Lemma 9.14.2. Let $p$ be a prime number. Let $F$ be a field of characteristic $p$. Let $t \in F$ be an element which does not have a $p$th root in $F$. Then the polynomial $x^ p - t$ is irreducible over $F$.

**Proof.**
To see this, suppose that we have a factorization $x^ p - t = f g$. Taking derivatives we get $f' g + f g' = 0$. Note that neither $f' = 0$ nor $g' = 0$ as the degrees of $f$ and $g$ are smaller than $p$. Moreover, $\deg (f') < \deg (f)$ and $\deg (g') < \deg (g)$. We conclude that $f$ and $g$ have a factor in common. Thus if $x^ p - t$ is reducible, then it is of the form $x^ p - t = c f^ n$ for some irreducible $f$, $c \in F^*$, and $n > 1$. Since $p$ is a prime number this implies $n = p$ and $f$ linear, which would imply $x^ p - t$ has a root in $F$. Contradiction.
$\square$

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