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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.

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  • $\begingroup$ What definition do you use for a separable field? Is it that every irreducible polynomial is separable? $\endgroup$ – mathcounterexamples.net Oct 17 '15 at 14:43
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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 $x,y \in F$).

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