College Algebra Exam Review 324

# College Algebra Exam Review 324 - 334 7. FIELD EXTENSIONS...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 334 7. FIELD EXTENSIONS – FIRST LOOK the proof, it sufﬁces to consider the case that K M ¤ H ¤ ¤ E and fe g ¤ S3 . Then we have dimE .M / 2 f2; 3g, and M Â M Â E , so either M D M , or M D E . In Exercise 7.4.7, the ﬁxed point subﬁeld is computed for each subgroup of AutK .E/ Š S3 . The result is Fix.Hi / D K.˛i /, Fix.A3 / D K.ı/ [and, of course, Fix.S3 / D K and Fix.fe g/ D E ]. Consequently, if H D Hi , then M D K.˛i / ¤ E , so M D M D K.˛i /. Similarly, if H D A3 , then M D K.ı/ ¤ E , so M D M D K.ı/. I We have proved the following: Theorem 7.4.9. Let K be a subﬁeld of C , let f .x/ 2 KŒx be an irreducible cubic polynomial, and let E be the splitting ﬁeld of f .x/ in C . Then there is a bijection between subgroups of AutK .E/ and intermediate ﬁelds K Â M Â E . Under the bijection, a subgroup H corresponds to the intermediate ﬁeld Fix.H /, and an intermediate ﬁeld M corresponds to the subgroup AutM .E/. This is a remarkable result, even for the cubic polynomial. The splitting ﬁeld E is an inﬁnite set. For any subset S Â E , we can form the intermediate ﬁeld K.S /. It is certainly not evident that there are at most six possibilities for K.S / and, without introducing symmetries, this fact would remain obscure. pp 3/ E D Q. 3 2; 2 p Q. 3 2/ r d rr d rr 3 2 2 rr d r d rr d rr dp p p 3 3 Q.! 2/ Q.! 2 d d 3 d 3d d d Q ¨ 3 ¨¨ 2 ¨¨ ¨ ¨ 2/ ¨ ¨¨ Q. 3/ ¨¨ Figure 7.4.1. Intermediate ﬁelds for the splitting ﬁeld of f .x/ D x 3 2 over Q. ...
View Full Document

## This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.

Ask a homework question - tutors are online