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Unformatted text preview: 9.4. SPLITTING FIELDS AND AUTOMORPHISMS (b) 429 jIsoK .M; L/j D ŒAutK .L/ W AutM .L/. Proof. According to Proposition 9.4.2, the map 7! jM is a surjection
of AutK .L/ onto IsoK .M; L/. Check that . 1 /jM D . 2 /jM if, and only if,
1 and 2 are in the same left coset of AutM .L/ in AutK .L/. This proves
part (a), and part (b) follows.
I Proposition 9.4.4. Let K Â L be a ﬁeld extension and let f .x/ 2 KŒx.
(a) If 2 AutK .L/, then permutes the roots of f .x/ in L.
(b) If L is a splitting ﬁeld of f .x/, then AutK .L/ acts faithfully on
the roots of f in L. Furthermore, the action is transitive on the
roots of each irreducible factor of f .x/ in KŒx. Proof. Suppose 2 AutK .L/,
f .x/ D k0 C k1 x C C kn x n 2 KŒx; and ˛ is a root of f .x/ in L. Then
f . .˛// D k0 C k1 .˛/ C
D .k0 C k1 ˛ C C kn .˛ n //
C kn ˛ n / D 0: Thus, .˛/ is also a root of f .x/. If A is the set of distinct roots of f .x/
in L, then 7! jA is an action of AutK .L/ on A. If L is a splitting ﬁeld
for f .x/, then, in particular, L D K.A/, so the action of AutK .L/ on A is
faithful. Proposition 9.4.1 says that if L is a splitting ﬁeld for f .x/, then
the action is transitive on the roots of each irreducible factor of f .x/. I Deﬁnition 9.4.5. If f 2 KŒx, and L is a splitting ﬁeld of f .x/, then
AutK .L/ is called the Galois group of f , or the Galois group of the ﬁeld
extension K Â L.
We have seen that the Galois group of an irreducible polynomial f is
isomorphic to a transitive subgroup of the group of permutations the roots
of f in L. At least for small n, it is possible to classify the transitive
subgroups of Sn , and thus to list the possible isomorphism classes for the
Galois groups of irreducible polynomials of degree n. For n D 3; 4, and 5,
we have found all transitive subgroups of Sn , in Exercises 5.1.9 and 5.1.20
and Section 5.5. ...
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 Fall '08
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 Algebra

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