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Adv Alegbra HW Solutions 149

Adv Alegbra HW Solutions 149 - 149 Solution If K k is a eld...

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149 Solution. If K / k is a fi eld extension and α K , then k (α) is the sub fi eld of K generated by k and α . In particular, F p (α) = F q because every nonzero element in F q is a power of α . (ii) Prove that the irreducible polynomial p ( x ) F p [ x ] of α has de- gree n . Solution. If deg ( p ) = d , then Proposition 4.30 says that [ F p (α) : F p ] = d ; it follows that | F p (α) | = p d . But | F p (α) | = | F p n | = p n , and so d = n . (iii) Prove that if G = Gal ( F q / F p ) , then | G | ≤ n . Solution. Every nonzero element of F q has the form α i for some i , and σ(α i ) = [ σ(α) ] i . Now σ permutes the roots of p ( x ) , the irreducible polynomial of α . Since deg ( p ) = n , there are at most n choices for σ(α) , and so | G | ≤ n . (iv) Prove that Gal ( F q / F p ) is cyclic of order n with generator the Frobenius F . Solution. If F j = 1 for some j < n , then a p j = a for all a F q . This says that the polynomial x p j x has p n > p j roots, and this is a contradiction. 5.23 Given f ( x ) = ax 2 + bx + c Q [ x ] , prove that the following statements are equivalent. (i) f ( x ) is irreducible. (ii) b 2 4 ac is not rational. (iii) Gal ( Q ( b 2 4 ac )/ Q ) has order 2. Solution. (i) (ii). A quadratic is irreducible if and only if it has no rational roots.
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