Unformatted text preview: 9.4. SPLITTING FIELDS AND AUTOMORPHISMS 431 I Proof. Exercise 9.4.3. Deﬁnition 9.4.10. A polynomial in KŒx is said to be separable if each of
its irreducible factors has only simple roots in some (hence any) splitting
ﬁeld. An algebraic element a in a ﬁeld extension of K is said to be separable over K if its minimal polynomial is separable. An algebraic ﬁeld
extension L of K is said to be separable over K if each of its elements is
separable over K .
Remark 9.4.11. Separability is automatic if the characteristic of K is zero
or if K is ﬁnite, by Theorems 9.3.1 and 9.3.4.
Theorem 9.4.12. Suppose L is a splitting ﬁeld for a separable polynomial
f .x/ 2 KŒx. Then Fix.AutK .L// D K . Proof. Let ˇ1 ; : : : ; ˇr be the distinct roots of f .x/ in L. Consider the
tower of ﬁelds:
M0 D K Â Â Mj D K.ˇ1 ; : : : ; ˇj /
Â Â Mr D K.ˇ1 ; : : : ; ˇr / D L: A priori, Fix.AutK .L// Ã K . We have to show that if a 2 L is ﬁxed
by all elements of AutK .L/, then a 2 K . I claim that if a 2 Mj for some
j 1, then a 2 Mj 1 . It will follow from this claim that a 2 M0 D K .
Suppose that a 2 Mj . If Mj 1 D Mj , there is nothing to show.
Otherwise, let ` > 1 denote the degree of the minimal polynomial p.x/
for ˇj in Mj 1 Œx. Then f1; ˇj ;
; ˇj 1 g is a basis for Mj over Mj 1 .
a D m0 C m1 ˇj C C m` 1 ˇj 1
for certain mi 2 Mj 1 .
Since p.x/ is a factor of f .x/ in Mj 1 Œx, p is separable, and the
` distinct roots f˛1 D ˇj ; ˛2 ; : : : ; ˛` g of p.x/ lie in L. According to
Proposition 9.4.1, for each s , there is a s 2 AutMj 1 .L/ Â AutK .L/
such that s .˛1 / D ˛s . Applying s to the expression for a and taking
into account that a and the mi are ﬁxed by s , we get
a D m0 C m1 ˛s C C m` `1
1 ˛s (9.4.2) for 1 Ä s Ä `. Thus, the polynomial .m0 a/ C m1 x C C m` 1 x l 1 of
degree no more than ` 1 has at least ` distinct roots in L, and, therefore,
the coefﬁcients are identically zero. In particular, a D m0 2 Mj 1 .
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