By J. Milne
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Example text
Pn) with Galois group Sn (acting by permuting the Xi ). Proof. We have shown that F (p1, . . , pn ) = F (X1, . . 12). 28. The Galois group of the general polynomial of degree n is Sn . S. MILNE Proof. , tn][X]. Consider the homomorphism F [t1, . . , tn ] → F [p1, . . , pn ], ti → pi . We shall prove shortly that this is an isomorphism, and therefore induces an isomorphism on the fields of fractions F (t1, . . , tn) → F (p1, . . , pn ), ti → pi . Under this isomorphism, f(X) corresponds to g(X) = X n − p1 X n−1 + · · · + (−1)n pn .
The formulas for the discriminant rapidly become very complicated, for example, that for 5 X + aX 4 + bX 3 + cX 2 + dX + e has about 60 terms. Fortunately, Maple knows them: the syntax is “discrim(f,X);” where f is a polynomial in the variable X. 3. Suppose F ⊂ R. Then D(f) will not be a square if it is negative. It is known that the sign of D(f) is (−1)s where 2s is the number of nonreal roots of f in C. Thus if s is odd, then Gf is not contained in An . This can be proved more directly by noting that complex conjugation will act on the roots as the product of s transpositions (cf.
The discriminants of f and g are equal. Proof. Compute everything in terms of the αi ’s. (Cf. ) Now let f be an irreducible separable quartic. Then G = Gf is a transitive subgroup of S4 whose order is divisible by 4. There are the following possibilities: G (G ∩ V : 1) (G : V ∩ G) S4 4 6 A4 4 3 V 4 1 D4 4 2 C4 2 2 FIELDS AND GALOIS THEORY (G ∩ V : 1) = [E : M], 31 (G : V ∩ G) = [M : F ]. Note that G can’t, for example, be the group generated by (12) and (34) because this is not transitive. The groups of type D4 are the Sylow 2-subgroups discussed above, and the groups of type C4 are those generated by cycles of length 4.