Bound on the automorphism group of a finite simple or separable extension

Suppose $𝐿 : 𝐾$ is a finite field extension which is simple (or separable, which amounts to the same thing). Then $| A u t (𝐿 : 𝐾) |$ is the number of distinct roots of the minimal polynomial $𝑚 𝜗 (𝑥) \in 𝐾 [𝑥]$ , field in particular

| A u t (𝐿 : 𝐾) | \leq [𝐿 : 𝐾]

with equality iff $𝐿 : 𝐾$ is separable and normal, i.e. Galois.¹

Proof

First consider the case $𝐿 = 𝐾 (𝛼)$ is simple. Note that $𝜎 \in A u t (𝐿 : 𝐾)$ is completely specified by $𝜎 (𝛼)$ , and
$𝑝 𝜎 (𝛼) = 𝜎 𝑝 (𝛼) = 𝜎 (0) = 0$
thus this choice of $𝜎 (𝛼)$ is from the roots of $𝑚 𝛼 (𝑥)$ . At the same time, by ^P1 each root indeed yields an automorphism.

The case $𝐿 : 𝐾$ is separable follows from the primitive element theorem.

As a corollary, automorphisms act faithfully and transitively on the roots of $𝑚 𝜗 (𝑥)$ .

tidy | en | SemBr

2009. Algebra: Chapter 0, §VII.1.2, p. 390 ↩

Bound on the automorphism group of a finite simple or separable extension

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Bound on the automorphism group of a finite simple or separable extension

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