algebra soln

8/6/2019 Algebra Soln

1/11

1 LINEAR ALGEBRA

PH.D. COMPREHENSIVE EXAM

November 21, 2008

(Corrections and Additions: December 1, 2008)

1 Linear Algebra

3. Let P be a projection operator on the finite dimensional vector space V. (This means P2 = P.) Let X = ker P =nullspace P. Let Y = imP = image P. Define an isomorphism : V = X Y.

Solution 1: Let F be the field over which V is defined. Define (v) = (v P v , P v), for v V. Note thatv P v ker P, since P(v vP) = P v P2v = P v P v = 0. Therefore, is a map from V into X Y.Also, is a homomorphism: ( F)(v1, v2 V)

(v1 + v2) = ((v1 + v2) P(v1 + v2), P(v1 + v2))

= (v1 P v1, P v1) + (v2 P v2, P v2)

= (v1) + (v2).

Thus, Hom(V, X Y). Next, define : X Y V by (x, y) = x + y. Clearly Hom(X Y, V), and(v) = v P v + P v = v holds for all v V. Thus, = 1V . Also, for x ker P and y imP,

(x, y) = (x + y) = (x + y P(x + y), P(x + y))

= (x + y P x P y , P x + P y) = (x, y).

The last equality holds since x ker P, and since y imP implies P y = y. (Proof: y imP y = P v forsome v V P y = P2v = P v = y.) Therefore, = 1XY . This proves that is an isomorphism.

Solution 2: As above, define (v) = (v P v , P v), for v V, and note that Hom(V, X Y). Let F be thefield over which V is defined, and consider the following diagram of F[]-module homomorphisms:

0 X

VP

Y 01X

1Y

0 X1 X Y

2 Y 0

Obviously, the rows are exact. Also, if x X, then (x) = (x, 0) = 11X(x), so the first box is commutative. Ifv V, then 2(v) = 2(v P v , P v) = P v = 1YP v, so the second box (hence the diagram) is commutative.By the short-5 lemma, is an isomorphism.

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1 LINEAR ALGEBRA

4. Let V be a finite dimensional vector space. Let T be a linear transformation from V to V.

(a) Show that

V T V T2V

and that

0 ker T ker T2 .

(b) Let V1 =i=1 TiV, and let V2 =

i=1 ker T

i. Show T V1 V1 and T V2 V2.

(c) Show V = V1 + V2 and V1 V2 = 0.

(d) Let Ti be the restriction ofT to Vi. Show that T1 is invertible and T2 is nilpotent.

Solution: (a) First note that T End(V) implies T V V. Suppose TnV Tn1V, and let v Tn+1V. Thenv = Tn+1u for some u V, so v = Tn(T u) = Tnu, where u = T u V (since T V V). Therefore,v TnV. This proves that Tn+1V TnV and, by induction, V T V T2V .

Since T is linear, 0 ker T. (Proof: T(v) = T(v + 0) = T(v) + T(0) T(0) = 0.) Suppose ker Tn1 ker Tn, and let v ker Tn. Then v ker Tn+1, since Tn+1(v) = T Tn(v) = T(0) = 0. This proves that

ker T

n

ker T

n+1

and, by induction, 0 ker T ker T

2

.(b) Let v V1 =

i=1 TiV. Fix n N. I claim that T v TnV. Indeed, v V1 implies v = T

n1u for someu V. Therefore, T v = Tnu TnV. Since n N was arbitrary, this shows that T V1 V1.

Let v V2 =

i=1 ker Ti. If v ker T, then T v = 0 V2. If v ker T

n for some n > 1, then0 = Tnv = Tn1T v. Thus, T v ker Tn1 V2. Therefore, T V2 V2.

(c) By part (a), 0 ker T ker T2 . I claim that, if ker Tn = ker Tn+1 for some n N, then ker Tn =ker Tn+j for all j N. For, suppose ker Tn = ker Tn+1 for some n N, and v ker Tn+j . Then0 = Tn+jv = Tn+1Tj1v implies Tj1v ker Tn+1 = ker Tn. This holds if and only if

TnTj1v = 0 Tn+1Tj2v = 0 Tj2v ker Tn+1 = ker Tn.

Continuing this way, after a finite number of steps, we reach v = Tjjv ker Tn, which proves ker Tn+j ker Tn. Combining this with part (a) gives ker Tn+j = ker Tn, as claimed.

Now, since V is finite dimensional and 0 ker T ker T2 , it is clear that ker Tn = ker Tn+1 musthold for some n N, and then, by the foregoing, ker Tn = ker Tn+1 = ker Tn+2 = . Let N be thesmallest positive integer such that ker TN = ker TN+1. Then V2 =

i=1 ker Ti = ker TN, and therefore,

V1 =

i=1 TiV = TNV.

Two ways to complete the proof are as follows:

(i) Ifv TNV ker TN, then TNv = 0 and v = TNu for some u V. Therefore, 0 = TNv = T2Nu,which implies u ker T2N = ker TN, so v = TNu = 0. This shows TNV+ker TN = TNVker TN.It remains to show V = TNV + ker TN. But this follows from dim(V) = dim(imf) + dim(ker f),which holds for any f End(V), in particular, f = TN.

(ii) Let F be the field over which V is defined. Then V (as well as any subspace of V) is a free (henceprojective) F[x]-module. In particular, TNV is projective, so the short exact sequence

0 ker TN

VTN

TNV 0

is split exact. In particular, V = ker TN TNV = V2 V1.(d) Note that ker T1 = {v V1 : T v = 0} = V1 ker T, and ker T V2. Then, since V1 V2 = 0, ker T1 = 0.

Therefore, T1 is invertible.

Let N be the positive integer found in part (c), so that V2 = ker TN. For v V2, T2v = T v. Thus, T V2 V2

implies T2V2 = T V2 V2, so T22 v = T

2v for all v V2. That is, T22 V2 = T

2V2. Inducting on k,Tk2 V2 = T

kV2, for all k N. In particular, TN2 V2 = T

NV2 = 0. Therefore, T2 is nilpotent.

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2 RING THEORY


November 21, 2008


2 Ring Theory

All rings are unitary, i.e., have an identity.

1. List the ideals of the rings:

(a) Z/12Z

(b) M2(R) (the ring of2 2 real matrices)

(c) Z/12Z M2(R).

Solution: (a) An ideal is, in particular, a subgroup of the additive group. Since Z12; +, 0 is cyclic, all

subgroups are cyclic. Thus, the subgroups of Z12; +, 0 are as follows: (0) = {0}

(1) = {0, 1, . . . , 11}

(2) = {0, 2, 4, 6, 8, 10}

(3) = {0, 3, 6, 9}

(4) = {0, 4, 8}

(6) = {0, 6}

Of course, 5, 7, and 11 are relatively prime to 12, so (1) = (5) = (7) = (11) = Z12. The remaininggenerators, 8, 4, and 9, are also redundant: (8) = (4), (9) = (3), and (10) = (2). It is easy to verify thateach of the 6 distinct subgroups listed above call them {Gi : 1 i 6} have the property that, ifr Z12 and a Gi, then ra Gi. Thus, each Gi, 1 i 6, is an ideal ofZ12.

(Remark: This also shows Z12 is a PID, hence the proper prime ideals are the maximal ideals, (2) and (3).)

(b) By the following lemma, since R is a field (and thus has no non-trivial proper ideals), the only ideals ofM2(R) are M2(0) and M2(R).

Lemma 2.1 Let R be a commutative ring with 1R = 0. Then the ideals ofM2(R) are precisely the subsetsM2(J) M2(R), where J is an ideal ofR.

Proof: Suppose J is an ideal and A1, A2 M2(J). Then A1 A2 has all elements in J, since J is, in particular, asubgroup. Therefore M2(J) is a subgroup ofM2(R); +,0. IfM= (mij) M2(R) and A = (aij) M2(J), thenMA = (

P2k=1mikakj). Clearly, all the elements ofMA are in J, since J is an ideal and aij J. Similarly, all the

elements ofAMare in J. This proves that M2(J) is an ideal ofM2(R) whenever J is an ideal ofR.

Now, let I M2(R) be an ideal ofM2(R), and let J be the set of all a R such that a is an element of some

A I. Clearly 0 J, since A I implies 0 = A A I. Let a, b J. Suppose a is the ijth entry ofA I and b is the klth entry ofB I. Let Mij be a matrix with 1 in the ij

th position and 0 elsewhere. Then

M1iAMj1 =

a 00 0

I and M1kBMl1 =

b 00 0

I. Therefore,

a b 0

0 0

I, so a b J, whence,

J is a subgroup. Fix r R. Then

r 00 0

a 00 0

=

ra 00 0

I, so ra J, and J is an ideal.

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2 RING THEORY

(c) By the following lemma, and parts (a) and (b), the ideals ofZ/12Z M2(R) are all sets of the form I J,where I {(0), (1), (2), (3), (4), (6)} and J {M2(0), M2(R)}.

Lemma 2.2 Let R and R

rings. Then the ideals ofR R

have the form I J, where I is an ideal ofRand J is an ideal ofR.

Proof: Let A R R be an ideal. I will show (i) A = A1 A2, for some A1 R and A2 R, and (ii) A1 and

A2 are ideals ofR and R, respectively.

(i) Define

A1 = {a R : (a, a) A for some a R} and A2 = {a

R : (a, a) A for some a R}.

That is, A1 (resp. A2) is the set of all first (resp. second) coordinates of elements in A. I claim that A = A1 A2.1

Fix a A1 and b A2. We show (a, b

) A. Since a A1, there is some a A2 such that (a, a

) A. Similarly,(b, b) A, for some b A1. Since A is an ideal, (1, 0) (a, a

) = (a, 0) A, and (0, 1) (b, b) = (0, b) A.Therefore, (a, 0) + (0, b) = (a, b) A, as claimed. This proves A1 A2 A. The reverse inclusion is obvious.

(ii) Fix a, b A1. Then, there exist a, b A2 such that (a, a

) A and (b, b) A. Now, since A is, in particular, a

subgroup of the additive group ofR R, we have (a, a) (b, b) = (a b, a b) A. Therefore, a b A1,so A1 is a subgroup ofR,+, 0. Fix r R. Then (ra,a

) = (ra, 1a) = (r,1) (a, a) A, since A is an ideal ofRR. Therefore, ra A1, which proves that A1 is an ideal ofR. The same argument, mutatis mutandis, proves thatA2 is an ideal ofR

. This completes the proof of (ii) and establishes the lemma.

1Although this may seem trivial, and to some extent it is trivial, there is something to verify here, since we could have (a, a) A, (b, b) A,so that (a, b) A1 A2, and yet (a, b) / A. That is, in general, A need not be equal to the product of all first and second coordinates ofelements ofA. (Consider, for example, the diagonal {(0, 0), (1, 1), (2, 2)}, which cannot be written as A1 A2.)

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2 RING THEORY

3. Let R be a commutative ring and let A,B, and C be R-modules with A a submodule ofB. Which (if any) of thefollowing conditions guarantee that the natural map A R C B R C is injective?

(a) C is free,(b) C is projective.

Solution: I will prove a slightly more general result (as the increased generality comes at no additional cost).

That is, if : A B is an R-module monomorphism, then, under conditions (a) or (b), the natural map 1C :A R C B R C in injective. (To answer the question above, take to be the inclusion map.)

We begin with the simplest case, where C = R.

Claim 1: If : A B is an R-module monomorphism, then the natural map 1R : A R R B R R isinjective.

Proof 1: Define A : A R R = A by A(a r) = ar, and B : B R R = B by B(b r) = br. (Seelemma 2.4 below for proof that these are, indeed, isomorphisms.) I claim that the following diagram ofR-module

homomorphisms is commutative:A R R

A A

1R

B R RB B

Fix a A and r R. By definition, ( 1R)(a r) = (a) 1R(r). Therefore, (B ( 1R))(a r) =B((a) r) = (a)r. Following the diagram in the other direction, ( A)(a r) = (ar) = (a)r, since isan R-module homomorphism. Therefore, (B ( 1R))(a r) = ( A)(a r). Since a A and r R werechosen arbitrarily, B ( 1R) and A agree on generators, so the diagram is commutative.

Now A is an isomorphism, so A and are both injective. Therefore, by lemma 2.6 (below), A is injective.Then, by commutativity of the diagram, B ( 1R) is injective. Finally, lemma 2.8 implies that 1R isinjective, which proves claim 1.

Next, let C be a free R-module. Then C =iIR, for some index set I. For ease of notation, in the sequel,

denotesiI.

Claim 2: If : A B is an R-module monomorphism, then the natural map 1PR : AR

R B R

Ris injective.

Proof 2: Define :

A

B by ({ai}) = {(ai)} for each {ai}

A. Then is an R-modulemonomorphism. This is seen as follows: ({ai}, {a

i}

A) (r R)

({ai} + {a

i}) = ({ai+ a

i}) = {(ai+ a

i)} = {(ai) + (a

i)} = {(ai)} + {(a

i)} = ({ai}) +({a

i}),

and (r{ai}) = ({rai}) = {(rai)} = {r(ai)} = r{(ai)} = r({ai}).

If({ai}) = {0}

B, then {(ai)} = 0, which implies (ai) = 0 for all i I, and therefore, {ai} = 0 A. Thus, is a monomorphism.

By (the proof of) lemma 2.4, the map :

A

(A R R) given by ({ai}) = {ai 1R} is an isomorphism.By (the proof of) lemma 2.5, there is an isomorphism :

(A R R) A R

R, defined on generators by

(a r, 0, . . . ) = a (r, 0, . . . ), (0, a r, 0, . . . ) = a (0, r, 0, . . . ), . . . , and

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2 RING THEORY

(a1 r1, a2 r2, . . . ) = ((a1 r1, 0, . . . ) + (0, a2 r2, 0, . . . ) + )

= (a1 r1, 0, . . . ) + (0, a2 r2, 0, . . . ) +

= a1 (r1, 0, . . . ) + a2 (0, r2, 0, . . . ) + .

(The sum has finitely many non-zero terms, since {ai} is non-zero for finitely many indices i I.)

Consider the following diagram ofR-module homomorphisms:

A

(A R R)

A R

R

1PR

B

(B R R)

B R

R

(1)

where and are the same maps as their bar-less counterparts, but defined on

B and

(B RR), respectively.I claim that diagram (1) is commutative. Indeed, for any {ai}

A, we have2

({ai}) = ({ai 1R}) = ((a1 1R, 0, . . . ) + (0, a2 1R, 0, . . . ) + )

= a1 (1R, 0, . . . ) + a2 (0, 1R, 0, . . . ) + .

( 1PR) ({ai}) = ( 1PR)(a1 (1R, 0, . . . ) + a2 (0, 1R, 0, . . . ) + )

= ( 1PR)(a1 (1R, 0, . . . )) + ( 1PR)(a2 (0, 1R, 0, . . . )) +

= (a1) (1R, 0, . . . ) + (a2) (0, 1R, 0, . . . ) + .

In the other direction, ({ai}) = ({(ai)}) = {(ai) 1R}, so

({ai}) = ({(ai) 1R}) = (((a1) 1R, 0, . . . ) + (0, (a2) 1R, 0, . . . ) + )

= (a1) (1R, 0, . . . ) + (a2) (0, 1R, 0, . . . ) + ,

which proves that and ( 1PR) agree on generators. Therefore, diagram (1) is commutative.

Now, and are R-module isomorphisms, and is an R-module monomorphism. Therefore, by lemma 2.6

below, is an R-module monomorphism, so commutativity of (1) implies that 1PR is also an R-

module monomorphism. Finally, since and are isomorphisms, so is . Whence, 1PR is injective, bylemma 2.7. This which proves claim 2 and completes part (a) of the problem.

Claim 3: If C is a projective R-module and : A B is an R-module monomorphism, then the natural map 1C : A R C B R C is injective.

Proof 3: Since Cis projective, there exists a free R-module F and a (projective) R-module D such that F = CD(lemma 2.3). Therefore,

A R F = A R (C D) = (A RC) (A RD) and B R F = B R (C D) = (B R C) (B RD),

where the isomorphisms are given by lemma 2.5.

By claim 2 above, the natural map 1F : A R F B R F is injective. Consider the diagram

A R C1 (A R C) (A R D)

A R (C D)

1C

1F

B R C1 (B R C) (B R D)

B R (C D)

(2)

2Again, the sums have finitely many non-zero terms, since {ai} is non-zero for finitely many indices i I. The same comment applies to thesums in the following two sets of equations.

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2 RING THEORY

By (the proof of) lemma 2.5 the map : (A R C) (A RD) A R (C D) given by (a1 c, a2 d) =a1 (c, 0) + a2 (0, d) is an R-module isomorphism. Given a c A R C, then,

( 1F) 1(a c) = ( 1F) (a c, 0) = ( 1F)(a (c, 0)) = (a) (c, 0).

Similarly, there exists : (B RC)(B RD) = B R (CD), with (b1c, b2d) = b1(c, 0)+ b2(0, d),and

1 ( 1C)(a c) = 1 ((a) c) = ((a) c, 0) = (a) (c, 0).

Therefore, ( 1F) 1 and 1 ( 1C) agree on generators ofA R C, so diagram (2) is commutative.

By lemma 2.6, ( 1F) 1 is injective, so, by commutativity, 1 ( 1C) is injective. It now follows fromlemma 2.8 that 1C is injective, which completes the proof of claim 3 and part (b) of the problem.

The following six lemmas are used in the answer to problem 3 given above. The first is a standard theorem about

projective modules, the proof of which is not hard, and can be found, e.g., in Hungerford. The next two lemmas

(2.4 and 2.5) are also standard, but I havent seen them proved in detail elsewhere and, as the solution given above

makes repeated use of the maps defined in proving these lemmas, I include detailed proofs below. The last threelemmas are trivial verifications.

Lemma 2.3 Let R be a ring. The following conditions on an R-module P are equivalent:

(i) P is projective;

(ii) every short exact sequence ofR-modules 0 Af

Bg

P 0 is split exact (hence B = A P);

(iii) there is a free R-module F and an R-module N such that F = N P.

Lemma 2.4 IfR is a commutative ring with 1R and A is a unitary R-module, then A R R = A.

Proof: Define : A R A by (a, r) = ar. Since is clearly bilinear, there exists a unique R-module homomorphism

: A R R A such that = , where : A R A R R is the canonical bilinear map. Define : A A R R by(a) = a 1R. Then it is easy to verify that is an R-module homomorphism and that

(a r) = (ar) = ar 1R = a r (a A)(r R).

Therefore, is the identity on generators ofA R R. Also, (a) = (a 1R) = a1R = a, so = 1A. Therefore, : A R R = A.

Lemma 2.5 Let R be a commutative ring with 1R and let A , B, Cbe unitary R-modules. Then A R (B C) =(A R B) (A R C).

Proof: Recall that if1 : A1 D and 2 : A2 D are (group) homomorphisms, then there is a unique homomorphism : A1 A2 D such that i = i (i = 1, 2), where i : Ai A1 A2 are the canonical injections. In other words,! Hom(A1 A2, D) such that the following diagram is commutative:

A1

1$$I

IIII

IIII

1 // A1 A2

A22oo

2zzuu

uuuuuuu

D

We can apply this theorem to the Abelian group (A R B) (A R C). Let 1 : A B A R (B C) be given by1(a, b) = a (b, 0), and let 2 : AC AR (B C) be given by 2(a, c) = a (0, c). It is easily verified that i are

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2 RING THEORY

bilinear and thus induce R-module homomorphisms 1 : A R B A R (B C) and 2 : A R C A R (B C)such that ii = i, where 1 : A B A R B and 2 : A C A R Care the canonical bilinear maps. Therefore,the two universal properties combine to give a unique homomorphism : (ARB) (ARC) AR (B C) such that

the following diagram is commutative:

A R B

1 ((QQQQ

QQQQ

QQQQ

Q

1 // (A R B) (A R C)

A R C

2vvmmmmm

mmmm

mmmm

2oo

A B

1

OO

1// A R (B C) A B

2

OO

2oo

Now, for all a A and b B, 1(a b) = (a b, 0) and 1(a b) = 11(a, b) = 1(a, b) = a (b,0), so,(a b, 0) = a (b,0). Similarly, (0, a c) = a (0, c), for all a A and c C. Therefore,

(a b, a c) = (a b,0) + (0, a c) = a (b, 0) + a (0, c) (a, a A, b B, c C).

Next, define : A (B C) (A R B) (A R C) by (a, (b, c)) = (a b, a c). Again, is bilinear, so there is aunique : AR (BC) (BC) (ARB) (ARC) such that = where : A (BC) AR (BC)is canonical. To complete the proof, we check that and given the appropriate identity maps:

(a b,0) = (a (b,0)) = (a b, a 0) = (a b,0), and

(0, a c) = (a (0, c)) = (a 0, a c) = (0, a c).

Thus, is the identity on generators of(A R B) (A R C), so = 1(ARB)(ARC). Finally,

(a (b, c)) = (a b, a c) = ((a b,0) + (0, a c)) = a (b, 0) + a (0, c) = a (b, c).

Thus is the identity on generators ofAR (BC), so = 1AR(BC). This proves that : (ARB) (ARC)=

A R (B C).

Lemma 2.6 Ifh HomR(A, B), g HomR(B, C), and f HomR(C, D) are injective, then f gh HomR(A, D)is injective.

Proof: fgh(a) = 0 gh(a) = 0 (since f is injective) h(a) = 0 (since g is injective) a = 0 (since h is injective).

Lemma 2.7 Ifg HomR(A, B), f HomR(B, C), ifg is surjective, and iff g is injective, then f is injective.

Proof: Suppose f(b) = 0. Since g is surjective, there is an a A such that g(a) = b. Then fg(a) = f(b) = 0, which impliesa = 0, since fg is injective. Therefore, b = g(a) = g(0) = 0 since g is a homomorphism.

Lemma 2.8 Ifg HomR(A, B), f HomR(B, C), and iff g is injective, then g is injective.

Proof: kerg = {a A : g(a) = 0} {a A : fg(a) = 0} = ker fg = {0}. The set containment holds since f is ahomomorphism, so f(0) = 0.

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3 GROUP S


November 21, 2008


3 Groups

4. Let G be a finite group, all of whose Sylow subgroups are normal. Prove that G is the product of its Sylowsubgroups.

Solution:3 Let P1, . . . , P r be the Sylow subgroups of G with orders |Pi| = peii , where p1, . . . , pr are distinct

primes and e1, . . . , er are positive integers. Since Pi G for each 1 i r, P1 Pr is a subgroup of G. Ifa Pi has order o(a), then o(a)|p

eii , by Lagranges theorem. Since p1, . . . , pr are distinct primes, o(a) p

ejj , for

j = i. Therefore, again by Lagranges theorem, a / P1 Pi1Pi+1 Pr. This proves that, for each 1 i r,Pi (P1 Pi1Pi+1 Pr) = (e). Two facts almost

4 immediately:

(i) P1 Pr = P1 Pr, and

(ii) |P1 Pr| = |P1| |Pr| = pe11 p

err .

Finally, P1 Pr < G and |P1 Pr| = pe11 p

err = |G| together imply that G = P1 Pr. Therefore, by (i),

G = P1 Pr.

3I believe my original solution to this problem was essentially correct, as the problem asks for proof that G is the product of its Sylow subgroups.Nonetheless, here it is again, with a minor addition: we can conclude that G is (isomorphic to) the direct product of its Sylow subgroups.

4See Theorem 8.6 and Corollary 8.7 of Hungerford, p. 61, for the proof of (i).

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4 F IE LDS


November 21, 2008


4 Fields

2. Let F be a field with pn elements (p a rational prime). Prove that F is a Galois extension of its prime subfield withcyclic Galois group. (You may assume any result not specific to finite fields.)

Solution 1: Consider f(x) = xpn

x Zp[x]. Since F is a field with pn elements, the multiplicative groupF = F\ {0} is cyclic of orderpn 1. By Lagranges theorem, then, ap

n1 = 1 holds for all a F. Therefore,

for all a F (including 0), apn

= a. That is, the pn elements ofF give pn distinct roots off(x). Since f(x) hasat most pn roots, these are all the roots, so F is a splitting field for f(x) over Zp. Since the roots are all distinct,f(x) is separable. This proves that F/Zp is a finite Galois extension.

To see that Gal(F/Zp) is cyclic, consider the map : F F given by (a) = ap. Since F has characteristic

p, (a b) = (a b)p

= ap

bp

. Also, (ab) = (ab)p

= ap

bp

. Thus, End(F). Fix a F. Thenb = ap

n1

F, so (b) = (apn1

)p = apn

= a, and is surjective. Also, a ker = {a F : ap = 0} implies

a = apn

= (ap)pn1

= 0, so ker = {0}, and is injective. Therefore, Aut(F). Finally, (r) = rp = r, forr Zp. To see this, note that Zp is a cyclic group of order p 1. Therefore, r

p1 = 1, for all r Zp , so rp = r,

for all r Zp. This proves that Gal(F/Zp).

If a F, then n(a) = n1(ap) = n2(ap2

) = = apn

= a. Therefore, n = 1F. Also, if k = 1F

for some 1 k < n, then apk

= a for all a F, which says that the polynomial xpk

x has pn roots,where pk < pn (nonsense). Therefore, n = 1F, and n is the smallest positive integer with this property, so() = {, 2, . . . , n = 1F} is a set ofn distinct elements in Gal(F/Zp). Finally, n = [F : Zp] = |Gal(F/Zp)|,since F/Zp is Galois. This proves that Gal(F/Zp) is cyclic with generator : a a

p.

Solution 2: Define(a) = a

n

.As above,

Gal

(F/Zp),and

()is a subgroup of order

n. By the fundamental

theorem of Galois theory, [F : Inv()] = n and Gal(F/Inv()) = (). On the other hand, Gal(F/Zp)implies Zp Inv(). Since n = [F : Zp] = [F : Inv()][Inv() : Zp] = n[Inv() : Zp], we have [Inv() :Zp] = 1, so Inv() = Zp. Equivalently, by the theorem on Galois extensions,5 F/Zp is Galois, with Galois groupGal(F/Zp) = ().

5See page 4 of Math 612 Notes: Galois Theory, William Lampe, July 2008.

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4 F IE LDS

3. Let F be a field of characteristic zero such that every finite extension of F has even degree. Show that

(a) every finite extension ofF has degree a power of 2,

(b) all proper minimal finite extensions ofF are quadratic.

(Hint: Sylow theory can be applied to Galois groups of normal extensions of F.)

Solution: (a) Suppose K is a finite extension of F. Let K be the normal closure of K over F. Since F hascharacteristic zero, K is separable and therefore Galois with |Gal(K/F)| = [K : F], which is even, byassumption. Thus, without loss of generality, we can assume |Gal(K/F)| = 2km, for some positive integersk and m, with m odd. Let G = Gal(K/F). By the first Sylow theorem, G has a Sylow 2-subgroup, H, oforder |H| = 2k. Therefore, by the fundamental theorem of Galois theory, [K : InvH] = |H| = 2k, whereInvH = {a K : (a) = a for all H}. Thus 2km = [K : F] = [K : InvH][InvH : F], whichimplies [InvH : F] = m, an odd integer. If InvH is a (proper) finite extension of F, then, by assumption,it must have even degree. The only way out is m = [InvH : F] = 1, which gives [K : F] = 2k. Finally,[K : F] = [K : K][K : F] implies [K : F] | 2k. Thus , [K : F] is a power of 2.

(b) Let E be a proper finite extension of F, so that [E : F] > 1. By part (a), [E : F] = 2k for some k 1. If[E : F] = 2, I claim that there is a proper intermediate field F < K < Esuch that 1 < [K : F] < [E : F].(This will establish that, if [E : F] = 2, then E is not a proper minimal finite extension of F, which willcomplete the proof.)

Suppose [E : F] = 2. Then [E : F] = 2k with k > 1. Let Ebe the normal closure ofE over F. Then, by part(a), |G| = |Gal(E/F)| = 2K = [E : F], where 1 < k K < . Consider the subgroup

FixE = { G : (a) = a for all a E}.

As a subgroup of G, |FixE| = 2j for some 0 j K. Also, |G|/|FixE| = [G : FixE] = [E : F] > 2, soFixE is a proper subgroup ofG.

Next, consider NG(FixE), and recall the following standard lemma about normalizers ofp-subgroups: ifH isa p-subgroup of a finite group G, then [NG(H) : H] [G : H] (mod p). By this lemma, we have

1 [NG(FixE) : FixE] [G : FixE] 0 (mod 2).

Therefore, 2 divides |NG(FixE)/FixE|, so NG(FixE)/FixE contains a subgroup of order 2, by Cauchystheorem. The third isomorphism theorem implies that this subgroup has the form J1/FixE, where FixEJ1 1, since any subgroup of order 2K1 is of index 2, hence normal, in

G.)

Finally, consider the subfield InvJr = {a E : (a) = a for all Jr}. Since Inv is order reversing on theposet Sub[G], , it is clear that FixE < Jr implies6 InvJr < InvFixE = E. Putting it all together, we have

1 < [InvJr : F] = [G : Jr] < [G : FixE] = [InvFixE : F] = [E : F].

That is, F < InvJr < E and 1 < [InvJr : F] < [E : F], as claimed.

6The equality InvFixE = E follows from the subgroup lemma of Professor Lampes 612 notes on Galois theory: If E/F is a finite Galoisextension and F < E < E, then E is Galois over E. Consequently, in the algebra of intermediate fields, E is closed (with respect to the closureoperator InvFix).

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