Adv Alegbra HW Solutions 149

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

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Unformatted text preview: 149 Solution. If K / k is a field extension and α ∈ K , then k (α) is the subfield of K generated by k and α . In particular, F p (α) = Fq because every nonzero element in Fq is a power of α . (ii) Prove that the irreducible polynomial p (x ) ∈ F p [x ] of α has degree 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 pn | = pn , and so d = n . (iii) Prove that if G = Gal(Fq /F p ), then |G | ≤ n . Solution. Every nonzero element of Fq 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(Fq /F p ) is cyclic of order n with generator the Frobenius F . j Solution. If F j = 1 for some j < n , then a p = a for all a ∈ Fq . j This says that the polynomial x p − x has pn > 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) √(x ) is irreducible. f b2 − 4ac is not rational. (ii) √ (iii) Gal(Q ( b2 − 4ac)/Q) has order 2. Solution. (i) ⇒ (ii). A quadratic is irreducible if and only if it has no rational roots. √ By the quadratic formula, b2 − 4ac is not rational. √ (ii) ⇒ (iii). If b2 − 4ac is irrational, then the splitting field of f (x ) √ is Q ( b2 − 4√ ), which is a proper extension field of Q. Thereac fore, | Gal(Q ( b2 − 4ac)| ≥ 2. On the other hand, we know that √ Gal(Q ( b2 − 4ac) is isomorphic to a subgroup of S2 ∼ F2 , by The= √ orem 5.21, and so | Gal(Q ( b2 − 4ac)| = 2. (iii) ⇒ (i). If f (x ) factors in Q [x ], then f (x ) = (x − a )(x − b) for a , b ∈ Q. Therefore, Q is the splitting field of f (x ) and so the Galois group has order 1. 5.24 Let E / k be a splitting field of a polynomial f (x ) ∈ k [x ]. If deg( f ) = n , prove that [ E : k ] ≤ n !. Conclude that E / k is a finite extension. 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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