Let $K$ be a commutative field with characteristic $p\in\mathbb{P}$ and the polynomial $$K[x]\ni f(x) : x^p-x+a,\quad 0\neq a\in K $$ where $f$ has no roots over $K$.

Show that $f$ is irreducible.

Quick side-step: it's generally incorrect to assume having no roots implies irreducibility. For instance, if we look at $x^2+1$ over $\mathbb{R}$, it has no roots, hence the fourth degree polynomial $(x^2+1)^2$ has no roots over $\mathbb{R}$, either, but clearly it's reducible.

If for some $g,h\in K[x]$ we have $f=gh$, then for irreducibility, either $g$ or $h$ has to be constant polynomial.

Some thoughts:
(1) - To show $(f)$ is the maximal ideal in $K[x]$, that would mean $K[x]/(f)$ is a field which in turn is sufficient (and necessary) for $f$ to be irreducible. Only, how do we show this?

(2) - Let $L$ be the splitting field of $f$ (such field always exists), we would have: $$f(x) = (x-\alpha _1)(x-\alpha _2)\ldots (x-\alpha _p) $$

The only thing to clearly conclude here is $\alpha _1\cdot\ldots\cdot \alpha _p = \pm a\in K$.

How to make progress?

  • $\begingroup$ This question has appeared many times with slightly varying details. A good general version is here. Other answers can be found here. Caveat: That question is phrased to be specifically about $K=\Bbb{F}_p$. Some but not all the answers there rely on that extra piece of information. This is very close to being a duplicate. I refrain from picking a duplicate target because A) I am personally involved with this theme, and B) I very much like stewbasic's answer. I don't recall seeing it before. $\endgroup$ Sep 29, 2016 at 8:25

1 Answer 1


Suppose $g(x)\in K[x]$ is a nonconstant monic factor of $f(x)$ and write $$ g(x)=(x-\alpha_1)\ldots(x-\alpha_k) $$ over a splitting field of $g(x)$. In particular $$ s=\alpha_1+\ldots+\alpha_k\in K. $$ Also $0=f(\alpha_i)=\alpha_i^p+\alpha_i-a$ for each $i$, so $$ s^p+s=\sum_i(\alpha_i^p+\alpha_i)=ka. $$ If $k<p$ then $$ (s/k)^p+(s/k)-a=(s^k+s)/k-a=0, $$ a contradiction as $f(x)$ is assumed to have no roots in $K$. Thus $k=p$, so $g(x)=f(x)$. Hence $f(x)$ is irreducible.

  • $\begingroup$ how do you get that $\alpha _1 +\ldots + \alpha _k\in K$? $\endgroup$
    – AlvinL
    Sep 29, 2016 at 7:48
  • 1
    $\begingroup$ @AlvinLepik: That sum is the coefficient of the degree $k-1$ term in $g(x)$. $\endgroup$ Sep 29, 2016 at 8:16
  • $\begingroup$ @JyrkiLahtonen oh of course, thanks! I'm not opposed to pulling rabbits out of hats, but there are so many hats to choose between. How should I know which one to pick ? :( $\endgroup$
    – AlvinL
    Sep 29, 2016 at 8:17
  • $\begingroup$ A very nice "averaging" argument :-) $\endgroup$ Sep 29, 2016 at 8:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.