Simple extension

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

Suppose is a finite field extension which is simple (or separable, which amounts to the same thing). Then is the number of distinct roots of the minimal polynomial , field in particular

with equality iff is separable and normal, i.e. Galois.¹

Proof

First consider the case is simple. Note that is completely specified by , and

thus this choice of is from the roots of . At the same time, by ^P1 each root indeed yields an automorphism.

The case is separable follows from the primitive element theorem.

As a corollary, automorphisms act faithfully and transitively on the roots of .

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Footnotes

1. 2009. Algebra: Chapter 0, §VII.1.2, p. 390 ↩