# Proving that a Finite Field Over Its Prime Field Is Galois

In the above, I don't understand why the author needs to use another theorem to show that $f$ is separable. Theorem 3.4.5 says that in a finite field, say $F$, the Frobenius automorphism gives us $F = F^p$, entailing that every irreducible polynomial in $F[x]$ is separable. My thinking is, if $E$ exhausts all possible $p^n$ distinct roots of $f$, then $f$ just cannot have repeated roots. This is pretty much the last statement in the paragraph. So, was the author being redundant?

• Why can $f$ not have repeated roots if all $p^n$ elements of $E$ are roots of $f$? I don't see how these two are related... – Servaes Oct 26 '15 at 15:02
• I agree with Andy Tam: Lagrange shows that $f$ has $p^n$ distinct roots in $E$ and its degree is $p^n$. Hence it must be separable. – Hagen Knaf Oct 26 '15 at 15:22
• Books are written by living breathing human beings, not by machines. Accordingly, we may find inefficiencies and infelicities, and even (gasp) errors in books, even highly regarded ones. – Lubin Oct 27 '15 at 14:11