ALGEBRA QUALIFYING EXAM, FALL 2017: SOLUTIONS


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1 ALGEBRA QUALIFYING EXAM, FALL 2017: SOLUTIONS Your Name: Conventions: all rings and algebras are assumed to be unital. Part I. True or false? If true provide a brief explanation, if false provide a counterexample (10 points each): 1. Let F, G : A B be isomorphic functors and A 1, A 2 be objects of A. Investigate each of the following for true/false: (i) F(A 1 ) = F(A 2 ) if and only if G(A 1 ) = G(A 2 ). (ii) F(A 1 ) = F(A 2 ) if and only if G(A 1 ) = G(A 2 ). (i) is false. Let A = B be the category with two objects A 1 and A 2 and four morphisms id A1 : A 1 A 1, id A2 : A 2 A 2, f : A 1 A 2 and g : A 2 A 1, such that f and g are isomorphisms. Let F : A A be the functor which sends both A 1 and A 2 to A 1, and all morphisms to id A1. Then F is isomorphic to the identity functor on A. (ii) is true. If α is an isomorphism from F to G, and F(A 1 ) = F(A 2 ), then G(A 1 ) = F(A 1 ) = F(A 2 ) = G(A 2 ), where the first isomorphism is α 1 A 1 and the third isomorphism is α A2. The converse is similar. 1
2 2 ALGEBRA QUALIFYING EXAM, FALL 2017: SOLUTIONS 2. If A is a commutative algebra over an algebraically closed field then all irreducible Amodules are 1dimensional. False. Consider the left regular module over the Calgebra C(x).
3 ALGEBRA QUALIFYING EXAM, FALL 2017: SOLUTIONS 3 3. If R is an artinian ring having no nonzero nilpotent elements then R is a direct sum of division rings. True. R is artinian, so J(R) is nilpotent. By assumption J(R) = 0. It follows that, R is left semisimple. by WedderburnArtin, R is a direct sum of matrix rings over division rings. Any matrix ring of size greater than 1 has nilpotent elements. Hence R is a direct sum of division rings.
4 4 ALGEBRA QUALIFYING EXAM, FALL 2017: SOLUTIONS 4. Let ϕ : X Y be a morphism of affine algebraic sets. Then ϕ is surjective if and only if ϕ is injective (recall that ϕ : F[Y ] F[X], f f ϕ). False. Note that I(im ϕ) = {g F[Y ] g(ϕ(x)) = 0 for any x X} = {g F[Y ] ϕ (g) = 0} = ker ϕ. Since V(I(im ϕ)) = im ϕ, we conclude that ϕ is injective if and only if the image of ϕ is dense, which is possible without ϕ being surjective, for example consider ϕ : V(xy 1) C, (x, y) x.
5 ALGEBRA QUALIFYING EXAM, FALL 2017: SOLUTIONS 5 5. If R, R are rings, F : RMod R Mod is a functor left adjoint to a functor G : R Mod RMod, and P is a projective R module, then FP is a projective R module. False without the assumption that G is exact. For example tensoring with Z/2Z over Z does not send projectives to projectives, even though this functor is left adjoint to Hom Z (Z/2Z, ).
6 6 ALGEBRA QUALIFYING EXAM, FALL 2017: SOLUTIONS Part II. Prove the following statements (10 points each): 1. Prove for any vector spaces U and V we have an isomorphism of vector spaces Hom F (U, V ) Hom F (V, U ), f f ι V, where ι V : V V is the natural embedding. Solution. We consider the linear map ι V : V V, v ϕ v, where ϕ v (f) = f(v) for any f V. Consider the linear maps α : Hom F (U, V ) Hom F (V, U ), and f f ι V β : Hom F (V, U ) Hom F (U, V ), g g ι U. It suffices to prove that α and β are mutual inverses. We show that β α = id, the identity β α = id follows by symmetry. We have β(α(f)) = α(f) ι U = (f ι V ) ι U = ι V f ι U, and to remains to prove that ι V f ι U = f. Well, (ι V f ι U )(u) = (f ι V ) (ι U (u)) = (f ι V ) (ϕ u ) = ϕ u f ι V, so we have to prove that ϕ u f ι V = f(u)(v). Well, ϕ u (f (ι V (v))) = ϕ u (ι V (v) f) = ι V (v)(f(u)) = f(u)(v).
7 ALGEBRA QUALIFYING EXAM, FALL 2017: SOLUTIONS 7 2. Let N be a normal subgroup of a finite group G and P be a Sylow psubgroup of G. Then P N is a Sylow psubgroup of N, and every Sylow psubgroup of N is of this form. Solution. Let S be a Sylow psubgroup of N. There exists a Sylow psubgroup Q of G which contains S. Now Q N S implies Q N = S. Thus every Sylow psubgroup of N is of the required form. Furthermore, let P be any Sylow psubgroup of G. Since Q and P are conjugate in G, and N is normal, Q H and P H have the same order. So P H is a Sylow psubgroup of N.
8 8 ALGEBRA QUALIFYING EXAM, FALL 2017: SOLUTIONS 3. Let V be an Rmodule. A family (V i ) i I of submodules of V is called a directed system of submodules if for any i, j I there exists k I such that V i V k and V j V k. Prove that V is finitely generated if and only if the union i I V i of any directed set of proper submodules is proper. Deduce that a finitely generated module has a maximal proper submodule. Solution. If V is infinitely generated, then the set of all finitely generated submodules of V is a directed set of proper submodules! The union is of course not proper, as every vector lies in a finitely generated submodule. Conversely, if V is finitely generated, and we have (V i ) a directed set of proper submodules whose union is V, then we get a contradiction, as each of the finitely many generators belongs to some V i, hence V i1 + + V in = V, which is impossible for a directed set of proper submodules. The final statement now comes from the Zorn Lemma.
9 ALGEBRA QUALIFYING EXAM, FALL 2017: SOLUTIONS 9 4. An element g of a finite group G is conjugate to g 1 if and only if χ(g) is a real number for every character χ. The irreducible characters form a basis for C(G), as do the indicator functions of the conjugacy classes of G. Hence g and h lie in the same conjugacy class of G if and only if χ(g) = χ(h) for all irreducible characters χ. Hence g and g 1 lie in the same conjugacy class of G if and only if χ(g) = χ(g 1 ) = χ(g) for all irreducible characters χ if and only if χ(g) = χ(g) for all characters χ.
10 10 ALGEBRA QUALIFYING EXAM, FALL 2017: SOLUTIONS 5. Let G be a finite group of automorphisms of the ring A, and R = A G = {a A ga = a for all g G} be the subring of invariants. Then the ring extension R A is integral. Solution. For any f A, the coefficients of the monic polynomial P f (t) := g G (t g f) belong to AG. On the other hand P f (f) = 0.
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