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Unformatted text preview: 450 9. FIELD EXTENSIONS SECOND LOOK Proposition 9.7.2. Let f 2 KOEx be an irreducible separable polynomial. Let E be a splitting field for f , and let 1 ;:::; n be the roots of f.x/ in E . The Galois group Aut K .E/ , regarded as a group of permutations of the set of roots, is contained in the alternating group A n if, and only if, the discriminant 2 .f / of f has a square root in K . Proof. The discriminant has a square root in K if, and only if, .f / 2 K . But .f / 2 K if, and only if, it is fixed by the Galois group, if, and only if, the Galois group consists of even permutations. n We do not need to know the roots of f in order to compute the dis criminant. Because the discriminant of f D P i a i x i is the discriminant of the monic polynomial .1=a n /f multiplied by a 2n 1 n , it suffices to give a method for computing the discriminant of a monic polynomial. Suppose, then, that f is monic....
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.
 Fall '08
 EVERAGE
 Algebra, Permutations

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