Appendix Homework and answers

Transcription

1 Appendix Homework and answers

2 Algebra II: Homework These are the weekly homework assignments and answers. Students were encouraged to work on them in groups. Some of the problems turned out to be more difficult than I intended. Homework. Homework (Abelian categories) 2. Homework 2 (Chain complexes) 3. Homework 3 (Ext ) 4. Homework 4 (Integral closure) 5. Homework 5 (Transcendence degree) 6. Homework 6 (Associated primes) 7. Homework 7 (Semisimplicity) 8. Homework 8 (Group rings) 9. Homework 9 (Character table) 0. Homework 0 (More characters) Answers to homework

3 MATH 0B: HOMEWORK. Homework 0 The following three problems are due next Thursday (/25/7)... Show that the category of finite abelian groups contains no nontrivial projective or injective objects. (Use the Fundamental Theorem: all finite abelian groups are direct sums of cyclic p-groups Z/p n.).2. Show that abelian categories have push-outs and pull-backs. I.e., Given an abelian category A and morphisms f : A C, g : B C there exists an object D (called the pull-back) with morphisms α : D A, β : D B forming a commuting square (f α = g β) so that for any other object X with maps to A, B forming another commuting square, there exists a unique morphism X D making a big commutative diagram. (Draw the diagram.) Hint: There is an exact sequence 0 D A B C.3. Let k be a field and let R be the polynomial ring R = k[x]. Let Q be the k vector space of all sequences: (a 0, a, a 2, ) where a i k with the action of X R given by shifting to the left: X(a 0, a, a 2, ) = (a, a 2, a 3, ) Then show that Q is injective. Hint: first prove that any homomorphism f : A Q is determined by its first coordinate. Date: January 8, 2007.

4 MATH 0B: HOMEWORK 2. Homework 02 The following three problems are due next Thursday (2/2/7). The strict deadline is :30pm Friday. 2.. Give a precise description of the ring R indicated below and show that it has the property that left R-modules are the same as chain complexes of abelian groups with three terms, i.e., d C 2 d 2 C C0 where C, C 2, C 3 are abelian groups and d, d 2 are homomorphisms so that d d 2 = 0 and a homomorphism of R-modules is a chain map f : C D, i.e. it consists of three homomomorphisms f i : C i D i so that the following diagram commutes. C 2 C C 0 f 2 f f 0 D 2 D D 0 R is a quotient ring: R = Z Z Z 0 Z Z / 0 0 Z Z [Hint: R is additively free abelian with five generators, three of which are idempotents e 0, e, e 2 and C i = e i C.] 2.2. Describe the chain complex corresponding to the free R-module R n (n finite) Assume without proof that the analogous statements hold for right R-modules, namely they are cochain complexes C 2 C C 0 where C i = Ce i (just as C i = e i C). Give a description (as a chain complex with three terms) of the injective R-module Hom Z (R n R, Q/Z). Date: January 25, 2007.

5 MATH 0B: HOMEWORK 3. Homework 03 The following problem is due next Thursday (2/8/7). deadline is :30pm Friday. The strict Compute Ext i Q[X](Q[X]/(f), Q[X]/(g)) using both the projective resolution of Q[X]/(f) and the injective coresolution of Q[X]/(g). [First take the projective resolution P of Q[X]/(f). Then Ext i Q[X](Q[X]/(f), Q[X]/(g)) = H i (Hom Q[X] (P, Q[X]/(g))) Then, find the injective (co)resolution Q of Q[X]/(g). The Ext groups can also be found by the formula Ext i Q[X](Q[X]/(f), Q[X]/(g)) = H i (Hom Q[X] (Q[X]/(f), Q )) The theorem that says that these two definitions are equivalent, we don t have time to prove. But working out this example might give you an idea on why it is true for i = 0,.] Date: February 2, 2007.

6 MATH 0B: HOMEWORK 4. Homework 04 The following problems are due Thursday (3//7). The strict deadline is :30pm Friday. () Show that unique factorization domains (UFDs) are integrally closed. (2) Show that the integral closure of Z[ 5] (in its fraction field) is Z[α] where α = (3) Combining these we see that Z[ 5] is not a UFD. Find a number which can be written in two ways as a product of irreducible elements. [Look at the proof of problem and see where it fails for the element α in problem 2.] (4) If E is a finite separable extension of K and α E show that the trace Tr E/K (α) is equal to the trace of the K linear endomorphism of E given by multiplication by α. [Show that the eigenvalues of this linear transformation are the Galois conjugates of α and each eigenvalue has the same multiplicity.] Date: February 5, 2007.

7 MATH 0B: HOMEWORK 5. Homework 05 The following problems are due Thursday (3/8/7). The strict deadline is :30pm Friday. () (Problems #3 on page 374) If L E are finitely generated field extension of K then show that the transcendence degree of E over K is equal to the product of the transcendence degree of E over L and the transcendence degree of L over K. (2) Let R = K[X, Y ]/(f) where f(x, Y ) = (X a)y 2 (X b) for some a b K. Find a transcendental element Z of R so that R is integral over K[Z]. [Use the proof of Noether Normalization.] Date: March 3, 2007.

8 MATH 0B: HOMEWORK 6. Homework 06 The following problems are due Thursday (3/22/7). The strict deadline is :30pm Friday. Assume R, M are both Noetherian. () Show that for any ideal I in R there are only finitely many minimal primes containing I. [Take a maximal counterexample.] (2) Suppose that p is a prime and n > 0. Let p (n) M := p n M p M Show that this is a p-primary submodule of M. (3) If φ : R S is a homomorphism of Noetherian rings and M is an S-module then show that ass R (M) = φ (ass S (M)) where φ : Spec(S) Spec(R) is the map induced by φ. Date: March 5, 2007.

9 MATH 0B: HOMEWORK 7. Homework 07 The following problems are due Thursday (4/2/7). The strict deadline is 3pm Friday. () Show that Z[/p]/Z is an Artinian Z-module. [Show that every proper subgroup is finite.] (2) Show that the center of the ring Mat n (R) is isomorphic to the center of R. [Show that the center Z(Mat n (R)) consists of the scalar multiples ai n of the identity matrix where a Z(R).] (3) Suppose that M is both Artinian and Noetherian. Then show that (a) r n M = 0 for some n > 0 (r n M := } rr {{ r} M) n (b) r i M/r i+ M is f.g. semisimple for all i 0. (4) Prove the converse, i.e., (a) and (b) imply that M is both Artinian and Noetherian. Date: April 4, 2007.

10 MATH 0B: HOMEWORK 8. Homework 08 The following problems are due Thursday (4/9/7). The strict deadline is :30pm Friday. () Let K be any field, let G be any finite group. Let V = K be the trivial representation of G. Let E = K[G] be the group ring considered as a representation (this is called the regular representation ) Find all G-homomorphisms f : V E (and show that you have a complete list). (2) If K is the field with two elements and G is the group with two elements then show that K[G] is not semisimple. [Using your answer to question, show that you have a short exact sequence 0 V E V 0 which does not split.] Date: April 5, 2007.

11 MATH 0B: HOMEWORK 09 one problem: 9. Homework 09 Calculate the character table of the quaternion group and find the 2-dim irreducible representation. Q = {, i, j, k, t, it, jt, kt} where t = i 2 = j 2 = k 2 is central and ij = k = jit, jk = i = kjt, ki = j = ikt.

12 MATH 0B: HOMEWORK 0. Homework 0 This is the last assignment. There is no final exam or any other work to do for this course! The following problems are due Thursday (5/3/7). The strict deadline is :30pm Friday. () If V, S are irreducible representations of G and S is one-dimensional, then show that V S is also irreducible. (2) Show that the irreducible representations of a product G H are the tensor products of irreducible representations of G, H. (3) Compute the character table of the alternating group A 4 and the induction-restriction table for the pair (S 4, A 4 ). Date: April 26, 2007.

13 Answers to homework

14 MATH 0B: HOMEWORK. Homework 0 The following three problems are due next Thursday (/25/7)... Show that the category of finite abelian groups contains no nontrivial projective or injective objects. (Use the Fundamental Theorem: all finite abelian groups are direct sums of cyclic p-groups Z/p n.) For any abelian group A, Hom(Z/n, A) is isomorphic to the set of all a A so that na = 0 with the isomorphism even by evaluating functions on a fixed generator of Z/n. If n = p k (p prime) and j 0 this gives Hom(Z/p k, Z/p j+k ) = p j Z/p j+k Z = Z/p k So, the short exact sequence (.) 0 Z/p k α Z/p 2k β Z/p k 0 gives the exact sequence 0 Hom(Z/p k, Z/p k ) α Hom(Z/p k, Z/p 2k ) β Hom(Z/p k, Z/p k ) which is isomorphic 0 Z/p k α Z/p k β Z/p k Therefore, α is an isomorphism and β = 0. This means that Z/p k is not projective for any k > 0. This implies that there are no nontrivial projectives in the category of finite abelian groups since any such projective P would have a direct summand of the form Z/p k for k > 0 by the fundamental theorem of finite abelian groups and any direct summand of a projective module is projective. (To see that suppose that P = X Y is projective and g : A B is an epimorphism. Then any morphism f : X B extends to a morphism f + 0 : P = X Y B which lifts to A. The restriction of the lifting to X gives a lifting of f making X projective.) Similarly, if we apply Hom(, Z/p k ) to the exact seqence (.) we get: 0 Hom(Z/p k, Z/p k ) β Hom(Z/p 2k, Z/p k ) α Hom(Z/p 2k, Z/p k ) Date: February 3, 2007.

15 2 MATH 0B: HOMEWORK which is isomorphic to 0 Z/p k β Z/p k α Z/p k So, α = 0 which implies that Z/p k is not injective for k > 0. Therefore there is no nontrivial injective object in the category of finite abelian groups..2. Show that abelian categories have push-outs and pull-backs. I.e., Given an abelian category A and morphisms f : A C, g : B C there exists an object D (called the pull-back) with morphisms α : D A, β : D B forming a commuting square (f α = g β) so that for any other object X with maps to A, B forming another commuting square, there exists a unique morphism X D making a big commutative diagram. (Draw the diagram.) Hint: There is an exact sequence 0 D A B C I will use the notation j A, j B for the inclusion maps of A, B into A B and p A, p B for the projections to A, B. Then j A p A + j B p B as part of the definition of a direct sum (or as a consequence as explained in the notes). Since A B is the coproduct of A and B, the morphisms f and g induce a morphism h = (f, g) : A B C so that h j A = f, h j B = g Let D A B be the kernel of this homomorphism. In other words D is the kernel in the exact sequence 0 D k A B h C Let α = p A k : D A and β = p B k : D B which is written k = ( α β). Then Claim. f α = g β : D C. In other words, f α g β = 0. This is a calculation: f α g β = (h j A ) (p A k) + (h j B ) (p B k) = h (j A p A + j B p B ) k = h k = 0 This calculation is often written in the matrix notation: ( ) α h k = (f, g) = f α g β β Claim 2. For any object X and any pair of morphisms α : X A, β : X B so that f α = g β there is a unique morphism γ : X D so that α = α γ and β = β γ.

16 MATH 0B: HOMEWORK 3 For existence, let φ = ( ) α β : X A B be the product of α and β. I.e., the unique morphism so that α = p A φ and β = p B φ. Then h φ = h (j A p A + j B p B ) φ = f α g β = 0 Therefore, there is a unique morphism γ : X D so that h γ = φ. Then α = p A φ = p A h γ = α γ and similarly, β = β γ. For uniqueness suppose that γ : X D is another morphism so that α = α γ and β = β γ. Then φ = h γ is another morphism so that p A φ = α and p B φ = β. By the universal property, φ = φ which implies that γ = γ. The conclusion is that D is the pull-back. If we reserve all the arrow in the above argument (or equivalently, repeat the same argument in the opposite category A op ) we see that the push-out of a diagram A f B is the cokernel of g C ( ) f : A B C g.3. Let k be a field and let R be the polynomial ring R = k[x]. Let Q be the k vector space of all sequences: (a 0, a, a 2, ) where a i k with the action of X R given by shifting to the left: X(a 0, a, a 2, ) = (a, a 2, a 3, ) Then show that Q is injective. Hint: first prove that any homomorphism f : A Q is determined by its first coordinate. Let p i : Q k be the projection to the i-th coordinate. This is linear map (i.e., a morphism of k-modules) and not a morphism of R-modules. Lemma.. For any R-module A there is an isomorphism φ : Hom R (A, Q) = Hom k (A, k) given by sending f : A Q to φ(f) = p f.

17 4 MATH 0B: HOMEWORK Proof. f can be reconstructed from g = φ(f) as follows. For any a A, let h(a) = (g(a), g(xa), g(x 2 a), g(x 3 a), ) If a is replaced by Xa then this entire sequence shifts to the left. Therefore h(xa) = Xh(a) making h into a homomorphism of R-modules h : A Q. To see that f = h look at the coordinates of f(a) f(a) = (f 0 (a), f (a), f 2 (a), ) g(a) is the first coordinate: g(a) = f 0 (a) and the other coordinates are given by shifting to the left and then taking the first coordinate: f n (a) = p 0 (X n f(a)) = p 0 (f(x n a) = g(x n a) = h n (a) Also, if we start with g and take h then φ(h) = g. So, φ if a bijection and thus an isomorphism. To show that Q is injective as an R-module suppose that (.2) 0 A α B β C 0 is a short exact sequence of R-modules. Then we want to show that induced sequence (.3) 0 Hom R (C, Q) β Hom R (B, Q) α Hom R (A, Q) 0 is exact. By composing with p 0 we get the sequence (.4) 0 Hom k (C, k) β Hom k (B, k) α Hom k (A, k) 0 which is exact since (.2) is a split exact sequence of k. Therefore the isomorphic sequence (.3) is also exact making Q into an injective R-module.

18 MATH 0B: HOMEWORK 2. Homework 02 The following three problems are due next Thursday (2/2/7). The strict deadline is :30pm Friday. 2.. Give a precise description of the ring R indicated below and show that it has the property that left R-modules are the same as chain complexes of abelian groups with three terms, i.e., d C 2 d 2 C C0 where C, C 2, C 3 are abelian groups and d, d 2 are homomorphisms so that d d 2 = 0 and a homomorphism of R-modules is a chain map f : C D, i.e. it consists of three homomomorphisms f i : C i D i so that the following diagram commutes. C 2 C C 0 f 2 f f 0 D 2 D D 0 R is a quotient ring: R = Z Z Z 0 Z Z / 0 0 Z Z [Hint: R is additively free abelian with five generators, three of which are idempotents e 0, e, e 2 and C i = e i C.] R is additively free abelian with five generators, e 0, e, e 2, α, β where the first three are diagonal matrices e 0 = diag(, 0, 0), e = diag(0,, 0), e 2 = diag(0, 0, ) and α is a matrix with a in the (0, )-entry and 0 s elsewhere and β is a matrix with in the (, 2)-entry and zeros elsewhere. Given an R-module C the corresponding chain complex is given as follows. Let C i = e i C. Since = e 0 + e + e 2 and e i are orthogonal idempotents in the sense that e i e j = δ ij e j, we get a decomposition C = e 0 C e C e 2 C = C 0 C C 2 Date: February 4, 2007.

19 2 MATH 0B: HOMEWORK Since α = e 0 αe, multiplication by α is zero on C 0 C 2 and has image in C 0. So it gives a homomorphism C C 0. Similarly, multiplication by β gives a homomorphism C 2 C. Since αβ = 0, the composition of these two homomorphisms is zero. Therefore, we have a chain complex structure on C. Conversely, suppose we have a chain complex C d 2 C d C 0. Then we have an action of R on C = C i by a 00 a a a 2 x 0 x = a 00x 0 + a 0 d(x ) a x + a 2 d(x 2 ) 0 0 a 22 x 2 a 22 x 2 where we write elements of C in column matrix form. We see that multiplication by e i is projection onto C i and multiplication by the matrix α sends (x 0, x, x 2 ) to (d(x ), 0, 0), i.e., it is projection to C followed by d : C C 0 followed by inclusion C 0 C. Similarly for β. So, the chain complex corresponding to this R-module is the original chain complex. Conversely, if we start with an R-module, form the chain complex and then make the chain complex back into an R-module by the above construction then we get back that same R-module. This is because d : C C 0 is α restricted to C and d 2 : C 2 C is β restricted to C Describe the chain complex corresponding to the free R-module R n (n finite). First take n =. Then R is free with 3 generators: e 0, e, e 2, α, β. Multiplying on the left with e 0 kills all but e 0, α. So, e 0 R = Z 2 with free generators e 0, α. Similarly, e R = Z 2 with free generators e, β and e 2 R = Z with free generator ɛ 2. So, the chain complex is: Z d 2 Z 2 d Z 2 As we already discussed, d 2 is left multiplication by β which sends the generator e 2 of C 2 = Z to the second generator β of C = Z 2. And d is left multiplication by α which sends the first generator e of C to the second generator α of C 0 = Z 2 and sends the second generator β of C to zero. If we take the direct sum of n copies of this we get the chain complex Z n d 2 Z 2n d Z 2n where d 2 is the inclusion into the first factor of Z 2n = Z n Z n and d is the projection to the second coordinate of C followed by inclusion

20 MATH 0B: HOMEWORK 3 into the second factor of C 0. In matrix notation: ( ) ( ) 0 0 In d 2 =, d = 0 0 I n 2.3. Assume without proof that the analogous statements hold for right R-modules, namely they are cochain complexes C 2 C C 0 where C i = Ce i (just as C i = e i C). Give a description (as a chain complex with three terms) of the injective R-module Hom Z (R n R, Q/Z). The right R-module R is the cochain complex C 2 = Re 2 = Z 2 with free generators e 2, β, C = Re = Z 2 with generators e, α and C 0 = Re 0 = Z with generator e0. d : C C 2, being right multiplication by β, is an isomorphism of the first factor of C to the second factor of C 2 and d 0 : C 0 C is an isomorphism of C 0 onto the second factor of C. If we hom into Q/Z we get (Q/Z) 2 d 2 (Q/Z) 2 d Q/Z Where d 2 is an isomorphism from the second factor of C 2 to the first factor of C and d is an isomorphism from the second factor of C to C 0. Hom Z (RR n, Q/Z) is a direct sum of n copies of this chain complex. So, it is (Q/Z) 2n d 2 (Q/Z) 2n d (Q/Z) n where ( ) 0 In d 2 =, d 0 0 = ( ) 0 I n

21 MATH 0B: HOMEWORK 3. Homework 03 answers The following problem was due Thursday (2/8/7). Compute Ext i Q[X](Q[X]/(f), Q[X]/(g)) using both the projective resolution of Q[X]/(f) and the injective coresolution of Q[X]/(g). [First take the projective resolution P of Q[X]/(f). Then Ext i Q[X](Q[X]/(f), Q[X]/(g)) = H i (Hom Q[X] (P, Q[X]/(g))) Then, find the injective (co)resolution Q of Q[X]/(g). The Ext groups can also be found by the formula Ext i Q[X](Q[X]/(f), Q[X]/(g)) = H i (Hom Q[X] (Q[X]/(f), Q )) The theorem that says that these two definitions are equivalent, we don t have time to prove. But working out this example might give you an idea on why it is true for i = 0,.] Actually, I did end up proving that theorem. 3.. projective resolution. Since Q[X] is a domain, the projective resolution of Q[X]/(f) is given by 0 Q[X] f Q[X] Q[X]/(f) 0 where f is multiplication by f. Since Hom R (R, M) = M, we get: Hom Q[X] (Q[X], Q[X]/(g)) = Q[X]/(g) The isomorphism φ x x is given by φ x () = x. The map induced by f is precomposition with f which corresponds to multiplication by f: (f ) (x) = (f ) (φ x ) = φ x (f ) = f x So, Ext Q[X](Q[X]/(f), Q[X]/(g)) is the cokernel of the map Q[X]/(g) Q[X]/(g) given by multiplication by f. This is Q[X] modulo the ideal I consisting of all Q[X]-linear combinations of f and g. In other words, I = (f, g). But Q[X] is PID. So this ideal is generated by one element, the greatest common divisor of f and g. Call this h. Then Ext Q[X](Q[X]/(f), Q[X]/(g)) = Q[X]/(h) Date: February 27, 2007.

22 2 MATH 0B: HOMEWORK Ext 0 Q[X](Q[X]/(f), Q[X]/(g)) = Hom Q[X] (Q[X]/(f), Q[X]/(g)) is the kernel of the map Q[X]/(g) f Q[X]/(g) This is K/(g) where K is the kernel of the composite map Q[X] Q[X]/(g) f Q[X]/(g) But this means that K is the set of all polynomials which, when multiplied by f become divisible by g. To figure out what this is you need to write f and g as products f = ha, g = hb where h = gcd(f, g). (So, a, b are relatively prime.) kf = kha is divisible by g = hb iff k is divisible by b. Therefore, K = (b) and Ext 0 Q[X](Q[X]/(f), Q[X]/(g)) = K/(g) = Q[X]/(h) where the isomorphism Q[X]/(h) = K/(g) is given by multiplication by b Injective resolution. To get an injective resolution of Q[X]/(g) we first need a lemma: Lemma 3.. The module Q[X]/(g) is isomorphic to its dual: Q[X]/(g) = Hom Z (Q[X]/(g), Q) = Q[X]/(g) Suppose that this is true. Then the short exact sequence: (3.) 0 Q[X] g Q[X] Q[X]/(g) 0 when dualized gives an injective coresolution of Q[X]/(g) = Q[X]/(g) : (3.2) 0 Q[X]/(g) Q[X] g Q[X] 0 This is exact since (3.) splits over Q. The injective module Q[X] consists of infinite sequences of rational numbers (a 0, a, a 2, ) where X acts by moving the sequence to the left: X(a 0, a, ) = (a, a 2, ) From the previous homework we know that Hom Q[X] (Q[X]/(f), Q[X] ) = Hom Z (Q[X]/(f), Q) which, by the lemma, is isomorphic to Q[X]/(f). When we apply Hom Q[X] (Q[X]/(f), ) to the injective coresolution (3.2) we get Q[X]/(f) g Q[X]/(f)

23 which has cokernel MATH 0B: HOMEWORK 3 Ext Q[X](Q[X]/(f), Q[X]/(g)) = Q[X]/(f, g) = Q[X](h) where h is the greatest common divisor of f and g and kernel where a = f/h. Ext 0 Q[X](Q[X]/(f), Q[X]/(g)) = (a)/(f) = Q[X]/(h) Proof of the lemma. Suppose first that g is irreducible of degree n. Then M = Q[X]/(g) is n dimensional over Q and is annihilated by multiplication by g. Then M is also n dimensional and is annihilated by multiplication by g. So, M = M. Next suppose that g is a power g = p k of an irreducible polynomial p of degree n. Then M is characterized by the fact that it is nk dimensional and is annihilated by p k but not by p k. But M also has this property. So, M = M. Finally suppose that g = p k i i where p i are distinct irreducible monic polynomials. Then M = Q[X]/(p k i i ) and M = Q[X]/(p k i i ) = Q[X]/(p k i i ) = M

24 MATH 0B: HOMEWORK 4. Homework 04 answers The following problems were due Thursday (3//7).. Show that unique factorization domains (UFDs) are integrally closed. Suppose that R is a UFD with quotient field QR. Then we want to show that any element x = a/b QR which is integral over R lies in R. By writing a, b as products of primes, we may assume that they are relatively prime. If x is integral over R then there are elements c,, c n R so that Multiplying by b n gives x n + c x n + + c n = 0 a n + c a n b + + c n b n = 0 So, b divides a n. This is impossible unless b is a unit in which case a/b R. 2. Show that the integral closure of Z[ 5] (in its fraction field) is Z[α] where α = Let R be the integral closure of Z[ 5] in its fraction field Q[ 5]. Then R contains α and R is integral over Z. So it suffices to show that any a + b 5 Q[ 5] which is integral over Z lies in Z[α]. You can prove this using the trace: Tr : R Z which is given by Tr(a + b 5) = (a + 5) + (a 5) = 2a 2a Z a = n/2 for some n Z and the norm: N : R Z given by N(a + b 5) = (a + 5)(a 5) = a 2 5b 2 Using the unique factorization of b we see that this is an integer iff b = m/2 where m n + 2Z. So, a + b 5 lies in either Z[ 5] or α + Z[ 5] in either case a + b 5 lies in Z[α]. So, R = Z[α] is the integer closure of Z and thus of Z[ 5] in Q[ 5]. Date: March 8, 2007.

25 2 MATH 0B: HOMEWORK 3. Combining these we see that Z[ 5] is not a UFD. Find a number which can be written in two ways as a product of irreducible elements. [Look at the proof of problem and see where it fails for the element α in problem 2.] The proof in problem fails for α = + 5 because, although the 2 numerator a = + 5 and the denominator 2 are relatively prime, the denominator divides the square of the numerator: ( + 5) 2 = = 2(3 + 5) This gives two factorizations of the same number into irreducible factors. To show that + 5, 2, are irreducible note that their norms are 4, 4, 4 respectively. Lemma 4.. Any element of Z[ 5] with norm ±4 is irreducible. Proof. This follows from the following two statements: () There is no element of Z[ 5] with norm ±2. (2) Any element of Z[ 5] with norm ± is a unit. The first follows immediately from the fact that 2 and 3 are not squares modulo 5. The second follows from that fact that if the norm of a+b 5 is ± then its inverse is given by ±(a b 5). Given these two facts, any element of Z[ 5] with norm ±4 must be irreducible since, if it factors, one factor must have norm ± since the only way that ±4 factors as a product of integers is ± 4 and ± If E is a finite separable extension of K and α E show that the trace Tr E/K (α) is equal to the trace of the K linear endomorphism of E given by multiplication by α. [Show that the eigenvalues of this linear transformation are the Galois conjugates of α and each eigenvalue has the same multiplicity.] The first proof ignores the hint. For any α E take the intermediate field K(α). Then n = [K(α) : K] is the degree of the minimal polynomial of α which is given by f(x) = (X α i ) = X n + c X n + + c n where c = α i and c n = ( ) n α i. A basis for K(α) over K is given by, α, α 2,, α n. Let x,, x m be a basis for E as a vector space over K(α). Then b ij = α i x j, i = 0, n, j =,, m forms a basis for E over K. The linear function α E : E E given by

26 MATH 0B: HOMEWORK 3 multiplication by α takes b ij to b i+,j if i n and n n α E (b n,j ) = α n x j = c k α n k x j = c k b n k,j So, the trace of α E is Tr(α E ) = k= m c = m α i j= We need to show that this is equal to Tr E/K (α). Let φ i : K(α) K, i =,, n be the n distinct embeddings of K(α) over K. Then φ i (α) = α i are the Galois conjugates of α. Each embedding φ i extends in exactly m ways to an embedding ψ ij : E K over K since E is a separable extension of K(α). Therefore, k= Tr E/K (α) = ψ ij (α) = m φ i (α) = m α i which is the same as the trace of α E. Now, a proof using the hint. Let g(x) K[X] be the minimal polynomial of the K-linear endomorphism α E : E E. Since g(α) = g(α E )() = 0 it follows that f(x) g(x). Conversely, g(x) f(x) since f(α E ) is multiplication by f(α) = 0. Therefore g(x) = f(x) (assuming they are chosen to be monic). So, deg(g) = deg(f) = n = [K(α), K]. Choose a basis x,, x m for E over K(α). Then E = K(α)x K(α)x 2 K(α)x m and each summand K(α)x j is invariant under the endomorphism α E. By the same argument as above, the minimal polynomial of α E restricted to each summand is f(x). So, the characteristic polynomial of α E is f(x) m. The eigenvalues of α E are the roots of the characteristic polynomial which are the roots of f(x) each with multiplicity m. Therefore, the trace of the matrix of α E is m α i = Tr E/K (α). This proof uses the following two facts about the characteristic polynomial of a linear endomorphism φ of a vector space V () If V = V V 2 where V, V 2 are both invariant under φ then the characteristic polynomial of φ is the product of the characteristic polynomials of φ V i. (2) The minimal polynomial of φ divided the characteristic polynomial of φ. In particular, they are equal if they have the same degree.

27 5. Homework 05 answers The following problems were due Thursday (3/8/7). () (Problems #3 on page 374) If L E are finitely generated field extension(s) of K then show that the transcendence degree of E over K is equal to the sum of the transcendence degree of E over L and the transcendence degree of L over K. Lang says: Let x,, x n be a transcendence basis for L over K and let y,, y m be a transcendence basis for E over L. Then show that x,, x n, y,, y m form a transcendence basis for E over K. So, we need to show (a) x,, x n, y,, y m are algebraically independent and (b) Every element of E is algebraic over K(x,, y m ). Proof of (): Suppose that f(x, Y ) K[X, Y ] so that f(x,, x n, y,, y m ) = 0 Rewrite f(x, Y ) as f(x, Y ) = f α (X)Y α α where f α (X) K[X] for all multiindices α. If we substitute X i = x i then each f α (x) K[x] L. So, f(x, Y ) is an element of L[Y ]. Since the y j are algebraically independent over L and f(x, y) = 0 it must be that f α (x) = 0 for all α. But, the x i are algebraically independent over K. So, f α (X) = 0 for all α. So, f(x, Y ) is the zero polynomial and the x i and y j together form an alg. ind. set. Proof of (2): We have to show that E is algebraic over K(x, y). So, take any element a E. Then a is algebraic over L(y) = L(Y ). So, there is a polynomial g(z) with coefficients in L(y) so that g(a) = 0. But each of the coefficients of g is a rational function in y,, y m with coefficients in L. The set of these coefficients forms a finite set S L so that a is algebraic over K(x, S, y). But each element of S is algebraic over K(x) and therefore also over K(x, y). So, a is also algebraic over K(x, y) as claimed. (2) Let R = K[X, Y ]/(f) where f(x, Y ) = (X a)y 2 (X b) for some a b K. Find a transcendental element Z of R so that R is integral over K[Z]. [Use the proof of Noether Normalization.] The proof of Noether normalization says that Z = X Y m for some large m. But m = is large enough. Then Z = X Y and X = Z + Y. We see that K[X, Y ] = K[Z, Y ]. The polynomial f, when written in terms of Y and Z becomes: f(z, Y ) = (Z + Y a)y 2 (Z b) = Y 3 + (Z a)y 2 Y (Z b) Since this is a monic polynomial in Y with coefficients in K[Z] we see two things: (a) Y is integral over K[Z] (b) R = K[Z, Y ]/(f) is a free module over K[Z] of rank 3. This implies that R is infinite dimensional as a vector space over K. Therefore, it cannot be an algebraic extension of K. So, Z is transcendental and R is integral over K[Z].

28 6. Homework 06 answers The following problems was due Thursday (3/22/7). Assume R, M are both Noetherian. () Show that for any ideal I in R there are only finitely many minimal primes containing I. [Take a maximal counterexample.] Let M = R/I. Then I p iff M p 0. (Proof: M p = 0 iff 0 in M p iff s / p s.t. s = 0 in M iff s I.) Therefore, the minimal primes containing I are exactly the minimal supporting primes of R/I which are the same as the minimal associated primes. We proved that there are only finitely many of these. (2) Suppose that p is a prime and n > 0. Let p (n) M := p n M p M Show that this is a p-primary submodule of M. Let N = p (n) M. Then, by definition, N is the inverse image of p n M p in M. So, M/N is isomorphic to an R-submodule of M p /p n M p. So, it suffices to show that p is the only associated prime of M p /p n M p. In M p each element s S = R\p act as isomorphisms (with inverse s R p ). This means that on M p and more generally on any R p -module, when considered as an R-module, the annihilator of any nonzero element will be disjoint from S, i.e., ann R (x) p. On the other hand p annihilates the entire module p k M p /p k+ M p. So, every nonzero element x will have ann R (x) = p. This means that p is the only associated prime (in R) of Q k = p k M p /p k+ M p. But, Q = M p /p n M p is an extension of the quotients Q k. So, p is the only associated prime of Q. Since M/N is a submodule of Q, p is also the only prime associated to M/N. So, N is p-primary. (3) If φ : R S is a homomorphism of Noetherian rings and M is an S-module then show that ass R (M) = φ (ass S (M)) where φ : Spec(S) Spec(R) is the map induced by φ. (φ (p) = φ (p)) Suppose first that M has only one associated prime in S, call it p. Then p is the set of zero divisors of M and every zero divisor acts nilpotently on M in the sense that a power of it annihilates M. Then q = φ (p) is the set of zero divisors of M in R. And each element of q acts nilpotently on M. This means that ass R (M) = {q}. So, the theorem is true in this case. Let 0 = Q i be a primary decomposition of 0 M as an S-submodule. Suppose that Q i is p i -primary. Then ass S (M/Q i ) = {p i } ass R (M/Q i ) = {q i } where q i = φ (p i ) by the argument above. This implies that 0 = Q i is also a primary decomposition of 0 as an R-submodule of M. So, ass R (M) = ass R (M/Q i ) = {q,, q n } = φ (ass S (M))

29 7. Homework 07 answers The following problems were due Thursday (4/2/7). () Show that Z[/p]/Z is an Artinian Z-module. [Show that every proper subgroup is finite.] The logic of the argument is as follows. This abelian group A = Z[/p]/Z has a sequence of elements a, a 2, satisfying the following two conditions: (a) The sequence generates the group. (b) If a subgroup B A does not contain a n+ then it has at most p n elements. Together, these two statements imply that any proper subgroup of A is finite. So, the DCC must hold for subgroups and A is Artinian. (a) The generating elements are: a n = /p n. By definition, any element of Z[/p] is an integer linear combination of these elements. So, they generate Z[/p] and, consequently, they also generate A. (b) Suppose that B is a subgroup of A which does not contain /p n+. Then the highest denominator in the elements of B is p n. So, the only possible elements of B are i/p n + Z where 0 i < p n. So, B p n. (2) Show that the center of the ring Mat n (R) is isomorphic to the center of R. [Show that the center Z(Mat n (R)) consists of the scalar multiples ai n of the identity matrix where a Z(R).] Let Z = Z(Mat n (R)) be the center of Mat n (R). Let x ij be the matrix with in the (i, j) position and zero everywhere else. Then, any matrix in Mat n (R) can be written as A = a ij x ij where a ij R. If A Z then, e.g., n n x 2 A = a 2j x j = Ax 2 = a i x i2 j= Comparing the coefficients of x 2 on both sides we see that a 22 = a. For j 2 the RHS has no x j terms. So, a 2j = 0. Changing, 2 to other indices we see that a ij = 0 if i j and the diagonal terms a ii must all be equal. To see that these diagonal entries must lie in the center of R, take any r R. Then (rx )A = ra x = A(rx ) = a rx So, ra = a r. i=

30 2 (3) Suppose that M is both Artinian and Noetherian. Then show that () r n M = 0 for some n > 0 (r n M := rr }{{ r} M) n Since M is Noetherian, it has at least one maximal submodule (assuming M 0). Therefore, the radical of M is a proper submodule. Repeating this process we get a descending sequence of submodules M rm r 2 M Since M is Artinian, the sequence stops: r n M = r n+ M. But this can happen only if r n M = 0 (2) r i M/r i+ M is f.g. semisimple for all i 0. By the finite intersection lemma, rm = N i is a finite intersection of maximal submodules N i. So, M/rM is a submodule of the semisimple module M/Ni, making it semisimple. Similarly each of the other subquotients r i M/r i+ M is semisimple. (4) Prove the converse, i.e., (a) and (b) imply that M is both Artinian and Noetherian. Suppose that n =. Then rm = 0 and M = M/rM is semisimple by assumption. But we know that semisimple modules are both Artinian and Noetherian. Now suppose that n 2. We may assume by induction that rm is Artinian and Noetherian. We also know that M/rM satisfies both chain conditions. Let N, N 2, be a sequence of submodules which is either increasing or decreasing. Then N i rm must eventually become stationary. Similarly, the image of N i in M/rM must also become stationary. Using the 5-lemma on the following diagram (just as we did for Noetherian modules over commutative rings), we see that the inclusion N i N j is an isomorphism and thus an equality for sufficiently large i, j. (i < j or i > j depending on whether N i is an ascending or descending chain.) 0 N i rm N N i + rm i rm = 0 N N j rm N j + rm j 0 rm = 0

31 8. Homework 08 answers The following problems were due Thursday (4/9/7). 8.. Let K be any field, let G be any finite group. Let V = K be the trivial representation of G. Let E = K[G] be the group ring considered as a representation (this is called the regular representation ) Find all G-homomorphisms (and show that you have a complete list). f : V E Let N K[G] denote the sum of all the elements of the group: N = σ G σ Then N is clearly invariant under multiplication by elements of G: τn = N τ G For any a K let f a : V E be given by f a (x) = axn. Then this is a linear map which is G-equivariant: f a (σx) = f a (x) = axn = axσn = σf a (x) To show that the f a are the homomorphisms V E, let f : V E be a homomorphism of G-modules. Let Then f() = a σ σ f() = f(τ) = τf() = a σ τσ Comparing the coefficients of τσ we see that a τσ = a σ for all σ, τ G. Taking σ = we see that a τ = a for all τ, i.e., the coefficients are the same. So, f() = an for some a K. But then f(x) = axn. So, f = f a If K is the field with two elements and G is the group with two elements then show that K[G] is not semisimple. [Using your answer to question, show that you have a short exact sequence 0 V E V 0 which does not split.] By problem there are only two homomorphisms V E, namely f 0 = 0 and f : V E which has image V 0 = {0, N}. The second point is that all one dimensional representations are isomorphic to the trivial representation. This is because Aut K (K) = K has only one element. If E were semisimple, then E = V 0 W where W E is another -dimensional submodule of E. But then W = V and we would get another homomorphism V E contradicting Problem. So, E is not a semisimple module. So, K[G] is not a semisimple ring.

32 9. Homework 09 answers The following problems were due Thursday (4/26/7). 9.. Calculate the character table of the quaternion group Q and find the 2-dimensional irreducible representation. If we mod out the central element t then we get Q/ t = Z/2 Z/2 with character table i j k Pulling this back to Q gives: χ χ 2 χ 3 χ 4 i j k t χ χ 2 χ 3 χ 4 χ Since characters determine the representation, the unique 2-dimensional irreducible representation is any representation with character χ 5. The module is the quaternions H viewed as a right C module. The units ±, ±i, ±j, ±k in H form a group isomorphic to the quaternion group Q and left multiplication by these elements commutes with right multiplication by elements of C and is therefore C-linear. To find matrices for the elements of Q we choose a basis for H, say v =, v 2 = j. Then the matrices for i, j, k are given by: ρ(i) = ( ) i 0, ρ(j) = 0 i ( ) 0, ρ(k) = 0 ( 0 ) i i 0 for example, iv = v i, iv 2 = v 2 ( i) making ρ(i) = D(i, i). Taking traces we see that this is the representation with character χ 5. So, it is the one we want.

33 0. Homework 0 This was the last assignment. It was due Thursday (5/3/7). 0.. If V, S are irreducible representations of G and S is one-dimensional, then show that V S is also irreducible. There are two proofs. The first proof uses the fact that tensor product distributes over direct sum and one dimensional characters have inverses. The second proof is a computation. First note that V C = V as G-modules if C is the trivial representation. This follows from the compuation: χ V C (σ) = χ V (σ)χ (σ) = χ V (σ) = χ V (σ) If χ is a one dimension character then χ is its inverse since for all σ G. Consequently χ(σ) = ζ = ζ (V S) S = V S S = V C = V is irreducible. This implies that V S must be irreducible. The second proof is just a computation. If V S = n i S i then χ V S 2 = n 2 i = χ V χ S, χ V χ S = χv (σ)χ S (σ)χ V (σ)χ S (σ) = n n Which implies that V S is irreducible. χv (σ)χ V (σ) = 0.2. Show that the irreducible representations of a product G H are the tensor products of irreducible representations of G, H. Suppose that φ,, φ a are the irreducible representations of G and ψ,, ψ b are the irreducible representations of H. Let π G : G H G, π H : G H G be the projection maps. Let φ i = φ i π G, ψ j = ψ j π H be the pull-backs of φ i, ψ j to G H. Claim : φi ψ j is irreducible. Proof: The restriction of φ i ψ j to G = G is irreducible: Res G H G φ i ψ j = φ i Claim 2: The irreducible representations φ i ψ j are distinct. Proof: φi ψ j restricts to φ i on G and ψ j on H.

34 2 Claim 3: G H has ab conjugacy classes. Proof: (σ, τ) and (σ τ ) are conjugate in G H iff σ, σ are conjugate in G and τ, τ are conjugate in H. These three together show that φ i ψ j give a complete set of irreducible representations of G H. Another proof uses the equation: χ φi, χ ψj φk = χ ψl φi, χ φk χ ψj, χ ψl = δ ik δ jl This equation shows simultaneously that φ i ψ j are irreducible and that they are distinct. Since there are ab of them, these are all the irreducible characters on G H Compute the character table of the alternating group A 4 and the induction-restriction table for the pair (S 4, A 4 ). First you need to find the conjugacy classes of A 4. By the orbitstabilizer formula, the size of the conjugacy class of σ in G is equal to the index of C G (σ) the stabilizer of σ in G. It is easy to see what these centralizers are: σ C(σ) A 4 : C(σ) A 4 (23) {, (23), (32)} 4 (2)(34) {, (2)(34), (3)(24), (4)(23)} 3 Since A 4 has 8 elements of cycle type (abc) and (23) has only 4 conjugates, it must mean that there is another conjugacy class of 3-cycles which also has 4 elements (elements of N G which are conjugate in G have the same number of conjugates in N). To find it you just conjugate (23) by any odd permutation, e.g., (23) to give (32). The quotient of A 4 by the Klein 4-group K = {, (2)(34), (3)(24), (4)(23)} is cyclic of order 3. So, its character table is: where ω = (23)K (32)K χ χ 2 ω w 2 χ 3 ω 2 ω

35 3 Pulling back to A 4, we get: (23) (32) (2)(34) χ χ 2 ω w 2 χ 3 ω 2 ω χ Here we use the fact that (2)(34) K to get χ i ((2)(34) = χ i () for i =, 2, 3 and χ 4 () = 3 is the only solution to the equation 2 = χ 4 () 2. The other values of χ 4 are obtained by column orthogonality. Looking at the character table of S 4 which I did in class we can get the induction restriction table: S 4 A 4 (2) (23) (2)(34) (234) χ χ 2 χ 3 χ 4 χ χ χ χ χ The induction-restriction table is obtained by looking at the values of the characters of S 4 on the conjugacy classes, (23), (2)(34). The restricted characters have the same value on the conjugacy classes (23) and (32).