TITLE: Is $x^p-x+1$ always irreducible in $\mathbb F_p[x]$?
QUESTION [17 upvotes]: It seems that for any prime number $p$ and for any non-zero element $a$ in the finite field $\mathbb F_p$, the polynomial $x^p-x+a$ is irreducible over $\mathbb F_p$. (It is of course obvious that there are no linear factors.)
Are there any general irreducibility criteria which can help to prove such a result?
(More generally, it seems that all irreducible factors over $\mathbb F_p$
of $$a+\sum_{j=0}^n{n\choose j}(-1)^jx^{p^j}$$
(where $a$ is of course a non-zero element of $\mathbb F_p$)
have a common degree given by a non-trivial power of $p$.)

REPLY [10 votes]: Here is a quick proof using Galois theory. 
Let $r$ be any of the roots in an algebraic closure of $\mathbb{F}_p$. Then $r^p=r-a$, hence by induction $r^{p^k}=r-ka$. Therefore the elements $r^{p^k}$ ($0\leq k\leq p-1$) are all different, hence $r$ is of degree $p$.
Added. More generally, it follows by Artin-Schreier theory that the polynomial
$$ f_{n,a}(x):=a+\sum_{j=0}^n{n\choose j}(-1)^jx^{p^j} $$
decomposes over $\mathbb{F}_p$ into irreducible factors of $p$-power degree. We proceed by induction on $n$. Let $n\geq 1$, and assume the statement for $n-1$ in place of $n$. Let us work in a fixed algebraic closure $\overline{\mathbb{F}_p}$. Let $r\in\overline{\mathbb{F}_p}$ be a root of $f_{n,a}$. Then $s:=r-r^p$ is a root of $f_{n-1,a}$, so $(\mathbb{F}_p(s):\mathbb{F}_p)$ is a power of $p$. Clearly, $\mathbb{F}_p(r)$ is an extension of $\mathbb{F}_p(s)$, because $s\in\mathbb{F}_p(r)$. This extension is of Artin-Schreier type, because $r$ is the root of the polynomial $x^p-x+s$ over $\mathbb{F}_p(s)$. Hence $(\mathbb{F}_p(r):\mathbb{F}_p(s))$ is equal to $1$ or $p$, and so $(\mathbb{F}_p(r):\mathbb{F}_p)$ is a power of $p$.
Remark. The degree of $(\mathbb{F}_p(r):\mathbb{F}_p(s))$ equals $p$ if and only if the trace of $s$ is nonzero. However, it seems difficult to show in this way that $(\mathbb{F}_p(r):\mathbb{F}_p)$ only depends on $n$ and $a$, let alone calculate it explicitly. David Speyer's elegant solution based on linear algebra seems to be the right approach here.