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# solutions04 - Problem Set#4 Solutions Chapter III 5.1.i...

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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, σ k = 1 i 1 <i 2 < ··· <i k n x i 1 x i 2 · · · x i k , we do the case n = 5: σ 1 = x 1 + x 2 + x 3 + x 4 + x 5 σ 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 σ 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 σ 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 σ 5 = x 1 x 2 x 3 x 4 x 5 . ii. How many monomial summands are there in σ 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 ) = n i =1 ( x - x i ) is irreducible in F 0 [ x ] . Suppose f ( x ) = g ( x ) h ( x ) in F 0 [ x ], with g ( x ) = i S ( x - x i ) , h ( x ) = i S c ( x - x i ) F 0 [ x ] for some subset S { 1 , 2 , . . . , n } . Supposing 1 S , 2 S , extend the F - automorphism φ = (12) S n of K to an F -automorphism of K [ x ] by setting φ ( x ) = x . By the preceeding paragraph in Grove, F 0 = F S n , so φ fixes all elements of F 0 [ x ]. But φg ( x ) = φ i S ( x - x i ) = i S ( φ ( x ) - x φ ( i ) ) = i S ( x - x φ ( i ) ) = i S ( x - x i ) = g ( x ) , so g ( x ) F 0 [ x ], a contradiction. Alternatively, one may use the second part to exercise 4.1. 5.3. Apply the process given on p. 105 to the following: Using the relations on p. 105, x 1 = σ 1 - x 2 - x 3 , x 2 2 = ( σ 1 - x 3 ) x 2 - ( σ 2 - x 3 σ 1 + x 2 3 ) , x 3 3 = σ 1 x 2 3 - σ 2 x 3 + σ 3 , 1

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we find g ( x 1 , x 2 , x 3 ) = x 2 1 + x 2 2 + x 2 3 = ( σ 1 - x 2 - x 3 ) 2 + x 2 2 + x 2 3 = σ 2 1 + 2
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solutions04 - Problem Set#4 Solutions Chapter III 5.1.i...

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