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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.
 Fall '11
 KeithCornell

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