Over ${\mathbb{F}}_p(t)$, the polynomial $x^p-x-t$ gives a cyclic Artin-Schreier extension

Suppose $t=b^p-b$ for some $b\in {\mathbb{F}}_p(t)$. Write $b=u\mathrm{/}v$ in lowest terms. At the pole $t=\mathrm{\infty }$, if $b$ has pole order $m>0$ then $b^p-b$ has pole order $pm$, a multiple of $p$; if $b$ has no pole there, neither does $b^p-b$. Both alternatives contradict the simple pole of $t$ at infinity. Therefore no such $b$ exists.

Let $\alpha$ be a root in an algebraic closure and let $m_{\alpha }$ have degree $d$ over $F$. Every conjugate of $\alpha$ is a root of $x^p-x-t$, hence has the form $\alpha +i$ with $i\in {\mathbb{F}}_p$ and therefore lies in $F(\alpha )$. The polynomial $m_{\alpha }$ thus splits in $F(\alpha )$, and it is separable because it divides a polynomial with derivative $-1$. By [L3], $F(\alpha )\mathrm{/}F$ is Galois. Its automorphisms inject into the additive group of ${\mathbb{F}}_p$ by $\sigma (\alpha )=\alpha +i$, so $d$ is $1$ or $p$. Step 1.1 excludes $d=1$, hence $d=p$ and $m_{\alpha }=x^p-x-t$. Thus this polynomial is irreducible.

Now [L2] makes $F(\alpha )\mathrm{/}F$ cyclic of degree $p$. Its full root set is $\alpha +i$ for $i\in {\mathbb{F}}_p$, since $(\alpha +i{)}^p-(\alpha +i)={\alpha }^p-\alpha =t$.