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. $\endgroup$ – Jyrki Lahtonen Sep 29 '16 at 8:25

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 $\alpha _1 +\ldots + \alpha _k\in K$? $\endgroup$ – Alvin Lepik Sep 29 '16 at 7:48
  • 1
    $\begingroup$ @AlvinLepik: That sum is the coefficient of the degree $k-1$ term in $g(x)$. $\endgroup$ – Jyrki Lahtonen Sep 29 '16 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 ? :( $\endgroup$ – Alvin Lepik Sep 29 '16 at 8:17
  • $\begingroup$ A very nice "averaging" argument :-) $\endgroup$ – Jyrki Lahtonen Sep 29 '16 at 8:29

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