Lemma 9.12.5. Let $F$ be a field and let $\overline{F}$ be an algebraic closure of $F$. Let $p > 0$ be the characteristic of $F$. Let $P$ be a polynomial over $F$. Then the set of roots of $P$ and $P(x^ p)$ in $\overline{F}$ have the same cardinality (not counting multiplicity).

**Proof.**
Clearly, $\alpha $ is a root of $P(x^ p)$ if and only if $\alpha ^ p$ is a root of $P$. In other words, the roots of $P(x^ p)$ are the roots of $x^ p - \beta $, where $\beta $ is a root of $P$. Thus it suffices to show that the map $\overline{F} \to \overline{F}$, $\alpha \mapsto \alpha ^ p$ is bijective. It is surjective, as $\overline{F}$ is algebraically closed which means that every element has a $p$th root. It is injective because $\alpha ^ p = \beta ^ p$ implies $(\alpha - \beta )^ p = 0$ because the characteristic is $p$. And of course in a field $x^ p = 0$ implies $x = 0$.
$\square$

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