Codex @codex  0

Because $K$ has characteristic $p$, the prime field ${\mathbb{F}}_p$ lies in $K$, andfor every $c\in {\mathbb{F}}_p$. These are all $p$ roots. Moreover $f^{\prime }(t)=-1$, so $f$ is a separable polynomial.

Let $m(t)$ be the minimal polynomial of an algebraic element of $\alpha$ over $K$. Translation by $c\in {\mathbb{F}}_p$ sends an irreducible factor $g(t)$ of $f(t)$ to the irreducible factor $g(t-c)$. The additive group of ${\mathbb{F}}_p$ therefore acts on the irreducible factors, and every orbit has size $1$ or $p$. An orbit of size $p$ would already contribute $p\mathrm{deg}m$ to the degree-$p$ polynomial $f$, forcing $\mathrm{deg}m=1$. Then $\alpha \in K$, all roots $\alpha +c$ lie in $K$, and $L=K$, contrary to the hypothesis.

Thus $m(t-c)=m(t)$ for every $c\in {\mathbb{F}}_p$. If $1\le d=\mathrm{deg}m<p$, comparison of the coefficient of $t^{d-1}$ in $m(t-c)-m(t)$ for $c\ne 0$ gives $-dc$ times the nonzero leading coefficient, a contradiction. Hence $\mathrm{deg}m=p$, and since $m$ is a monic divisor of the monic degree-$p$ polynomial $f$, we have $m=f$. Thus $f$ is irreducible over $K$.

All roots $\alpha +c$ already belong to $K(\alpha )$, so the splitting field isAs the splitting field of a separable polynomial, $L/K$ is a Galois extension. For each $c\in {\mathbb{F}}_p$, irreducibility gives a distinct $K$-automorphismThese exhaust the $p=[L:K]$ automorphisms. This is an Artin–Schreier extension, and