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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 its 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 groupswhose 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 Eisensteins 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....
View Full Document

Page1 / 7

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online