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 , #m/thm/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 .

#state/tidy | #lang/en | #SemBr

Footnotes