Let $k$ be a field of characteristic $p$ and $a \in k$. Assume that the equation $x^p - x + a$ does not have any solutions in $k$. Show that the polynomial $x^p - x + a$ is irreducible over $k$.
Below is my proof of this:
Let $\alpha$ be a root of $x^p - x + a$, so that $\alpha^p - \alpha + a$ = 0. Then $(\alpha + 1)^p - (\alpha + 1) + a = \alpha^p + 1 - \alpha - 1 + a = \alpha^p - \alpha + a = 0$ $\Rightarrow$ $\alpha + 1$ is a root of $x^p - x + a$. Thus, letting $\beta = \alpha + 1$, by the same logic as before, we would have that $\beta + 1 = \alpha +2$ is a root of $x^{p} - x + a$. Continuing inductively in this fashion, we see that we can obtain all $p$ roots of $x^p - x + a$ in this way $\Rightarrow$ $\alpha + i$ ($1 \leq i \leq p$) are roots of the polynomial $x^p - x + a$.
First, we note that, since $x^p - x + a$ does not have any solutions in $k$, $a \neq 0$, for if $a = 0$, the element $0 \in k$ would be a solution to $x^p - x + a$. Also, we see that $x^p - x + a$ is separable, since $\frac{d}{dx}(x^p - x + a) = px^{p-1} - 1 = 0 - 1 = -1 \neq 0$ (keeping in mind that we are in a field of characteristic $p$). Thus, letting the splitting field of $x^p - x + a$ over $k$ be denoted by $L$, we see that the finite algebraic field extension $L/k$ is normal and separable (where we've used that a finite field extension $F$ of a field $K$ is a finite normal extension of $F$ if $K$ is a separable splitting field over $F$) $\Rightarrow$ $L/k$ is a Galois extension. Thus, $|Aut(L/k)| = [L:k] = p$, where $[L:k] = p$ follows from the fact that if we obtain one root $\alpha$ of $x^p - x + a$, we obtain all $p$ roots of $x^p - x + a$ by the argument given in the last paragraph. Thus, the Galois group of $L$ over $k$ is of order $p$ $\Rightarrow$ the Galois group of $L$ over $k$ is cyclic of order $p$ $\Rightarrow$ the Galois group of $L$ over $k$, which is isomorphic to $\mathbb{Z}/p\mathbb{Z}$, is a transitive subgroup of $S_p$ $\Rightarrow$ $x^p - x + a$ is irreducible over $k$, where the final implication follows from Theorem 2.9 in https://kconrad.math.uconn.edu/blurbs/galoistheory/galoisaspermgp.pdf.
I'm aware that this question has been asked multiple times on the network. But I'd like a verification of my proof provided above, using Galois theory. In particular, I'm worried that I've made too much of a jump in saying that the finite extension described is normal.
Thank you for your time!
Your proof is essentially the same as the argument in the first paragraph of the proof of Theorem 5.10 here, except more attention is given there to explaining why the field extension $L/k$ (your notation) has degree $p$, along the lines of what Jyrki Lahtonen is saying in the comments above. The "jump" in your proof is not in explaining why $L/k$ is normal (a splitting field over $k$ of a polynomial in $k[x]$ is a normal extension), but in explaining why $[L:k] = p$ since when you don't (yet) know the polynomial $x^p-x+a$ is irreducible you can't be sure that adjoining a root of it to $k$ will give you an extension of degree equal to $p$.