$x^p-x+a$ irreducible for nonzero $a\in K$ a field of characteristic $p$ prime

Question

Is it true that $f(x)=x^p-x+a\in K[x]$ is irreducible for nonzero $a\in K$ a field of characteristic $p$ prime?

I've seen variants of this question around, but they don't seem to answer the question as worded. (It's possible I have not searched well enough or misunderstood the techniques already given)

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I almost understand the case for finite fields:

If $p=2$, then to show irreducibility we need only show that it has no roots.

For $p>2$, I can show that it's separable since the formal derivative is $-1$. Separability also follows from $f(\alpha)=0\Rightarrow f(\alpha+1)=0$. This also shows that if $\alpha$ were a root in $\mathbb{F}_p$ then $0$ would be a root, a contradiction since we assumed $a\neq 0$; hence the polynomial has no roots in the prime subfield of $K$. [...] But then? Arguments I have seen seem to use the additional fact that $f(x)\in \mathbb{F}_p$.

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I imagine it will come down to some kind of argument with coefficients (depending on roots, maybe using elementary symmetric polynomials) but other nifty ways I'm not seeing also appreciated.

Answer 1

Let $K$ be any field of characteristic $p$ with more than $p$ elements. Then $x^p-x$ has only the $p$ elements of the prime field as roots. Pick $b$ not in the prime field and let $a=-(b^p-b)$. Then $x^p-x+a$ has $b$ as root (and $b+k$ for $k$ in the prime field).

Answer 2

No, in general it is not true that the Artin-Schreier polynomial $x^p-x+\alpha$ is irreducible in any field of characteristic $p$. For instance, if $K$ is algebraically closed, the polynomial is obviously never irreducible.

Answer 3

Or, just take $K=\mathbb Z_p[y]/\langle y^p-y+1\rangle$ for the most basic counterexample, since $y^p-y+1$ is prime in $\mathbb Z_p[y]$. Then But $x^p-x+1$ has a root $y$ in $K$.