Suppose that $K$ is a field of characteristic $p>0$ and that $f,g\in K[X]$ such that $f(X)=g(X^{p^r})$ for some $r\in \mathbb{N}$ and let $\overline{K}$ an algebraic closure of $K$, so we can write $g(X)=\prod_{k=1}^n (X-a_k)$ for some $a_k\in \overline{K}$, therefore $f(X)=\prod_{k=1}^n (X^{p^r}-a_k)$.
All is well by now, however in a book of field theory I've read than choosing some $p^r$-th roots of the $a_k$, say $c_k^{p^r}=a_k$ we have the identity $f(X)=\prod_{k=1}^n (X-c_k)^{p^r}$ as a polinomial in $\overline{K}[X]$, this equality is true and very easy to verify.
My question is the following: over $\overline{K}$ the polynomial $X^{p^r}-a_k$ have $p^r$ roots, and as we can choose any of these $p^r$-th roots to write $f$ as a polynomial over $\overline{K}$ I guess that these $p^r$-th roots cannot be different, otherwise $f$ will be ill-defined as a polynomial over $\overline{K}$. For example suppose that $c$ and $\tilde c$ are two different roots of $X^{p^r}-a$, and that $g(X)=X-a$, then we will had that $f(X)=(X-c)^{p^r}=(X-\tilde c)^{p^r}$ what doesn't make any sense if $c\neq \tilde c$.
Is my reasoning correct? That is, its true that in a field $K$ of characteristic $p>0$ and fixing some algebraic closure $\overline{K}$, some $r\in \mathbb{N}$ and some $a\in \overline{K}$ all roots of the polynomial $X^{p^r}-a$ are equal?
I will close my own question with a formal proof.
Lemma: let $K$ a field with $\operatorname{char}(K)=p>0$, $\overline{K}$ an algebraic closure and $\beta \in \overline{K}$. Then there is a unique $\alpha \in \overline{K}$ such that $\alpha ^p=\beta$.
Proof: suppose there are $\alpha,\alpha_1\in \overline{K}$ such that $\alpha ^p=\alpha _1^p=\beta$, then by the binomial theorem we will have that $$ (X-\alpha )^p=X^p-\alpha ^p=X^p-\beta =X^p-\alpha _1^p=(X-\alpha _1)^p $$ Therefore $\alpha =\alpha _1$ as $\overline{K}[X]$ is a UFD.∎