# Equivalent definitions of perfect field.

With $T$ some field, I'd like to prove the following implication:

If all irreducible polynomials are separable (have distinct roots), then $T$ is either of characteristic $0$, or it's characteristic is $p>0$ and the frobenius endomorphism $x \mapsto x^p$ is an automorphism

I already have the opposite implication, but I'd appreciate a hint on how to the one I wrote in this question.

If $T$ has characteristic $p>0$ and $a \in T$ is not a $p$th power in $T$, then any irreducible factor of $x^p-a$ is inseparable.
Suppose that $T$ has characteristic $p>0,$ and that $x \mapsto x^p$ is not an automorphism. Since a ring homomorphism from a field into itself is injective or zero, then $x \mapsto x^p$ is not surjective.
Let $a \in T$ be such that $a \neq b^p$ for every $b \in T$. Let $f(x)=x^p-a$. We prove that $f$ is irreducible and inseparable. If $\alpha$ is a root of $x^p-a$, then $x^p-a = (x-\alpha)^p$, so $\alpha$ is a multiple root of $f$ (with multiplicity $p$), hence $f$ is inseparable. Now, let $g(x)$ be an irreducible factor of $f(x)$. Note that $\alpha \not\in T$, otherwise $a=\alpha^p$, contrary to the assumption. Then $g(x)=(x-\alpha)^k$ for some $1<k \leq p$. Using the binomial theorem, we have $$g(x)=(x-\alpha)^k=x^k - k\alpha x^{k-1} + \cdots + (-\alpha)^k.$$ Therefore, $k\alpha \in T$. Since $\alpha \not\in T$, then $k=p$, so $g=f$. Hence, $f$ is irreducible.