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Unformatted text preview: Problem Set #4 Solutions Chapter III. 5.1.i. Write out the elementary symmetric polynomials for n = 3 , 4 , 5 . Using the form given on p. 103, k = X 1 i 1 <i 2 < <i k n x i 1 x i 2 x i k , we do the case n = 5: 1 = x 1 + x 2 + x 3 + x 4 + x 5 2 = x 1 x 2 + x 1 x 3 + x 1 x 4 + x 1 x 5 + x 2 x 3 + x 2 x 4 + x 2 x 5 + x 3 x 4 + x 3 x 5 + x 4 x 5 3 = x 1 x 2 x 3 + x 1 x 2 x 4 + x 1 x 2 x 5 + x 1 x 3 x 4 + x 1 x 3 x 5 + x 1 x 4 x 5 + x 2 x 3 x 4 + x 2 x 3 x 5 + x 2 x 4 x 5 + x 3 x 4 x 5 4 = x 1 x 2 x 3 x 4 + x 1 x 2 x 3 x 5 + x 1 x 2 x 4 x 5 + x 1 x 3 x 4 x 5 + x 2 x 3 x 4 x 5 5 = x 1 x 2 x 3 x 4 x 5 . ii. How many monomial summands are there in k ? The number of monomial summands equals the number of ways of choosing k elements out of n elements, i.e., n k . 5.2. The general polynomial f ( x ) = Q n i =1 ( x x i ) is irreducible in F [ x ] . Suppose f ( x ) = g ( x ) h ( x ) in F [ x ], with g ( x ) = Y i S ( x x i ) , h ( x ) = Y i S c ( x x i ) F [ x ] for some subset S { 1 , 2 ,...,n } . Supposing 1 S , 2 6 S , extend the F automorphism = (12) S n of K to an Fautomorphism of K [ x ] by setting ( x ) = x . By the preceeding paragraph in Grove, F = F S n , so fixes all elements of F [ x ]. But g ( x ) = Y i S ( x x i ) = Y i S ( ( x ) x ( i ) ) = Y i S ( x x ( i ) ) 6 = Y i S ( x x i ) = g ( x ) , so g ( x ) 6 F [ x ], a contradiction....
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This note was uploaded on 08/06/2008 for the course MATH 220 taught by Professor Morrison during the Spring '08 term at UCSB.
 Spring '08
 MORRISON
 Algebra, Polynomials

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