How do I prove that $x^p-x+a$ is irreducible in a field with $p$ elements when $a\neq 0$?

Question

Let $p$ be a prime. How do I prove that $x^p-x+a$ is irreducible in a field with $p$ elements when $a\neq 0$?

Right now I'm able to prove that it has no roots and that it is separable, but I have not a clue as to how to prove it is irreducible. Any ideas?

Answer 1

Greg Martin and zyx have given you IMHO very good answers, but they rely on a few basic facts from Galois theory and/or group actions. Here is a more elementary but also a longer approach.

Because we are in a field with $p$ elements, we know that $p$ is the characteristic of our field. Hence, the polynomial $g(x)=x^p-x$ has the property $$g(x_1+x_2)=g(x_1)+g(x_2)$$ whenever $x_1$ and $x_2$ are two elements of an extension field of $\mathbb{F}_p$. By little Fermat we know that $g(k)=k^p-k=0$ for all $k\in \Bbb{F}_p$. Therefore, if $r$ is one of the roots of $f(x)=x^p-x+a$, then $$f(r+k)=g(r+k)+a=g(r)+g(k)+a=f(r)+g(k)=0,$$ so all the elements $r+k$ with $k \in \Bbb{F}_p$ are roots of $f(x)$, and as there are $p$ of them, they must be all the roots. It sounds like you have already shown that $r$ cannot be an element of $\Bbb{F}_p$.

Now assume that $f(x)=f_1(x)f_2(x)$, where both factors $f_1(x),f_2(x)\in \Bbb{F}_p[x]$. From the above consideration we can deduce that $$ f_1(x)=\prod_{k\in S}(x-(r+k)), $$ where $S$ is some subset of the field $\Bbb{F}_p$. Write $\ell=|S|=\deg f_1(x)$. Expanding the product we see that $$ f_1(x)=x^\ell-x^{\ell-1}\sum_{k\in S}(r+k)+\text{lower degree terms}. $$ This polynomial was assumed to have coefficients in the field $\Bbb{F}_p$. From the above expansion we read that the coefficient of degree $\ell-1$ is $|S|\cdot r+\sum_{k\in S}k$. This is an element of $\Bbb{F}_p$, if and only if the term $|S|\cdot r\in\Bbb{F}_p$. Because $r\notin \Bbb{F}_p$, this can only happen if $|S|\cdot1_{\Bbb{F}_p}=0_{\Bbb{F}_p}$. In other words $f_1(x)$ must be either of degree zero or of degree $p$.

Answer 2

I think the following idea works. Let $f(x) = x^p-x+a$. They key observation is that $f(x+1)=f(x)$ in the field of $p$ elements. Now factor $f(x) = g_1(x) \cdots g_k(x)$ as a product of irreducibles. Sending $x$ to $x+1$ must therefore permute the factors $\{ g_1(x), \dots, g_k(x) \}$. But sending $x$ to $x+1$ $p$ times in a row comes back to the original polynomial, so this permutation of the $k$ factors has order dividing $p$. It follows that either every $g_j(x)$ is fixed by sending $x$ to $x+1$ - which I think is a property that no nonconstant polynomial of degree less than $p$ can have, but that needs proof - or else there are $k=p$ factors, which can only happen in the case $a=0$.

Answer 3

$x \to x^p$ is an automorphism sending $r$ to $r-a$ for any root $r$ of the polynomial. This operation is cyclic of order $p$, so that one can get from any root to any other by applying the automorphism several times. The Galois group thus acts transitively on the roots, which is equivalent to irreducibility.

Answer 4

$x^p-x+a$ divides $x^{p^p}-x$. If $f$ is an irreducible divisor of $x^p-x+a$ of degree $d$ then $\mathbf{Z}_p[x]/f$ will be a subfield of the field with $p^p$ elements so $p^p = (p^d)^e$ and so $d=1$ or $e=1$. since $x^p-x+a$ has no roots $e=p.$

Answer 5

A little bit late, but I have no doubt people will still come back to this question. To the proof:

If $r$ is a root of $f$ then $r+j$ is a root of $f$, for $j\in \mathbb F_p$. If $r\in \mathbb F_p$, then $0$ is a root of $f$, but $f(0)=a\neq 0$. Therefore, $f$ has no roots in $\mathbb F_p$. Let $r$ be a root of $f$. Then $\mathbb F_p(r)$ is Galois over $\mathbb F_p$, since all roots are of the form $r+j$, and $f'(x)=-1\neq 0$ so $f$ is separable.

There exists an automorphism of $\mathbb F_p(r)$ given by $\sigma(r)=r+1$. Therefore, $\sigma^j(r)=r+j$, and therefore $r+j$ is a root of the minimal polynomial of $f$, for all $j\in \mathbb F_p$. Therefore, $f$ is irreducible.