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Unformatted text preview: 334 7. FIELD EXTENSIONS – FIRST LOOK the proof, it sufﬁces to consider the case that K M
¤ ¤ 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
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.
E D Q. 3 2; 2 p Q. 3 2/ r
3 Q.! 2/ Q.! 2 d d 3 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. ...
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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