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 Â 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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