We will prove something stronger that if then is irreducible over . Notice that if (in some larger field) is a root of then is a root. This is because . Consequently, this means are the roots of (since degree is ). Therefore, is a splitting field over of . Now say that where are irreducible over . Let be any root of and be any root of . Remember by above and are splitting fields over of . Thus, . This means, obviously, that and so . In other words, all polynomials in the factorization of have the same degree. But a prime! Thus, either all are linear or else is its own irreducible polynomial factorization. It cannot be that all are linear for that would imply that the roots of lie in which is impossible because for all by Fermat's theorem. Thus, is irreducible.