

and similarly b Q( a + b). Thus the reverse inclusion holds.2007 C3.


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66 2007 QUAL

Let F be a Galois extension of a eld K of degree 27. Prove that there exist Galois extensions of K contained in F of degree 3 and 9.

Theorem (Fundamental Theorem of Galois Theory) . If F is a nite-dimensional Galois extension of K , then there is a one-to-one correspondence between the set of all intermediate elds of the extension and the set of all subgroups of the Galois group Aut K F (given by E E = Aut E F ) such that:

(1) the relative dimension of two intermediate elds is equal to the relative index of the corresponding subgroups; in particular,Aut K F has order [F : K ].

(2) F is Galois over every intermediate eld E , but E is Galois over K if and only if the corresponding subgroup E = Aut E F is normal in G = Aut

K F ; in this case G/E is (isomorphic to)

the Galois group Aut K E of E over K .

Lemma. Let G be a group of order pn , where p is prime. For each k = 0 , . . . , n , G has a normal subgroup of order pk .

Proof of 2007 C3. Denote G = Aut K F . As F K is a nite-dimensional Galois extension, the Fundamental Theorem may be usedthroughout this proof. Note rst that |G| = [F : K ] = 33. Usingthe lemma, there exist H i (i = 1 , 2) such that |H i| = 3 i and H i G.Lagranges theorem indicates that [ G : H 1] = 9 and [G : H 2] = 3. TheGalois correspondence gives intermediate elds

K H 1, H 2 F such that (Fundamental Theorem) H i K is Galois. ALso,

[H 1 : K ] = [G : H 1] = 9[H 2 : K ] = [G : H 2] = 3

as desired.

Proof of the Lemma. Induct on n. When n = 0, G (and theproblem) is trivial. Assume a group of order pn1 has normal subgroupsof order pi for i = 0 , . . . , n 1.Let |G| = pn , as a p-group G has non-trivial center. Lagrangestheorem says that |C (G)| divides p

n

, so in particular p divides |C (G)|.Cauchys theorem gives an element a C (G) of order p. Thus a 0(prime). Choose u F and let f K [x] be its irreducible polynomial,say f (x) = ni=0 a ix

i with an = 1.By way of contradiction, suppose that f is not separable. Then the

rst lemma indicates that

f (x) =n

i=1

ia ixi1 = 0

which implies that ia i = 0 for i = 1 , . . . , n . Certainly then, each isuch that a i = 0 must be divisible by p. But this means that f is apolynomial in x p, say f (x) = g(x p) where g K [x]. Using the secondlemma, each of gs coefficients has a unique pth root; now since K hascharacteristic p, it follows that f (x) = g(x) p. For this reason, f isreducible in K [x]: a contradiction to the fact that f is an irreduciblepolynomial.

(iii) Prove that if K is of characteristic zero then F is separable over K .


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79/217

PART C 79

In other words, each element of K is a root of f . Accordingly, F is asplitting eld of f over Z3.

Since char F = 3, the derivative f (x) = 1 is relatively prime tof (x). The multiple roots theorem implies that f has no multiple rootsin F . If E F denotes the set of roots of f , then of course |E | = 3 22.Verify Z3 E . Now, since F = Z3(E ) it follows that F = E whence|F | = 3 22. Now

322 = |F | = |K |[F :K ] = 32[F :K ]

shows that [F : K ] = 2.If F 1 is another extension of K with degree 2, then

[F 1 : Z3] = [F 1 : K ] [K : Z3]implies [F

1: Z

3] = 4, so

|F

1|= 3 22. Show that F

1is also a splitting

eld of f over Z3, and use the fact that any two splitting elds of f overZ3 are isomorphic.


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81/217

2005 Qual

Part A

2005 A1 unnished.

2005 A2.(a) Exhibit 2 distinct Sylow 2-subgroups of S 5 and an element of S 5

that conjugates one into another. Consider the permutationsa = (1 2 3 4) , b = (12)(34)

from S 5 (the symbol 5 is xed by both). Observing that (1234) =4, (12)(34) = 2 , (1234)1 = (1 4 3 2) and

ba = (1 2)(3 4)(1 2 3 4) = (2 4 3)a1b = (1 4 3 2)(1 2)(3 4) = (2 4 3)

shows that(1234), (12)(34) D8.

As |S 5| = 5! = 23

15, a subgroup H < S 5 is a Sylow 2-subgroup if and only if |H | = 8. Certainly then, (1234), (12)(34) is a Sylow2-subgroup.For any S 5

(1234) 1 = (1) (2) (3) (4) , (12)(34) 1 = (1) (2))( (3) (4) .In particular,

(45) (1234), (12)(34) (45)1 = (1235), (12)(35) .These Sylow 2-subgroups are indeed distinct since 5 is xed by one butnot the other.

(b) How many elements of order 7 are there in a simple group of order 168?

Solution. Let G be a simple group of order 168 = 7 24. Sylowstheorems imply that n7 1 (mod 7), but the hypothesis that G issimple forces n7 = 1. Another well-known theorem states that n7 24,and it follows that the only possibility is n7 = 8.

Three facts:81


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82 2005 QUAL

(1) Each Sylow 7-subgroup of G contains exactly 6 elements of order 7.

Justication: A Sylow 7-subgroup P of G has exactly 7elements since 7 is the highest power of 7 dividing 168. Asa group of prime order, P is cyclic. By Lagranges theorem,each non-identity element of P generates P ; there are exactly6 such elements in P .

(2) Every element of order 7 belongs to some Sylow 7-subgroup of G.

Justication: Any a G with order 7 generates a cyclicsubgroup a of G with order 7. A such, a is contained in aSylow 7-subgroup P of G; in particular a P . (More is true,a actually is a Sylow 7-subgroup of G.)

(3) Distinct Sylow 7-subgroups of G have trivial intersections.Justication: If P and Q are distinct Sylow 7-subgroups of G and x is a non-identity element of P Q, then x generatesboth P and Q whence P = Q. Contradiction.

Taken together, it is clear that G contains 8 6 = 48 elements of order7.

2005 A3.(1) Prove that D8 is not isomorphic to D4 Z2.(2) If

|G

|= pn then G contains normal subgroups of order pi for

each i = 0 , . . . , n .

1. D8 has generators x, y such that |x| = 8 , |y| = 2 and xy = yx1.Yet D4 Z2 has generators ( a, 0), (b,0), (e, 1) and (ab, 0) = ( ba1, 0).The latter contains no element of order 8.2. Induct on n. When n = 0, G (and the problem) is trivial.

Assume a group of order pn1 has normal subgroups of order pi fori = 0 , . . . , n 1.Let |G| = pn , as a p-group G has non-trivial center. Lagrangestheorem says that

|C (G)

|divides pn , so in particular p divides

|C (G)

|.

Cauchys theorem gives an element a C (G) of order p. Thus a 0 if and only if f has three real roots. What conclusion can you draw if D > 0?

Proof. It was not stated, but shall be assumed that f K [x].Suppose that f factors in some splitting eld over F K as

f (x) = ( x u1)(x u2)(x u3) F [x].Recall that

D = ( u1 u2)2(u1 u3)2(u2 u3)2.


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99/217

PART C 99

The irreducible polynomial of over Q is 8, which has degreedeg8 = (8) = 4. Since the eld of rationals has characteristic zero,

the algebraic extension F 8 Q is Galois. According to the Fundamen-tal Theorem of Finite Dimensional Galois Theory

|G| = [F 8 : Q] = [Q( ) : Q] = deg 8 = 4 .Each G is completely and uniquely determined by its action ( ) = i for some i {0, . . . , 7}. Similarly, ( ) = j and since 1 = 1 =1 it follows that i is a unit modulo 8.

The map dened byG i Z8

is easily seen to be a monomorphism of groups, so G is isomorphic toa subgroup of the units modulo 8. But

|G

|=

|Z8

|, so it must be the

case that G is isomorphic to the group of units modulo 8 itself. Notethat

G Z8 Z2 Z2.


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101/217

2004 Qual, version 2 (October)

Part A

2004v2 A1 unnished.

2004v2 A2.(a) Solvable quotient implies solvable group. Let G be a nite group

and H a normal subgroup of G. Show that if H and G/H are solvable,then G is solvable also.

Proof. Let G denote G/N . By hypothesis there are solvable series

e = N 0 N 1 . . . N n = N e = G0 G1 . . . Gm = G.

By the standard theorem characterizing the subgroups of a quotient,for each i = 1 , . . . m there exists a subgroup Gi of G containing N suchthat

Gi = Gi /N,so actually

e = G1/N G2/N . . . Gm /N = G/N.

By the Third Isomorphism Theorem,

Gi+1 / Gi = Gi+1 /N Gi /N Gi+1 /G i .

Accordingly, the factors in the penultimate display are abelian. Puttingthe two series together gives

e N 1 . . . N n = N = G0 G1 . . . Gm = G,

which is a solvable series in G.

(b) Solvable factors implies solvable product. If H, K are solvablenormal subgroups of G with HK = G and H K = e , then G issolvable.

Lemma. If G H K , then G/H K .101


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102 2004 QUAL, VERSION 2 (OCTOBER)

Proof. Consider the canonical projection homomorphism : H K K . Note im = K, ker H . There is also the canonicalinjection homomorphism : H H K , with im H . Observeim = ker . By the rst isomorphism theorem,

H K

/ /K

H K ker / /im

i.e. H K/ ker im . Now, H K G and ker = im H impliesH K/ ker G/H

and since im = K , it follows that G/H K .Proof of (b). The standard theorem on internal direct products

indicates that G H K , so it suffices to prove that H K is solvable.Using part (a) G here plays the role of G in (a) H here plays the role of H in (a), solvable K here plays the role of G/H in (a), solvable

hence G is solvable.

2004v2 A3. Let S be a multiplicative subset of the commutativering R and let I be an ideal of R. Show that S 1Rad( I ) = Rad( S 1I ),where Rad( J ) = {x R : xn J for some positive integer n}.

Proof. Verify that S 1Rad( I ) Rad( S 1I ).Conversely, choose yt Rad( S 1I ). Then

yntn = P

xs for some x

I, s S, n 1. Therefore there exists a S such that s (yn s xt n ) =0, that is to say s yn s = s xt n I. Since I is an ideal we now have(s )n yn sn I so that s ys Rad( I ).

Let z s ys and t = s st . Then tzt = tyt and hence t(yt zt ) =?,so that yt = zt , where z Rad( I ).2004v2 A4. Duplicates 1995 A3.

2004v2 A5. Let G be a nite simple group having a subgroup H of index n. Show that G is isomorphic to a subgroup of S n .

Remark. This is similar to [2002 A3(a)].

Proof. Using [G : H ] = n, write

G/H = {a iH : i = 1 , . . . , n }.


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PART B 103

The group G acts on the set G/H by left multiplication,

g

a iH = ( ga i)H,

note (ga i)H = a j H for some j {1, . . . , n }. This action induces agroup homomoprhism : G S n whereby(g)(i) = j g a iH = a j H.

Recall from undergraduate algebra that ker G. As G is simple,either ker = G xor ker = e . Certainly is not the trivial homo-morphism, unless G is the trivial group (in which case there is nothingto prove). By the First Isomorphism Theorem

G

/ /S n

G/ e / /im

i.e. G G/ e im < S n .Part B

(1) Decompose the following as a direct sum of cyclic groups(a) Z12 Z19(b) Hom( Z12, Z9)

(2) Let A, B be nitely generated abelian groups. Prove that

Hom(A, B ) is nitely generated.Lemma. Zn Zm = Zgcd( m,n )

Proof. See JMDs solution for 2009 B3, or do it NMs way: Firstshow that for any abelian group A and all m > 0, A Zm = A/mA . Usebasic properties of . Next use cyclic group theory to show Zn /m Zn =Zgcd( m,n ) .

Lemma. Hom Z(Zm , Zn ) = Zgcd( m,n ) .

Proof. Hungerford IV.4.Ex1(a,b). Use NMs method.

1(a). The greatest common divisor of 9 and 12 is 3, so Z12 Z9 = Z3by the Lemma.

1(b). By the second Lemma, Hom( Z12, Z9) = Z3.

2. Suppose A is generated by {x1, . . . , x n}. All elements of A havethe form xm 11 . . . x m nn where m i Z . Dene a homomorphism Zn Aof abelian groups by extending ei xi to be Z-linear, where ei is the


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ith standard basis element of Zn . This msp is surjective. Let K = kerthis map, so there is an exact sequence

0 K Zn A 0.Apply the left exact contravariant functor Hom Z(, B )

0 Hom(A, B ) Hom(Zn , B ) Hom(K, B ).This shows that Hom( A, B ) is a submodule of Hom(Zn , B ); if we canshow that the latter is nitely generated then since the underlying ringZ is Noetherian it will follow that Hom( A, B ) is nitely generated.

Dene a homomorphism of abelian groups Hom( Zn , B ) B n byextending ( (e1), . . . , (en )) to be Z-linear. Verify that this isan isomorphism, using the structure theorem Hungerford II.2.1 to getsurjectivity. Since B is nitely generated, so too is B

n

and Hom( Zn

, B ).

Part C

PLEASE CIRCLE THE Four PROBLEMS YOU WANT TO BEGRADED.

2004 C1.Let F be the splitting eld over Q of the polynomial x4 4x2 1.Let g(x) = x3 + 6 x2 12x 12. Does g(x) have a root in F ? Prove your answer.Lemma (Eisensteins Criterion) . For a unique factorization D and

its quotient eld F , if

f (x) =deg f

i=0

a ixi D[x]

with deg f > 1 and p D is irreducible such that p | adeg f , but p|a i for 1 i < deg f and p2 | a0, then f is irreducible over F .Theorem (Rational Root Theorem) . If r/s Q is a root of the

polynomial

f (x) =deg f

i=0

a ixi Q[x]

then s|adeg f and r |a0.Solution. Eisensteins criterion ( D = Z , p = 3) indicates that g

is irreducible over Q . In particular, g has no roots in Q . The rationalroot theorem implies that if r/s Q is a root of f , then r/s = 1.


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PART C 105

Neither of 1 are roots of f , so f has no rational roots and hence anyfactorization of f in Q[x] into non-unitsf (x) = h(x)k(x)

must satisfy deg h = 2 = deg k.Exploring the purported factorization f = hk further, since f is

monic with constant term 1 it must be the case thath(x)k(x) = ( x2 + ax + 1)( x2 + bx 1)

for some a, b Q . Expanding the right-hand side produces the systemof equations

b + a = 0ab = 4ba = 0

which is inconsistent. Consequently, there exists no such factorizationof f . At this point all possible factorizations of f in Q[x] into non-unitshave been shown to be impossible, and it follows that f is irreducibleover Q .

Substituting y = x2, solving for y using the quadratic formula, andre-substituting shows that the roots of f are

2 + 5, 2 + 5, 2 5, 2 5.Note thatF = Q 2 + 5, 2 5

has degree 4 over Q . The roots of f are distinct, which means that f and also F are separable over Q . A standard theorem (Hungerford V3.11) indicates that F is algebraic and Galois over Q . Another theorem(Hungerford V 3.14) shows that F is normal over Q .

If g has a root u F , then g must split in F and accordingly the

splitting eld E of g over Q is contained in F . Now [E : Q] divides3! = 6. But

[F : Q] = [F : E ] [E : Q]

implies that 6 |4, which is false. Therefore g has no root in F .2004v2 C3.


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Let F 12 be a cyclotomic extension of Q of order 12. Determine Aut QF 12 and all intermediate elds. The eld F 12 is a splitting eld of

x12

1 over the rationals. If denotes a primitive twelfth root of unity,then generates set of all twelfth roots of unity. The latter is exactlythe set of roots of x12 1, so

F 12 = Q().

The extension F 12 Q is algebraic and consequently, since char Q = 0,is Galois. Let G denote Aut QF 12. The Fundamental Theorem of Finite-Dimensional Galois Theory indicates that F 12 is Galois over every inter-mediate eld E , and there is a one-to-one correspondence between in-termediate elds E and subgroups H of G given by F 12 E E G.Furthermore, if Q L M F 12 then [M : L] = [L : M ] and, inparticular,

|G| = [Q() : Q] = (12) = 4 .Any G is completely and uniquely determined by the image

() = i for some i {0, . . . , 11}. Now 1() = j similarly, andsince 1 = 1 F 12 = 1 it follows that i is a unit in Z12. The functiondened by

G i Z12is evidently a monomorhpism of groups, which places G in isomorphismwith a subgroup of Z12. But the group of units modulo 12 has order

(12) = 4, which is exactly the order of G, so it must be the case that

G Z12. Note that each element of Z12 has order 2, so thatZ12 Z2 Z2.

The only proper subgroups of the latter are {0} Z2 and Z2 {0},hence the only intermediate elds are those given by the Galois corre-spondence, which I do not care to write down in this document.

2004v2 C4.Construct a eld with 49 elements and give the rules for its addition

and multiplication. If a is a generator, what is the multiplicative inverse of 1 + a in terms of your set of minimal generators?

Proof. The ground will be K = Z7. Consider f (x) = x2 3K [x]. Since degf = 2 it suffices to observe that f has no roots in K to conclude that f is irreducible over K . Thus in K [x] the ideal (f ) ismaximal, and so

K [x](f )

= {r0 + r1x + ( f ) : r i K }


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PART C 107

is a eld with 49 elements. Note also that this eld is isomorphic toF =

{r 0 + r 1a : r i K, a 2

3 = 0

}.

In F , multiplication and addition is done according to the rule a2 = 3.The extension I will use is K F .

After experimenting with a few cases, one is led to(1 + a)(3 + 4 a) = 3 + 4 a + 3 a + 4 a2 = 3 + 12 + (4 + 3) a = 1 .


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109/217

2003 Qual

Part A

2003 A1. A group cannot be the union of two proper subgroups.

Proof. Let G be a group and suppose H,K < G satisfy H K =G. Choose k G H , ( k K ). For any h H , the elementhk / H . As H K = G, it must be that hk K . But k1 K , so(hk)k1 = h K . This shows G K , so G = K .

2003 A2 unnished.

2003 A3. Let G1 and G2 be two non-trivial non-isomorphic simple groups. Prove that any proper trivial normal subgroup of G1 G2coincides with G1 or G2.

Lemma. If : G H is a group homomorphism and N H ,then 1(N ) G.Proof. The pullback of N is dened as

1(N ) = {a G : (i) N }.Let a 1(N ) and g G. Then

(gag1) = (g)(a)(g1) = (g)(a)(g)1 N so gag1 1(N ). As a, g were chosen arbitrarily it follows that

g1(N )g1 1(N )for all g G, which is what it means for 1(N ) G.

Proof of 2003 A3. Let N G1 G2. There are canonical injec-tions (monomorphisms of groups) i : Gi G1G2 for i = 1 , 2. By thelemma, 1i (N ) Gi for each i. As Gi are simple and non-isomorphic,exactly one of the following cases holds:

(1) 11 (N ) = G1 and 12 (N ) = {eG2}(2) 11 (N ) = {eG1}and 12 (N ) = G2109


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110 2003 QUAL

Since G1 G2 G2 G1, assume without loss of generality that (1)is true. ThenN = 111 (N ) = 1(G1) = G1 {eG2} G1.

2003 A4. A free abelian group is either trivial or isomorphic to Z .

Definition. A group G is free on a set X when there is an injectivemap of sets : X A and for any group H and map of sets f : X H there exists a unique group homomorphism f : G H such thatf = f .

G!f

A A A A A

A A

X

O O

f / /H

Proof. Let A be abelian and free on X . If X = then A is trivialand there is nothing to prove.

If X = {x}is a singleton, then every reduced word in A has theform xn for some n Z . Dene a map A xn n Z and verifythat it is an isomorphism of groups.If |X | 2, then there exist distinct elements x, y X . The groupA contains the reduced word

1 = xyx1y1 Aand it follows that

xy = yx

whence A is not abelian. This is a contradiction to the hypothesis,so if A is to be abelian then |X | {0, 1}, i.e. A is either trivial orisomorphic to Z .

2003 A5. Describe up to isomorphism all groups of order 121.Prove your answer.

Solution. Since 121 = 112 and 11 is prime, G is abelian. By theFundamental Theorem of Finite Abelian Groups,

G A {Z121 , Z11 Z11}For the proof that |G| = p2 implies G abelian, see [1995 A5].


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PART B 111

2003 A6. Prove that if S is a Sylow subgroup of a nite groupG,then N G (N G (S )) = N G (S ).

Proof. Note that S is a Sylow subgroup of any subgroup K < Gcontaining S . By Sylows second theorem, any conjugate of S is aSylow subgroup of G. If S is a conjugate of S contained in a subgroupK < G , then S is a Sylow subgroup of K as well. In particular, we arethinking of K = N G (S ); here S N G (S ) implies (by Sylows secondtheorem again) that S is the only Sylow subgroup of N G (S ).

Given x N G (N G (S )) it must be shown that x N G (S ). For thisit suffices to show that xSx 1 = S . By the denition of x

xN G (S )x1 = N G (S ).Since S < N G (S ) it follows that

xSx 1 < xN G (S )x1 = N G (S ).So xSx 1 is another Sylow subgroup of N G (S ), which necessitatesxSx 1 = S . Thus N G ((N G (S )) < N G (S ). The other containmentis obvious, hence equality holds.

Part B

2003 B1 unnished.

2003 B2 unnished.

2003 B3. Let R be a commutative ring with unique maximal idealM . Let A be the smallest subring of R containing the multiplicativeidentity 1 of R. Show that A is ring isomorphic to either Z or Z/p n Zfor some p, n N with p prime. (Hint: consider the idempotents of R.)

Proof. Dene : Z A by (1) = 1 R , and extend to be aring homomorphism. Because of As denition (Z) = A. The rstisomorphism theorem for rings yields Z / ker = A. If is injectivethen A = Z , and if is not injecive then since the only ideals of Zare mZ (m Z) we have A = Z /m Z . It remains to show that m = pnfor some prime p and some n

0. For simplicity let us identify A and

Z /m Z .Idea: let m = pn 11 pn rr . If r > 1 then ( p1), ( p2) are maximal idealsin A. End of rigor. Do something like this: there exist s1, . . . , s r Z

such that s1 p1 + + sr pr = 1. Therefore (not being careful) ( p1) + + ( pr ) = R. Whatever.

2003 B4 unnished.


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112 2003 QUAL

2003 B5. Let

0

A

B

C

0

be an exact sequence of R-modules.(a) What part of the sequence

0 Hom R (C, D )

Hom R (B, D )

Hom R (A, D ) 0is exact for every left R-module D?

Answer. The answer is the rst part

0 HomR (C, D )

HomR (B, D ).Given that im = C , it must be checked that if , : C D satisfy() = () then = . The new assumption can be reworded as = , i.e. for all b B we have (b) = (b). As bruns through all elements of B , im = C guarantees that (b) runsthrough all elements of C ; it follows that = .

(b) Show that the remaining part of the new sequence is exact if and only if the original sequence is split exact. The remaining part of thenew sequence is the end

HomR (B, D )

HomR (A, D ) 0.Given that is injective, it must be checked that for all f : A Dthere exists g : B D such that f = g . A priori there is no reasonwhy this should be true.Proof. Assume that the original sequence is split exact; in partic-

ular assume there exists : B A such that = 1 A . Given f asabove, observe that g = f does the job.Conversely, assume that0 HomR (C, D )

HomR (B, D )is exact for all R-modules D. Specialize to the case D = A and consider

f = 1 A .

2003 B6. Let K be a eld and let M and N be nitely generated modules over the polynomial ring K [x]. Suppose that N has invariant factors (x 2)(x 3)2, (x 2)(x 3)2(x 5) and M has invariant factors (x 1), (x 1)(x 2)2, (x 1)(x 2)2(x 3).


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PART C 113

(a) Give the elementary divisors of M N . Recall that K [x] is aprincipal ideal domain (if I is an ideal of K [x] choose f I of minimal

degree, then I = ( f ) using the division algorithm). The structuretheorem IV 6.12 indicates that

M K [x]

(x 1)K [x]

(x 1)(x 2)2K [x]

(x 1)(x 2)2(x 3)N

K [x](x 2)(x 3)2

K [x](x 2)(x 3)2(x 5)

.

Elementary divisors are the prime-power divisors of the invariant fac-tors. Accordingly the elementary divisors of M are

x

1, x

1, (x

2)2, x

1, (x

2)2, x

3

and those of N are

x 2, (x 3)2, x 2, (x 3)2, x 5.The elementary divisors of M N are given by concatenating theselists.

(b) Give the invariant factors of M N . Using the power gridon the elementary divisors gives

(x 2)(x

2)2(x

1)(x

3)

(x 2)2(x 1)(x 3)2(x 2)2(x 1)(x 3)2(x 5)

as the invariant factors of M N .

Part C

2003 C2. Let F be a eld extension of K .Dene the Galois group G of F over K .

Definition. The Galois group isG = Aut K F := Aut( F ) HomK (F, F ).

Now dene what one means by a stable subeld of this extension.

Definition. An intermediate eld F E K is stable wheneverim |E = E for all G.


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114 2003 QUAL

Prove that if E is a stable intermediate eld, then the Galois groupof F over E is a normal subgroup of G.

Proof. Certainly Aut E F G. Let G and Aut E F . Givenu E , since E is stable it is true that im |E = E = im 1|E . Bychoice |E = 1 E . So

1 (u) = 1(e ) = 1(e ) = e E.

Conversely, prove that if H is a normal subgroup of G, then the xed eld of H is a stable subeld of the extension F K .

Proof. For all H and any G there exists H suchthat = . Given

u H = {u F : (u) = u H }observe that (u) = (u) = (u)

so (u) H .

2003 C3. Let F K be elds.(i) Dene the maximal algebraic extension of K in F .

Definition. The maximal algebraic extension of K in F is the in-termediate eld K A F that is algebraic over K and not propertycontained in any other algebraic extension of K which is contained inF . In other words, if

K A E F and E is algebraic over K , then E = A.

Such an extension always exists by Zorns lemma.(ii) Suppose that for every extension F K , the maximal algebraic

extension of K in F is K . Prove that K is algebraically closed.

Proof. Let K be an algebraic closure of K . In particular K isalgebraic over K . In the extension K K , then, A = K . But A = K by hypothesis, so K = K is algebraically closed.

(iii) Now suppose K is algebraically closed and let F K . Prove that the maximal algebraic extension of K in F is K .

Lemma. K is an algebraic closure of K if and only if for all alge-braic extensions E K there exists a K -monomorphism E K .

This is proved after the proof of (iii) .


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PART C 115

Proof. Here the setup isK = K A F.

But, by the lemma, K is an algebraic closure of itself if and only if for all algebraic extensions E K there exists a K -automorphismE K . In particular there exists a K -monomorphism A K , and itfollows that A = K .

Proof of the lemma. ( ) K is algebraic over K by denition,suppose that E is also algebraic over K . If E is algebraically closed,then K K E and the proof halts. If not, then let E be an algebraicclosure of E . Since E E K and E K is algebraic it follows thatE is an algebraic closure of K . Thus there exists a K -isomorphism : E K K , which gives a K -monomorphism

|E : E

K .

( ) Assume K K is algebraic and let E be an algebraic closureof K . Note that E is an algebraic closure of K . By hypothesis thereis a K -monomorphism : E K , but since K E we have theinclusion map K E . It follows that K = E .


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117/217

2002 Qual

Part A

2002 A1. Show that if a positive integer d divides the order n of a nite cyclic group G, then G has a unique subgroup of order d.

Proof. Say G = a . Ill show that an/d is the one and onlysubgroup of order d. From undergraduate group theory

|an/d | =n

(n,n/d )=

nn/d

= d,

so | an/d | = d. Suppose H < G has order d. All subgroups of a cyclicgroup are cyclic, so H = am where m is the least positive integersuch that am H . Also note that m|n, so (m, n ) = m and

d = |am | = |a(m,n )| =n

(m, n )=

nm

which implies m = n/d . Thus H = an/d .

2002 A2. Can there be a simple group of order 200?

Proof. Let G be a group of order 200 = 23 52. Sylows thirdtheorem givesn2 1 (mod 2), n5 1 (mod 5)

and a well-known theorem indicates that

n2|25, n2|8.Accordingly, n5 = 1. Say P is the unique Sylow 5-subgroup of G.

Sylows second theorem states thatSyl5(G) = {xP x 1 : x G}

but it was just shown that xP x 1 = P for all x G. Hence P G,and the latter cannot be simple.

2002 A3.117


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118 2002 QUAL

(a) If H is a subgroup of G and [G : H ] = n, then H contains a normal subgroup K G and [G : K ] divides n!.

Proof. The group G acts on the set

G/H = {a iH : i = 1 , . . . , n }by left multiplication

g a iH = ( ga i)H = a j H.Dene a group homomorphism : G S n by

(i) = j (gai)H = a j H.

The kernel K = ker is a normal subgroup of G, and

g K i : (ga i)H = a iH.Taking a i = e gives g K implies gH = H , which is to say g H .Thus K < H , as desired.

By the rst isomorphism theorem

G / /

proj

S n

G/K / /im

the group G/K is isomorphic to im < S n . Therefore, by Lagrangestheorem,

[G : K ] = |G/K | = |im | n!(b) If G is nite and n = p is the least prime dividing |G|, then H is normal.Proof. Let : G S p and K = ker be as before, and denote[H : K ] = k so that

[G : K ] = [G : H ][H : K ] = pk.

Since G/K im , pk| p! and hence k|( p 1)!. All prime divisors of ( p 1)! are less than p, while all prime divisors of k are greater thanor equal to p (since p is the smallest prime dividing |G|). This forcesk = 1, and consequently H = K G.


119/217

PART B 119

2002 A4. Show that each nite p-group is solvable.

Definition. A sequence of subgroups

G = G0 > G 1 > > G t = eis a solvable series provided that

(1) Gi Gi+1(2) Gi /G i+1 is abelian.

Theorem (II 8.5). G is solvable if and only if G has a solvable series.

Proof. If n = 0 then G is trivial and there is nothing to prove.

For n 1, Sylows rst theorem gives subgroups Gi of G with order pi for i = 1 , . . . , n , each of which is normal in some subgroup of order pi+1 . Putting these together, there is a sequence of subgroups

e G1 G2 . . . Gn1 Gn .

Certainly Gn = G, since |Gn | = pn = |G|. It is also clear thatGi+1 /G i = p(i+1) i = p

whence the factors of this series are cyclic, hence abelian. Summarily,there exists a solvable series in G.

2002 A6. Let R be a commutative unitary ring. An ideal I or Ris said to be primary if a, b R and ab I imply a I or bn I forsome n N. Assume I is primary and S is a multiplicative subset of R such that S I = . Show that S 1I is a primary ideal of S 1R.

Proof. Let as bt S 1I . In other words abst = iu for some i I, uS . This means there exists v S such that v(abu ist ) = 0. That isto say (ab)(vu) I , and so either ab I or (vu)n I for some n (I is primary). Well if (vu)n I then ( vu)n I S which contradictssomething. So it must be the case that ab I . Again either a I orb

m

I for some m (I is primary). If a I thenas S

1

I . If bm

I then ( bt )m = b

m

tm S 1I , which completes the proof.

Part B

2002 B4.(i) Duplicates 2007 B1(a).


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120 2002 QUAL

(ii) True or false: If 0 A B C 0 is exact, then A R BA R C implies B C . The answer to this is an emphatic no. Forexample, Q Z Q Q Z Z but Q Z as abelian groups.

Proof that Q Z Q Q Z Z . The sequence

0 Z

Q

Q / Z 0is exact. Apply the left-exact covariant functor Z Q , and writefor Z to get a new exact sequence

0 Z Q 1Q

Q Q 1Q

Q/ Z Q .Facts:

(1) Z Q Q

This is true by virtue of the more general fact that R RM M . Dene a middle-linear map by (n,p/q ) np/q . Theuniversal property gives a unique homomorphism : Z Q Q agreeing with the middle-linear map on the simple tensors. has inverse : p/q 1 p/q , whence is injective. It isclearly surjective.

(2) Q/ Z Q 0All simple tensors do this

p/q + Z r/s = p/q + Z qr/qs= pq/q + Z r/qs

= 0 r/qs = 0 .Composing with the isomorphisms gives

0 Q( 1Q )Q Q

( 1Q )0,still exact. Now

im ( 1Q ) = ker( 1Q) = Q Q .So ( 1Q) : Q Q Q .Summarily

Q Z Q Q Q Z Z .

2002 B5. True or false?(i) A torsion-free module is projective. FALSE Q is torsion-free

(since it is a domain) as a Z-module, but it is not a projective Z-module.(ii) A projective module is torsion-free. FALSE Use an R that is

not a domain. As a module for itself, R = Z4 is free on {1}. ObviouslyR is not torsion-free since ann R (2) = (2) 0.


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PART C 121

(iii) A torsion-free module is free. FALSE As a Z-module, Q istorsion free but has no basis. This assertion is easy to show using least

common multiples and clever cancellation: no two elements of Q arelinearly independent over Z , and Q Z so any purported basis of Qmust have size at least 2.

2002 B6. True or false?(i) A free module is torsion-free. FALSE As a module for itself,

R = Z4 is free on {1}, but the annihilator of 2 R is nonzero.(ii) A projective module is free. FALSE Recall from undergrad-uate group theory that Z6 Z2 Z3. Since Z6 is free on {1}as amodule for itself it follows that Z2 is a projective Z6-module. But Z2has no basis over Z6 (easy to verify).

(iii) A free module is projective. TRUE Let P be free on X . Givena diagram

X / /P

A f / /B / /0

dene a map of sets g : X A by f (x) = a f 1((x)), well-denedbecause f is surjective. The universal property of free gives a uniquehomomorphism g : P A agreeing with g on X .

X / /g

P

!g ~ ~ ~ ~

~ ~

A f / /B / /0

f g(x) = (x) x X

That f g = is true by virtue of the setup.

Part C

2002 C1.(i) Prove that if f K [x] is a polynomial of degree n, then there

exists an extension of K in which f has a root.

Proof. Let p K [x] be an irreducible factor of f , say of degreer ; it suffices to construct an extension of K in which p has a root. Theideal ( p) is maximal in K [x], so the quotient

F =K [x]( p)

= a0 + a1x + . . . a r 1xr 1 + ( p) : a i K

is a eld. Identifying K with its image under the monomorphism a a +( p), it can be assumed that F is an extension of K . Let u = x +( p).


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122 2002 QUAL

Then, writing p(x) = ri=1 bixi , the ith term in the expression of p(u)

has the form

bii

j =1

i j

x i( p)i j = bixi + ( p).

Thus, p(u) = p(x) + ( p) = 0 + ( p).

(ii) Consider the example f (x) = x3 5x 2 Q[x], let u be a real root of this polynomial. What is the natural basis of Q(u) and write the element x4 3x + 1 as a linear combination of the basis elements.

Solution. As a vector space over Q ,Q(u) = {a0 + a1u + a2u2 : a i Q , f (u) = 0}.

Polynomial division givesx4 3x + 1 = xf (x) + 5 x2 x + 1 ,

so that in Q(u),u4 3u + 1 = 5 u2 u + 1 .

2002 C3.F is an algebraic closure of K if and only if F is algebraic over

K and for any algebraic eld extension E 1 of another eld K 1 and isomorphism of elds : K 1 K , extends to a monomorphism E 1 F .

Theorem. Let : A L be an isom

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