Proof

To show that for every prime power p^a there exists a field with exactly p^a elements, it suffices to construct a finite field L in which X^q - X factors into linear factors (here q = p^a): for then the subfield of elements x satisfying x^q = x has q elements.

So we start with f = X^q - X in Z/pZ[X]. If f factors into linear factors, we are done. If not, then choose an irreducible factor f₁ of f and consider the field F₁ = Z/pZ[X]/(f₁). In this field, X + (f₁) is a zero of f₁, so f₁, whence by a previous lemma, also f, has a linear factor in F₁[X₁]. (Notice the new role of the indeterminate X; we index it by 1 in order to distinguish it from the previous X.)

If f does not completely factor into linear factors in F₁[X₁], then choose an irreducible factor f₂ of f (irreducible in F₁[X₁]), and construct F₂ = F₁[X₁]/(f₂), etc.

Since the number of linear factors increases in every step, this process must terminate and produces a field containing Z/pZ in which f factors into linear factors.

Finally, by the Characterization theorem, we only have to show that L is isomorphic to Z/pZ[X]/(f) for some irreducible polynomial f in Z/pZ[X]. Let a be a primitive element of L^* (for existence, see the previous theorem). Now consider the morphism Z/pZ[X] -> L, h(X) -> h(a). This morphism is clearly surjective but not injective. Let f be a polynomial of minimal positive degree in the kernel. By applying division with remainder to any polynomial in the kernel, it easily follows that every element in the kernel is a multiple of f. Hence the kernel equals the ideal (f) and, by the First isomorphism theorem, L is isomorphic to Z/pZ[X]/(f). As L is a field, f has to be irreducible.