A finite extension is normal if every irreducible polynomial in having a root in splits completely over . It is separable if the minimal polynomial over of every element of has no repeated root. It is Galois if
Put , with , and letFix a -embedding . If is the minimal polynomial of over , an extension of to is determined by the image of , which must be a root of the polynomial obtained by applying to the coefficients of . Conversely, each distinct root gives one extension. Since is algebraically closed, there is at least one such root, and there are at most distinct roots.
Starting from the unique embedding of , induction therefore givesIf is separable, each is separable, so every transformed polynomial has exactly distinct roots. Every inequality is then an equality. Taking proveswith equality in the separable case.
After fixing , every -automorphism of is a -embedding . HenceIf is normal, every -embedding of into has image . It is consequently a -automorphism of , so the two sets, and therefore their cardinalities, are equal.
If is separable, part (a) givesIf it is also normal, part (b) givesThuswhich is exactly the defining condition for to be Galois.
Take . Write , so the extension isThe polynomial is irreducible over , for example by Eisenstein's criterion at in , and has as a root. Thus the extension has degree five.
The element has order five. Hence all roots of areand already lie in . The extension is therefore the splitting field of and is normal. Its derivative is in characteristic eleven, so the polynomial is separable. By part (c),
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