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.


2 Answers 2


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 $\mathbb{Z}_{13}$) $$x^{2014}-x^{1000}+1=(-6)^{1007}-(-6)^{500}+1=(-6)^{11}-(-6)^8+1=0.$$

  • $\begingroup$ 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? $\endgroup$
    – Rick
    Jul 9, 2019 at 19:55
  • $\begingroup$ I found it with a computer. For algorithms for factoring polynomials over $\mathbb{Z}_p$, see en.wikipedia.org/wiki/…. $\endgroup$ Jul 9, 2019 at 19:58
  • $\begingroup$ 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. $\endgroup$
    – Rick
    Jul 9, 2019 at 20:01
  • 2
    $\begingroup$ 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. $\endgroup$ Jul 9, 2019 at 20:20


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.

  • 3
    $\begingroup$ This works for elements, not for polynomials: $1$ and $x^{12}$ are different polynomials even though they induce the same function. $\endgroup$
    – lhf
    Jul 9, 2019 at 18:33
  • $\begingroup$ @Mark Fischler I know what you wrote is Fermat's theorem but how to use it here? $\endgroup$
    – Rick
    Jul 9, 2019 at 18:34
  • 1
    $\begingroup$ There’s really no way to use Little Fermat in a case like this, far as I know. $\endgroup$
    – Lubin
    Jul 10, 2019 at 0:37

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.