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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!

michiganbiker898
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    Many proofs here. – Jyrki Lahtonen Jan 31 '20 at 18:33
  • @JyrkiLahtonen Thanks for the comment. I'm aware of those proofs. I was just wondering about mine. (= – michiganbiker898 Jan 31 '20 at 18:45
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    That's fine! I was just checking which of those proofs might be close to yours. And then missus called me to do some chores :-) May be zyx's proof is close to yours in that the key is the realization that the Galois group must be cyclic of order $p$. – Jyrki Lahtonen Jan 31 '20 at 18:56
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    But IMO you should be more detailed in how you conclude that there are $p$ automorphisms. The key is that any automorphism must be of the form $\alpha\mapsto \alpha+i$ for some $i=1,2,\ldots,p-1$. And any such automorphism is of order $p$. Counting the roots does not suffice. For example the zeros of $(x^2-2)(x^2-8)$ are clearly all of the form $n\sqrt2$ with integer $n$, so adjoining one of those roots gives you all four. This does not imply that there would be four automorphisms. – Jyrki Lahtonen Jan 31 '20 at 19:01
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    So: 1) $\alpha$ is not in $k$. 2) With $\alpha$ we get the other roots $\alpha+i$, so $L=k(\alpha)$ is the splitting field. 3) Separability implies $L/k$ is Galois of degree $>1$. 4) A non-identity $k$-automorphism $\sigma$ of $L$ must come from $\sigma(\alpha)=\alpha+i$, $0<i<p$. 5) $\sigma$ has order $p$. 6) $Aut(L/k)$ acts transitively on the set of roots. 7) it is irreducible. Compressing some steps is possible. But I'm a bit worried about the potential for circularity in your argument. The fact that $L$ is a splitting field doesn't imply $[L:k]=p$. – Jyrki Lahtonen Jan 31 '20 at 19:08
  • @JyrkiLahtonen I see. Thanks a lot for your comments. I do agree that I should be more careful about showing there are $p$ automorphisms. I appreciate your time. (= – michiganbiker898 Jan 31 '20 at 21:23

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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 $p$.

KCd
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