# Prove that $x^{5}-x+a$ is an irreducible polynomial [duplicate]

Question. Prove that $$f(x)=x^{5}-x+a$$ is irreducible in $$\Bbb Z[x]$$ if $$5\nmid a$$.

My approach. If we let $$f(x)=(x+a_0)(x^4+b_3x^3+b_2x^2+b_1x+b_0)$$, we know that $$a=a^5-a$$. So $$5 \mid a$$ and it's contradictory to condition.

But can we let $$f(x)=(x^2+a_1x+a_0)(x^3+b_2x^2+b_1x+b_0)$$? I can't prove we can't. I've used Eisenstein's criterion, but it was helpless. I couldn't use it anywhere. Can somebody help me? (I don't want a generalized proof.)

• The polynomial in the title is different from the polynomial in the first line of your question – Zubin Mukerjee Feb 5 '19 at 3:53
• Sorry. It's a typo. – coding1101 Feb 5 '19 at 3:54

Here's an approach: we show that it's irreducible over $$\mathbb{Z}/5$$, and that implies it's irreducible over $$\mathbb{Z}$$.

As a function on the integers mod $$5$$, $$f$$ is simply the constant function $$a$$. Of course, since we can interpolate a polynomial of degree $$\le 4$$ to match any function on those five points, that isn't going to solve our problems on our own. We thus consider possible factorizations.

First, writing $$f$$ as the product of a degree-1 polynomial and a degree-4 polynomial. Since every polynomial of degree $$1$$ has a root, this is impossible; the product would then have a root, and $$f$$ doesn't.

Second, writing $$f$$ as a product of a degree-2 polynomial $$g$$ and a degree-3 polynomial $$h$$. Multiplying and dividing by constants, we can assume WLOG that $$g$$ has leading coefficient $$a$$. Then, completing the square, we write $$g(x)\equiv a\left((x-b)^2+c\right)$$.
As before, $$g$$ must not have any roots. That forces either $$c\equiv 2$$ or $$c\equiv 3$$.

Case 1: $$c\equiv 2$$. We have $$g(x)\equiv \begin{cases}2a&x-b\equiv 0\\3a&x-b\equiv\pm 1\\a&x-b\equiv \pm 2\end{cases}$$. Dividing, $$h(x)\equiv\begin{cases}3&x-b\equiv 0\\2&x-b\equiv\pm 1\\1&x-b\equiv \pm 2\end{cases}$$.
What's an interpolating polynomial that matches those values? To get $$3$$ at $$y=0$$ and $$2$$ at $$y=\pm 1$$, we can simply take $$3-y^2$$. That would give us a value of $$3-4\equiv -1$$ at $$\pm 2$$, so we need to add $$2$$ there to make it work. For something that's zero at $$y=-1,0,1$$, we need a factor of $$y^3-y$$ - but then $$2^3-2\equiv 1$$ and $$3^3-3\equiv -1$$. Those values don't match; we'll have to multiply by $$y$$ again to get an even function with equal values at $$\pm 2$$. Specifically, $$y^4-y^2\equiv 2$$ at $$\pm 2$$. Combine the parts, and
$$3-y^2+(y^4-y^2) \equiv \begin{cases}3&y\equiv 0\\2&y\equiv \pm 1\\1&y\equiv \pm 2\end{cases}$$. Thus $$h(x)\equiv 3-2(x-b)^2+(x-b)^4$$ as functions on $$\mathbb{Z}/5$$.
But polynomial interpolation is unique; there's only one polynomial (up to equivalence mod 5) of degree $$\le 4$$ that matches a complete set of values. We have $$h$$ equivalent to a fourth-degree polynomial, so it can't have degree $$3$$, and there's no factorization.

Case 2: $$c\equiv 3$$. We have $$g(x)\equiv \begin{cases}3a&x-b\equiv 0\\4a&x-b\equiv\pm 1\\2a&x-b\equiv \pm 2\end{cases}$$. Dividing, $$h(x)\equiv\begin{cases}2&x-b\equiv 0\\4&x-b\equiv\pm 1\\3&x-b\equiv \pm 2\end{cases}$$.
The interpolating polynomial this time is $$2+2y^2-(y^4-y^2)$$, so $$h(x)\equiv 2+3(x-b)^2-(x-b)^4$$ as a function. By the same argument, that equivalence is also true as a polynomial, and there's no factorization.

With all cases covered, we're done - $$f$$ is irreducible over $$\mathbb{Z}/5$$, and therefore also irreducible over $$\mathbb{Z}$$.

[Edit - just saw that comment go up. This is a grubby special case of the linked material, without the deeper insights]

• Is it enough to write "$f(x)$ doesn't have any roots" without any explanation in the contests? – coding1101 Feb 5 '19 at 5:15
• The explanation is the first sentence of the previous paragraph. Also, you already had the case of a potential linear factor down. I didn't see the need to elaborate. – jmerry Feb 5 '19 at 5:21
• Oh, sorry. I didn't see that sentence. – coding1101 Feb 5 '19 at 5:22

If $$5\nmid a$$, then $$x^5-x+a$$ is an irreducible polynomial over $$\Bbb F_5$$. Then it is irreducible over $$\Bbb Z$$, and then by Gauss's lemma, over $$\Bbb Q$$.

Proving that it is irreducible over $$\Bbb F_5$$ is essentially Artin-Schreier theory. If $$\xi$$ is a zero of $$x^5-x+a$$ in an extension field $$k$$ of $$\Bbb F_5$$ then $$\xi^5=\xi-a$$. But $$\xi^5=\varphi(\xi)$$ where $$\varphi$$ is the Frobenius automorphism. Then $$\varphi^k(\xi)=\xi-ka$$, and $$\varphi^k(\xi)=\xi$$ iff $$5\mid k$$. Therefore $$\xi$$ defines a degree $$5$$ extension of $$\Bbb F_5$$, and so $$x^5-x+a$$ is irreducible over $$\Bbb F_5$$.

One can replace $$5$$ here by any prime.