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Adv Alegbra HW Solutions 152

Adv Alegbra HW Solutions 152 - 152 Solution By Exercise...

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152 Solution. By Exercise 3.67, it suf fi ces 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), contra- dicting p ( x ) being irreducible. 5.32 (i) If k = F p ( t ) , the fi eld 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 fi eld). Then t = α p and x p t = x p α p = ( x α) p . If p ( x ) is the irreducible polynomial of
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