In characteristic $p$, a degree-$p$ extension is cyclic exactly when it is generated by a root of $x^p-x-a$ with $a\in F$ and that polynomial irreducible

For the forward direction, assume $K\mathrm{/}F$ is cyclic and choose a generator $\sigma$ of its Galois group. Since $[K:F]=p$ and $\mathrm{char}F=p$, one has ${\mathrm{Tr}}_{K\mathrm{/}F}(-1)=-p=0.$ By [L1], choose $\alpha \in K$ with $-1=\alpha -\sigma (\alpha ),$ so $\sigma (\alpha )=\alpha +1$.

For the converse direction, assume ${\alpha }^p-\alpha =a\in F$ and $x^p-x-a$ is irreducible. For each $i\in {\mathbb{F}}_p$ one has $(\alpha +i{)}^p-(\alpha +i)={\alpha }^p-\alpha =a,$ so the roots of the polynomial are exactly $\alpha +i$ for $i\in {\mathbb{F}}_p$. Thus all roots lie in $F(\alpha )$, the derivative is $-1\mathrm{\ne }0$, and [L2] makes $F(\alpha )\mathrm{/}F$ Galois. Because the polynomial is irreducible of degree $p$, this extension has degree $p$.

In characteristic $p$, one has $(\alpha +1{)}^p-(\alpha +1)={\alpha }^p-\alpha ,$ so the element $a:={\alpha }^p-\alpha$ is fixed by $\sigma$ and therefore lies in $F$. Also $\alpha \mathrm{\notin }F$ because step 1.1 gives $\sigma (\alpha )\mathrm{\ne }\alpha$. Since $[K:F]=p$ is prime and $F\subseteq F(\alpha )\subseteq K,$ one must have $F(\alpha )=K$. The polynomial $x^p-x-a\in F[x]$ has root $\alpha$ and degree $p=[F(\alpha ):F]$, so it is the minimal polynomial of $\alpha$ over $F$ and is irreducible.

The rule $\tau (\alpha )=\alpha +1$ permutes the root set and fixes $F$, so it extends to an $F$-automorphism of $F(\alpha )$. Its $p$-th power fixes $\alpha$ and each smaller positive power moves $\alpha$, so $\tau$ has order $p$. A degree-$p$ finite Galois extension has at most $p$ automorphisms, hence exactly the cyclic group generated by $\tau$. Therefore $F(\alpha )\mathrm{/}F$ is cyclic of degree $p$.

Steps 2.1 and 2.2 prove the equivalence, and the displayed root set in step 1.2 proves the final sentence.