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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.

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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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