Adv Alegbra HW Solutions 152

Adv Alegbra HW - 152 Solution By Exercise 3.67 it suffices to prove that p(x = 0 But if p(x = 0 then p(x = g(x p for some g(x ∈ k[x by part(ii

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Unformatted text preview: 152 Solution. By Exercise 3.67, it suffices to prove that p (x ) = 0. But if p (x ) = 0, then p (x ) = g (x ) p for some g (x ) ∈ k [x ], by part (ii), contradicting p(x ) being irreducible. 5.32 (i) If k = F p (t ), the field of rational functions over F p , prove that x p − t ∈ k [t ] has repeated roots. (It can be shown that x p − t is an irreducible polynomial.) Solution. Let α be a root of x p − t (in a splitting field). Then t = α p and x p − t = x p − α p = (x − α) p . If p (x ) is the irreducible polynomial of α , then p (x ) | (x − α) p , and so it has repeated roots. (One can show that t p − x is, in fact, irreducible in k [t ].) (ii) Prove that E = k (α) is a splitting field of x p − t over k . Solution. Absent. (iii) Prove that Gal( E / k ) = {1}. Solution. Absent. ...
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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