field of prime characteristic Say $F$ is a field of characteristic $p$ and let $f(x) = x^p - a \in F[x]$. Show that $f$ is irreducible over $F$ or $f$ splits in $F$.
Well, my solution would be since $Char F = p$, then $(x - a^{1/p})^p = x^p - a$. Therefore, $f$ splits in $F$. Is that correct?
 A: Let $\alpha$ be a root of $f$ in some splitting field, then $\alpha ^p=a$. So we have $$f(x)=x^p-a=x^p-\alpha^p=(x-\alpha)^p $$ Now suppose $f=g_1\cdots g_k$ be a factorization of $f$ in $F[x]$ into monic irreducible factors $g_i$. So each $g_i$ is the minimal monic polynomial of $\alpha $ over $F$. So $\deg (g_i)|p$ for each $i$. This shows that either $k=1$ and $f=g_1 $ in which case $f$ is irreducible and  , or $g_i(x)=x-\alpha$ for all $i$ and so $f$ splits over $F$.
A: it is correct for each $a\in F$ when $F$ is a perfect field. 
A: It's clear that $f$ splits over any extension of $F$ in which it has a root. Suppose that it does not have a root in $F$, i.e., that $a$ is not a $p$-th power in $F$, and let $K$ be a field containing a root $\alpha$ of $f$, so $\alpha^p=a$. Let $g$ be a monic irreducible factor of $f$ in $F[x]$. Since $f(x)=(x-\alpha)^p$ in $K[x]$, we must have $g=(x-\alpha)^r$ for some $r$ between $1$ and $p$. In fact, since $\alpha\notin F$, $r>1$, and, assuming $f$ is not irreducible, we must have $\deg(g)<p$. Now, since $(x-\alpha)^r\in F[x]$, in particular, the coefficient of $x^{r-1}$ is in $F$. Using the binomial theorem, we see that this coefficient is $\pm r\alpha$. But because $r$ is prime to $p$, it is a unit in $F$, and we see that $\alpha=(\pm r)^{-1}(\pm r\alpha)\in F$. This is a contradiction. So we must have $g=f$, and $f$ is irreducible. 
