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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?

darij grinberg
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MathTeacher
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    When reading recently an article about the Artin-Schreier theorem, some properties of the so-called *Artin extensions* were used, and, if no mistakes occur here, those are intimately related to the polynomials of the form $x^p-x+a$. Is there indeed any error that occur? and is there any reference to know more in this direction? Thanks in advance. – awllower Nov 13 '11 at 14:05
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    @awllower: [This question](http://math.stackexchange.com/q/50041/11619) may get you started? – Jyrki Lahtonen Nov 13 '11 at 16:39
  • @JyrkiLahtonen: Thanks for the question. I really appreciate this. – awllower Nov 13 '11 at 16:46
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    The original version of the question (which I've now edited) omitted the requirement that $p$ be a prime. This requirement was necessary: The polynomial $x^4-x+a$ over $\mathbb{F}_4$ is reducible for every $a \in \mathbb{F}_4$ ! – darij grinberg Sep 18 '16 at 03:20
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    This is also a particular case of http://math.stackexchange.com/questions/136164 . – darij grinberg Sep 18 '16 at 03:34

8 Answers8

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$x \to x^p$ 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.

zyx
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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 $p$.

darij grinberg
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Jyrki Lahtonen
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    Well done, it is a good proof. – awllower Nov 13 '11 at 14:01
  • I love your proof. One thing is bothering me, though. And It's probably really obvious. How do you know that if $|S|\cdot r\in F_p$, then $|S|$ must be a multiple of $p$ in order for $|S|\cdot r$ to be in $F_p$? I know it has to do with the fact that $r\notin F_p$, and I have some intuition for it, but I don't know how to prove it. – MathTeacher Nov 14 '11 at 04:21
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    @MathMastersStudent: If $|S|$ is not a multiple of $p$, then $|S|\cdot 1$ is an invertible element of $F_p$. So if $|S|\cdot r= b$ with $b\in F_p$, then $r=b|S|^{-1}$ would be in the prime field $F_p$ as well contradicting known facts. – Jyrki Lahtonen Nov 14 '11 at 07:20
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    Great answer as usual, Jyrki: +1. Just to show you how pathologically nit-picking some guys are, I would write $|S|\cdot 1_{F_p}=0_{F_p}$ rather than $|S|=0_{F_p}$, since $S$ is an integer and the integers are not included in a finite field... – Georges Elencwajg Jul 24 '13 at 08:05
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    This argument also applies to any field of characteritic $p$. – Lao-tzu Dec 31 '14 at 12:41
  • @Lao-tzu: Correct. Provided that the polynomial $x^p-x-a$ has no zeros in the said field. When we are looking at it over the prime field this happens, iff $a\neq0$. For other fields of characteristic $p$ it is more complicated. – Jyrki Lahtonen Dec 31 '14 at 13:08
  • @Jyrki Lahtonen: Yes, that's what I mean (I should have written as you did). – Lao-tzu Dec 31 '14 at 13:50
  • I thought so, @Lao-tzu. Added the remark just in case a future visitor starts wondering :-) – Jyrki Lahtonen Dec 31 '14 at 13:59
  • Good proof, Sir ! – Bhaskara-III Jun 17 '15 at 00:25
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$f(x)$ is separable since its derivative is $f'(x) = -1 \ne 0$.

Suppose $\theta$ is a root of $f(x) = x^p - x + a$. Using the Frobenius automorphism, we have: \begin{align} f(\theta + 1) &= (\theta + 1)^p - (\theta + 1) + a\\ &= \theta^p + 1^p - \theta - 1 + a\\ &= \theta^p - \theta + a\\ &= f(\theta) = 0 \end{align}

Thus, by induction, if $\theta$ is a root of $f(x)$, then $\theta + j$ is also a root for all $j \in \mathbb F_p$.

By above, if $f(x)$ were to have a root in $\mathbb F_p$, then $0$ would a be a root too, but this contradicts $a \ne 0$. Thus, $f(x)$ has no roots in $\mathbb F_p$. (This can also be shown using Fermat's little theorem.)

Suppose $\theta$ is a root of $f(x)$ in some extension of $\mathbb F_p$. We know that $\theta + j$ is also a root for all $j \in \mathbb F_p$. Since $f(x)$ is of degree $p$, these are all of the roots of $f(x)$.

Clearly, $\mathbb F_p(\theta) = \mathbb F_p(\theta + j)$ for all $j \in \mathbb F_p$. Thus, all $\{\theta + j\}$ have the same degree over $\mathbb F_p$. Since $f(x)$ is separable, it follows that $f(x)$ must be the product of all minimal polynomials of $\{\theta + j\}$. Suppose the minimal polynomials have degree $m$. We have $p = km$ for some $k$. Since $p$ is prime, either $m = 1$; hence $\theta \in \mathbb F_p$, a contradiction. Or $k = 1$; hence $f(x)$ is irreducible because it's the minimal polynomial.

darij grinberg
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Ayman Hourieh
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  • Ah, I already accepted it. I want to accept it another time but it will get as unaccepted :P Thank You for this wonderful proof :) :) :) –  Aug 03 '13 at 10:03
  • A good one! ${}$ – Jyrki Lahtonen Aug 03 '13 at 10:06
  • Happy to help! I have it in my notes. I don't actually remember if I came up with it myself or found it somewhere. – Ayman Hourieh Aug 03 '13 at 10:11
  • What ever may be the source, Your Answer is good :) –  Aug 03 '13 at 10:16
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    You cannot conclude $f(x)$ is separable simply by $f'(x)\ne 0$ without assuming $f$ is not irrreducible. For example, $f(x)=x(x+1)^2$ is not separable but $f'(x)=(x+1)^2+2x(x+1)\ne 0$. – Bach Jan 28 '20 at 17:42
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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 $a=0$.

Greg Martin
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  • This probably needs the separability of $f$, because if some of the $g_1,g_2,\ldots,g_k$ could be equal, the concept of a permutation and its order would be somewhat muddled. – darij grinberg Sep 18 '16 at 03:25
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One more proof, similar to Greg Martin's: Suppose $\alpha$ is a root of $f(x)=x^p-x+a$ in some splitting field; then \begin{equation*} (\alpha+1)^p - (\alpha+1) + a = \alpha^p + 1 - \alpha - 1 + a = \alpha^p - \alpha + a = 0, \end{equation*} so that $\alpha+1$ is also a root. It follows that the roots of $f$ are $\alpha+i$ for $0\le i < p$. If $f$ factors in $\mathbb{F}[x]$, say $f = gh$, then the sum of the roots of $g$ is $k\alpha + r$ where $\deg g = k$ and $k, r\in\mathbb{F}_P$. Since $g\in \mathbb{F}_p[x]$ it follows that $\alpha\in \mathbb{F}_p$. But that implies that $f$ splits in $\mathbb{F}_p$, which is not the case (for example, neither $0$ nor $1$ is a root). Thus $f$ is irreducible.

rogerl
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  • This is a very helpful and simple answer. Can I ask why it is not the case why f splits in Fp? – P-S.D Apr 01 '17 at 23:20
  • Also, by f splits in Fp, do you mean that Fp is the splitting field of f? Why would Fp being the splitting field of f imply that 0 or 1 is a root? – P-S.D Apr 01 '17 at 23:26
  • @P-S.D If $\alpha\in\mathbb{F}_p$, then the $p$ elements $\alpha+i$, $0\le i

    – rogerl Apr 02 '17 at 14:53
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$x^p-x+a$ 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 $e=p.$

Jyrki Lahtonen
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Utkarsh
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    Doesn't the first sentence "$x^p - x + a$ divides $x^{p^p} - x$" already assume that $x^p - x + a$ is irreducible? Also this reasoning would show that any polynomial of prime degree over a finite field with no roots is irreducible, which is not true. E.g. an irreducible quadratic times an irreducible cubic over $\mathbb{F}_2$ doesn't divide $x^{2^5} - x$ – math54321 Apr 04 '20 at 05:16
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The supposition of Greg Martin is truth, if a polinomyal $f$ with $deg(f)=n$ satisfies the property, then $n\ge p$, by contradiction argument, just write the expansion with the Newton's formula and analyse the coeficient of $x^{n-1}$ term, you get $\binom n 1a_{n}+a_{n-1}=a_{n-1}$, if $n\lt p$, this equation is an absurd.

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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.

Or Kedar
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