Let $F$ be a field of characteristic $p$. Show that $F$ is perfect field if and only if for every element $a\in F$ there is an element $b\in F$ such that $a=b^p$
I've been trying to proof it, but I can't get anywhere, I tried to use that $f$ is separable if and only if $f(x)\neq g(x^P)$ for all $g \in F[x]$
thanks.
Here are some hints to get you going.
For one implication, suppose that some $a \in F$ does not have a $p$th root in $F$ (i.e. there is no $b \in F$ with $a = b^p$), and consider the polynomial $f(x) = x^p - a$.
For the other implication, suppose that every element in $F$ has a $p$th root in $F$, and show that any polynomial of the form $g(x^p)$ is reducible (hint: recall that $(x + y)^p = x^p + y^p$ for any $x,y \in F$).