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fe-sols - MATH 250A FINAL EXAM SOLUTIONS FALL 2007 JUSTIN...

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Unformatted text preview: MATH 250A FINAL EXAM SOLUTIONS, FALL 2007 JUSTIN BLANCHARD 1. Let f ( x 1 ,...,x n ) and g ( x 1 ,...,x n ) be symmetric polynomials of degree d < n . Suppose that f ( x 1 ,...,x n- 1 , 0) = g ( x 1 ,...,x n- 1 , 0). Show that f = g . Solution. In the notation of Lang § IV.6, consider the rings A [ x 1 ,...,x n ] S n ∼ = A [ s 1 ,...,s n ] and A [ x 1 ,...,x n- 1 ] S n- 1 ∼ = A [ s 1 ,...,s n- 1 ], and the map ( a priori just between polynomial rings) given by p ( x 1 ,...,x n ) 7→ p ( x 1 ,...,x n- 1 , 0). Notice s k = ∑ <i 1 < ··· <i k <n +1 x i 1 ...x i k 7→ 0 + ∑ <i 1 < ··· <i k <n x i 1 ...x i k = s k if k < n , whereas s n 7→ 0. Hence this gives a map between symmetric polynomial rings with kernel ( s n ); in particular it’s injective below degree n . This can also be shown directly without reference to the elementary symmetric polynomials: if p 7→ 0 then x n | p ; if p is also symmetric, x 1 ...x n | p . 2. List the finite groups of order 8. Show that you have them all, and that no two on your list are isomorphic. Solution. See midterm solutions. 3. Let G be a finite group. Show that there is a field L that is finite over Q and a field K ⊂ L such that L/K is Galois, with Galois group G . Solution. It suffices to find L/ Q whose Galois group contains a subgroup H isomor- phic to G : then K = L H is Galois over Q and G ( L/K ) = H ∼ = G . In particular, it suffices to find L such that G ( L/ Q ) ∼ = S n for n ≥ | G | . This reduces in two ways to exercises in Lang. On one hand, we may take S n ⊂ S p ( p prime); then the result follows from Example 6 in Lang § VI.2 and Exercise IV.17 (which respectively say that an irreducible polynomial in Q [ x ] of prime degree p with exactly two nonreal roots has Galois group over Q isomorpic to S p , and that one may find such a polynomial). A more direct approach (exercise VI.14) is to construct a polynomial with a Galois group (over Q ) of S n ; however, this problem relies on the machinery of reduction modulo a prime to compute Galois groups—whose validity has not been proved in class or the assigned readings. The former method is easy: in particular, Exercise IV.17(b) follows if you apply Eisenstein’s criterion to p dn g n ( x/p n ). (Of course, only a complete proof received credit. We also saw some cute variations of the proof while grading.) 4. (a) Show that an exact sequence 0 → A → B → C → 0 in an abelian category is split if and only if the induced sequence 0 → Hom( C,A ) → Hom( C,B ) → Hom( C,C ) → 0 is exact. (b) Use this to prove that if A,B,C are finite abelian groups and B is isomor- phic to A ⊕ C , then any short exact sequence 0 → A → B → C → 0 is split....
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fe-sols - MATH 250A FINAL EXAM SOLUTIONS FALL 2007 JUSTIN...

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