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Unformatted text preview: 9.5. THE GALOIS CORRESPONDENCE 435 Suppose now that K is inﬁnite (which is always the case if the characteristic is zero). L is generated by ﬁnitely many separable algebraic
elements over K , L D K.˛1 ; : : : ; ˛s /. It sufﬁces to show that if L D
K.˛; ˇ/, where ˛ and ˇ are separable and algebraic, then there is a such
that L D K. /, for then the general statement follows by induction on s .
Suppose then that L D K.˛; ˇ/. Let K Â L Â E be a ﬁnite–
dimensional ﬁeld extension such that E is Galois over K . Write n D
dimK L D jIsoK .L; E/j (Corollary 9.4.17). Let f'1 D id; '2 ; : : : ; 'n g be
a listing of IsoK .L; E/.
I claim that there is an element k 2 K such that the elements 'j .k˛ C
ˇ/ are all distinct. Suppose this for the moment and put D k˛ C ˇ .
Then K. / Â L, but dimK .K. // D jIsoK .K. /; E/j
n D dimK L.
Therefore, K. / D L.
Now, to prove the claim, let
x .'i .˛/ 'j .˛/ C .'i .ˇ/ 'j .ˇ/ :
1Äi <j Än The polynomial p.x/ is not identically zero since the 'i are distinct on
K.˛; ˇ/, so there is an element k of the inﬁnite ﬁeld K such that p.k/ ¤ 0.
But then the elements k'i .˛/ C 'i .ˇ/ D 'i .k˛ C ˇ/, 1 Ä i Ä n are
I Corollary 9.5.2. Suppose K Â L is a ﬁnite–dimensional ﬁeld extension,
and the characteristic of K is zero. Then L D K.˛/ for some ˛ 2 L. Proof. Separability is automatic in case the characteristic is zero. I Proposition 9.5.3. Let K Â L be a ﬁnite–dimensional separable ﬁeld
extension and let H be a subgroup of AutK .L/. Put F D Fix.H /. Then
(a) L is Galois over F .
(b) dimF .L/ D jH j.
(c) H D AutF .L/. Proof. First note that jH j Ä jAutK .L/j Ä dimK .L/, by Corollary 9.4.18,
so H is necessarily ﬁnite. Then F D Fix.AutF .L//, by Proposition 9.4.9,
so L is Galois over F and, in particular, separable by Theorem 9.4.14.
By Proposition 9.5.1, there is a ˇ such that L D F .ˇ/. Let f'1 D
id; '2 ; : : : ; 'n g be a listing of the elements of H , and put ˇi D 'i .ˇ/. ...
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- Fall '08