# Irreducibility of a polynomial in a finite field

The question asks me to show that $$\mathbb{Z}_{13}[x]/\langle x^{2014}-x^{1000}+1 \rangle$$ is a field.

Now what I know is that since $$\mathbb{Z}_{13}$$ is a field, so if I show that $$x^{2014}-x^{1000}+1$$ is irreducible, then $$\langle x^{2014}-x^{1000}+1 \rangle$$ will be a maximal ideal and so $$\mathbb{Z}_{13}[x]/\langle x^{2014}-x^{1000}+1 \rangle$$ will be a field.

But I don't see how to show the polynomial is irreducible in $$\mathbb{Z}_{13}[x]$$. For the degree $$\le 3$$, I could use whether there is a root or not but I don't know for such a big degree.

Maybe there is a duplicate for this question, but please help anyway. Thank you.

This is not true. The polynomial $$x^{2014}-x^{1000}+1$$ has $$x^2+6$$ as a factor, and in particular is not irreducible. You can quickly verify this by observing that if $$x^2=-6$$ then (working in $$\mathbb{Z}_{13}$$) $$x^{2014}-x^{1000}+1=(-6)^{1007}-(-6)^{500}+1=(-6)^{11}-(-6)^8+1=0.$$

• is there a sure shot method to find irreducibility in an arbitrary $\mathbb{Z}_p[x]$ or we have to find some factor as you did. How to come up with such a factor?
– Rick
Jul 9, 2019 at 19:55
• I found it with a computer. For algorithms for factoring polynomials over $\mathbb{Z}_p$, see en.wikipedia.org/wiki/…. Jul 9, 2019 at 19:58
• Yeah, I saw that before I asked the question, but thought there might be some general method like reduction etc which I couldn't find. Thanks for the answer.
– Rick
Jul 9, 2019 at 20:01
• In this particular case, you could do it by noticing that even though $x^{2014}-x^{1000}+1$ does not have a root in $\mathbb{Z}_{13}$, $y^{1007}-y^{500}+1$ does, and so this gives a quadratic factor. If there weren't a quadratic factor, though, I don't think there would have been any shortcut. Jul 9, 2019 at 20:20

HINT

Use the fact that for $$p=13$$ prime, $$x^{p-1}\equiv 1 \pmod p$$. The resulting t3nth degree polynomial is much easier to work with.

• This works for elements, not for polynomials: $1$ and $x^{12}$ are different polynomials even though they induce the same function.
– lhf
Jul 9, 2019 at 18:33
• @Mark Fischler I know what you wrote is Fermat's theorem but how to use it here?
– Rick
Jul 9, 2019 at 18:34
• There’s really no way to use Little Fermat in a case like this, far as I know. Jul 10, 2019 at 0:37