College Algebra Exam Review 331

College Algebra Exam Review 331 - 7.5 SPLITTING FIELDS OF...

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Unformatted text preview: 7.5. SPLITTING FIELDS OF POLYNOMIALS IN C Œx 341 pp E D Q. 2; 3/ 2 2 p d d 2 d d d dp p Q. 3/ Q. 2/ d d Q. 6/ d 2d 2 2 d d Q pp Figure 7.5.1. Lattice of intermediate fields for Q  Q. 2; 3/. Theorem 7.5.11. Suppose K is a subfield of C , f .x/ 2 KŒx, and E is the splitting field of f .x/ in C . A subgroup N of AutK .E/ is normal if, and only if, its fixed field Fix.N / is Galois over K . In this case, AutK .Fix.N // Š AutK .E/=N . This result is also proved in a more general setting in Section 9.5. Example 7.5.12. In Example 7.5.10, all the field extensions are Galois, since the Galois group is abelian. Example 7.5.13. In Example 7.4.10, the subfield Q.ı/ is Galois; it is the splitting field of a quadratic polynomial, and the fixed field of the normal subgroup A3 of the Galois group S3 . Example 7.5.14. Consider f .x/ D x 4 2, which is p irreducible over Q p 4 by the Eisenstein criterion. The roots of f in C are ˙ p2; ˙i 4 2. Let E denote the splitting field of f p C ; evidently, E D Q. 4 2; i /. in p The intermediate field Q. 4 2/ is of degree 4 over Q and E D Q. 4 2; i / p is of degree 2 over Q. 4 2/, so E is of degree 8 over Q, using Proposition 7.3.1. Therefore, it follows from the equality of dimensions in the Galois correspondence that the Galois group G D AutQ .E/ is of order 8. p Since E is generated as a field over Q by 4 2 and i , a Q–automorphism of E is determined by its action on these two elements. Furthermore, for p any automorphism of E over Q, . 4 2/ must be one of the roots of f , and .i/ must be one of the roots of p C 1, namely, ˙i . There are exactly x2 eight possibilities for the images of 4 2 and i , namely, p p 4 4 2 7! i r 2 .0 Ä r Ä 3/; i 7! ˙i: ...
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