Algebra Qualifying Exam Solutions January 18, 2008 Nick Gurski 0 A B C 0


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1 1. Show that if B, C are flat and Algebra Qualifying Exam Solutions January 18, 2008 Nick Gurski 0 A B C 0 is exact, then A is flat as well. Show that the same holds for projectivity, but not for injectivity. Solution: The claim in the first sentence follows immediately from the fact that the fd(c) 0 if and only if given any resolution 0 P 1 P 0 C 0 with P 0 flat then P 1 is flat as well. But fd(c) 0 if and only if C is flat. The same holds when flat is replaced by projective by a similar argument. The short exact sequence 0 Z Q Q/Z 0 gives a counterexample when flat is replaced by injective as both Q and Q/Z are injective since they are divisible but Z is not injective. 2. Let A be an abelian category and S a set. Let A S denote the category of Sindexed families of objects in A with Sindexed families of maps between them. If A has all products, show that the product functor A S A is left exact for every set S. Show that if A is the category of Rmodules, the product functor is exact, but that the product functor is not exact for every abelian category A and every set S. Solution: First, recall that a functor is left exact when it preserves kernels. The kernel of a map f is just the equalizer of f and the zero map, and the product of zero maps is once again the zero map. Therefore the product functor Π : A S A is left exact if the equalizer of a product of maps is the same as the equalizer of the product. Since both are limits, it is easy to check that the equalizer of the product has the same universal property as the product of the equalizers. When A is the category of Rmodules, the product functor also takes surjective maps to surjective maps: if each f s : A s B s is surjective, then Π s S f s : Π s S A s Π s S B s is clearly surjective as well. This implies that the product functor is exact. If A is the abelian category of sheaves of abelian groups on a space X, the product of epimorphisms need not again be an epimorphism as we show below. Given any short exact sequence of sheaves, 0 F G H 0, the sheaf H is isomorphic to the sheafification of the presheaf U G(U)/F(U). Thus this sequence can be written as 0 F G (G/F) + 0, where the quotient is taken in the category of presheaves and the superscript + means sheafification. The claim is then proved by exhibiting a collection of inclusions of sheaves F n G n
2 ) + ( ) +. such that (ΠG n /F n is not isomorphic to Π G n /F n That these two groups are not always isomorphic can be easily seen from the direct construction of the sheafification functor as a set of functions satisfying local conditions. 3. Let R be a ring and S a multiplicatively closed subset of R. If is an exact sequence of Rmodules, show that 0 A f B g C 0 0 S 1 A S 1 f S 1 B S 1 g S 1 C 0 is also exact. Now show that the converse is not true: there are sequences which are not exact but for which 0 A f B g C 0 0 S 1 A S 1 f S 1 B S 1 g S 1 C 0 is exact for some multiplicatively closed subset S in R. Solution: For the first claim, the standard proof can be found in Atiyah and MacDonald. For the second, let R = Z/6 and S = {1, 3}. It is easy to compute that S 1 Z/6 = Z/2. We then have the sequence 0 Z/3 Z/3 2 0 Z/6 Z/2 0 which is visibly not exact. Applying the functor S 1 yields the sequence which is exact. 0 0 Z/2 Z/ Classify all groups of order 8 up to isomorphism. Solution: For the abelian groups, there are the three isomorphism classes of Z/8, Z/4 Z/2, and Z/2 Z/2 Z/2. For the nonabelian groups of order 8, we prove that there is a unique group, up to isomorphism, with an element of order 4 and a unique group, up to isomorphism, with no element of order 4. Assume that x G has order 4. Then x is normal in G since it has index two. Now we show that there is an element y / x of order 2. If not, then every such y has order 4 or 8; we rule out 8 as it forces G to be cyclic. If y has order 4, then y 2 has order 2, thus y 2 x. We know that y 2 is not 1, x, or x 3 by order consideration as these force y to have orders 2, 8, or 8, respectively. Thus y 2 = x 2. The same reasoning applies to both xy and x 1 y, and therefore forces (x 1 y) 2 = (xy) 2 = x 2 = y 2. This gives the equation y = xyx. This can be shown to for x and y to commute, which then shows x 2 = 1, contradicting that x has order 4. Therefore there must exist a y / x with order 2. But this gives subgroups H = x and K = y such that H is normal in G and H K = 1. If we show that HK = G, then G must be the semidirect product of H and K; since the automorphism group of the
3 cyclic group of order 4 has order 2, this gives precisely one nonabelian group of order 8 with an element of order 4. But the fact that HK = G follows immediately since HK is a subgroup of G with order greater than 4. Now we show that there is a unique nonabelian group of order 8 without an element of order 4. First, note that G has nontrivial center. Let x be in the center and let y 1 be any other element. Then H = {1, x, y, xy} is a subgroup of G and is therefore normal. Once again, we pick a z / H and let K = z. It is clear that H K = 1, and by the same argument as above HK is a subgroup of G with order greater than 4, hence HK = G. Therefore G is a semidirect product, so we must compute the group of automorphisms of H. It is clear that H is isomorphic to Z/2 Z/2 which has automorphism group Z/2 by exchanging coordinates. Thus there is a single isomorphism class of nonabelian groups of order 8 with no element of order Let k be a field and α, β k be two elements of that field. Compute Tor k[x] (k, k) where k[x] acts on the first copy of k by p(x) σ = p(α)σ and on the second copy of k by p(x) σ = p(β)σ for σ k. Solution: The k[x]module structure on the first copy of k above implies that k[x] k given by x α is a k[x]module map. Now consider the short exact sequence below. The long exact sequence for Tor reduces to 0 k[x] (x α) k[x] k 0 0 Tor k[x] 1 (k, k) k[x] k k[x] k k k 0 since k[x] is free and therefore all the other terms vanish. Thus we see that Tor k[x] 1 (k, k) is the kernel of the map k[x] k k[x] k above. It is now simple to check that the diagram below commutes, where both vertical arrows are the canonical isomorphisms. k[x] k k (x α) 1 (β α) k[x] k k Thus we see that if β = α, the map k[x] k k[x] k in question is the zero map so Tor k[x] 1 (k, k) = k, and if β α the map k[x] k k[x] k in question is an isomorphism so Tor k[x] 1 (k, k) = 0. Now all that remains is to compute Tor k[x] 0 (k, k) = k k. First, note that a b = (ab) 1 for a, b k. Thus to compute k k, we need only determine which elements a 1 are equal to 0. This occurs exactly there is a polynomial p(x) for which p(α) = a and p(β) = 0. Thus we see immediately that if α = β, a 1 0 for every a, so in this case k k = k. If α β, then the polynomial p(x) = x β α β has the propery that p(β) = 0 and p(α) = 1, showing that 1 1 = 0 so that k k = 0.
4 6. Let K be a field of characteristic 2 and let L/K be an extension of degree 2. Show that L = K( d) for some d K which is not a square in K. Solution: Let α L be any element not in K. Then the minimal polynomial for α is p(x) = x 2 + bx + c since L/K is a degree 2 extension. Completing the square (note that we use here the fact that the characteristic is not 2) gives (α + b 2 )2 + c b2 4 = 0. Let d = b2 4 c. Then we have α = d b 2 and thus L = K(α) = K( d). 7. Prove directly that if R is Noetherian, then so is the ring of formal power series R[[x]]. Solution: Let I be an ideal of R[[x]]. If f = Σ n=k a nx n, we set l(f) = a k. The l(f) s generate an ideal in R which is finitely generated, say by c 1,..., c m. Let g i I be such that l(g i ) = c i ; then the g i s generate an ideal I I. Define d i by g i (x) = c i x di + higher degree terms, and let d be the maximum of the d i. Define G i (x) = x d di g i (x). Let g I be any element. Without loss of generality, we can assume that g(x) = Σ n=k a nx n for k > d, otherwise we subtract elements of I from g to ensure this. By construction, we know that a k = Σ m l=1 α lc l. Therefore ( ) g(x) Σ m l=1 α lg l (x) x k d is a power series with lowest term of the form a k+1 x k+1. Continue this process, raising the degree of g(x) by subtracting off elements of I. Collecting terms together, this procedure gives a collection of power series C i (x) for which g(x) = ΣC i (x)g i (x) so that I = I and therefore I is finitely generated. 8. Compute, up to equivalence, all of the extensions of abelian groups of Z/p by Z/p. Now compute, up to equivalence, all of the extensions of Z/p 2 by Z/p and determine the behavior of the map Z/p Z/p 2 which sends 1 to p on equivalence classes of extensions. Solution: Equivalence classes of extensions of A by B are in bijection with elements of Ext 1 (A, B), so we compute Ext 1 (Z/p, Z/p) and Ext 1 (Z/p 2, Z/p), and then determine the map Ext 1 (Z/p 2, Z/p) Ext 1 (Z/p, Z/p) induced by Z/p Z/p 2. Consider the following short exact sequence. The long exact sequence for Ext then gives 0 Z p Z Z/p 0 0 Hom(Z/p, Z/p) Hom(Z, Z/p) Hom(Z, Z/p) Ext 1 (Z/p, Z/p) 0 since Z is projective. All three of the Homgroups are isomorphic to Z/p. The first map is then an isomorphism and the second map is the zero map. That forces the third map to be an isomorphism as well, so Ext 1 (Z/p, Z/p) = Z/p. Now consider the exact sequence 0 Z p2 Z Z/p 2 0
5 and the corresponding long exact sequence for Ext. 0 Hom(Z/p 2, Z/p) Hom(Z, Z/p) Hom(Z, Z/p) Ext 1 (Z/p 2, Z/p) 0 Once again all the Homgroups are isomorphic to Z/p so the same argument gives that Ext 1 (Z/p 2, Z/p) = Z/p. Now the two short exact sequences used above are related by the following commutative diagram. 0 p Z Z Z/p 0 0 Z 1 p Z Z/p 2 0 p 2 Using the naturality of the connecting homomorphism, we get the following commutative square. Hom(Z, Z/p) = Ext 1 (Z/p 2, Z/p) 1 Hom(Z, Z/p) = Ext 1 (Z/p, Z/p) But since all of the labeled maps are isomorphisms, the induced map is also an isomorphism. 9. Let I, J be ideals of the ring R. Show that Ext 1 (Z/p 2, Z/p) Ext 1 (Z/p, Z/p) Tor R 1 (R/I, R/J) = I J IJ and use this to prove that if I + J = R then I J = IJ. Solution: We have the short exact sequence 0 I R R/I 0 which gives the following by using the long exact sequence for Tor. 0 Tor 1 (R/I, R/J) I R/J R/J R/I R/J 0 Now I R/J = I/IJ, and Tor R 1 (R/I, R/J) is then the kernel of the map I/IJ R/J given by sending r + IJ to r + J. This is clearly I J IJ. Now assume I + J = R. The Rbilinear map R/I R/J R/(I + J) given by (r 1 + I, r 2 + J) r 1 r 2 + I + J can easily be shown to induce an isomorphism R/I R/J R/(I + J) using the universal property of the tensor. Since we assumed I +J = R, the long exact sequence above becomes the short exact sequence below. 0 Tor 1 (R/I, R/J) I R/J R/J 0
6 Fix a decomposition 1 = α + β where α I and β J. We define a map R/J I R/J by r + J α (r + J). This map is clearly Rlinear. A straightforward computation shows that this map is the inverse to the map I R/J R/J in the short exact sequence above so Tor 1 (R/I, R/J) = 0 which is equivalent to I J = IJ. 10. Show that 2 is a square in the ring of 7adic integers. Solution: Since the ring of 7adic integers is the completion of Z under the topology induced by the ideal (7), we can view it as the ring with elements the sequences (a n with a i Z/7 i such that a i is congruent to a i 1 in Z/7 i 1. The number 2 is then the constant sequence (2, 2, 2,...), and to show that it is a square we will find a 7adic integer such that a 2 i 2 mod 7 i. This can be done inductively as follows. First note that 4 2 2(7), so we set a 1 = 4. Assume we have constructed a 1,..., a n such that a i a i 1 (7 i 1 ) and a 2 i 2(7i ). We now construct a n+1 as a n + t7 n for some t. To have a 2 n+1 2(7n+1 ), we must have or equivalently a 2 n + 2ta n7 n + t 2 7 2n 2(7 n+1 ) 2ta n 7 n 2 a 2 n (7n+1 ). Inductively, we know that 2 a 2 n = b7 n so we define t to be the unique integer between 0 and 6 such that 2ta n b(7). Starting with a 1 = 4, we get a 2 = 10 and a 3 = 108 as the next two terms.
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