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I have the following question:

Let $F$ be a field of characteristic $p$ (where $p$ is prime) such that every irreducible polynomial in $F[x]$ is separable. Show that for every $a\in F$, there is $b\in F$ such that $a=b^p$

I have shown that in a splitting field $E$, for the polynomial $x^p -a$, there can be only one root, and therefore, this polynomial is reducible. I'm stuck on how to finish the proof.

Here's a recap of the work I've done. If $E$ is the splitting field of $x^p-a$ over $F$, then char$E$ = char$F$ = p. If $b$ and $c$ are both roots of this polynomial in $E$, then it must be the case that $b=c$, so in the splitting field, $x^p -a = (x-b)^p = x^p-b^p$ by the ``Freshman's Dream". Hence, we know by assumption that $x^p -a$ is actually reducible in $F$. However, this is where I'm stuck.

Any help would be appreciated, especially in determining a next step. TIA.

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Write $X^p-a$ has a product $X^p-a=P_1(X)....P_n(X)$ where $P_i(X)$ is irreducible, suppose that $deg(P_1)>1$, there exists an extension $E$ of $F$ which is the splitting field of $P_1$ let $b$ be a root of $P_1$ in $E$, we deduce that $P_1(b)=0$ and $b^p-a=P_1(b)...P_n(b)=0$, this implies that $X^p-a=(X-b)^p$ since $P_1$ divides $X^p-a=(X-b)^p$ it is not separable contradiction.

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