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 let
Fix 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 gives
If is separable, each is separable, so every transformed polynomial has exactly distinct roots. Every inequality is then an equality. Taking proves
with equality in the separable case.
After fixing , every -automorphism of is a -embedding . Hence
If 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) gives
If it is also normal, part (b) gives
Thus
which is exactly the defining condition for to be Galois.
Take . Write , so the extension is
The 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 are
and 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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