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I have the following polynomial: $f(x)=x^4+1$. I have to prove that it is irreducible over $\mathbb{Z}[x]$ using reduction criterion.

The Reduction Criterion says that:

Let $\mathfrak{m}$ be maximal in Dedekind domain A and $f(X)\in A[X] $. If $f$ reduced modulo $\mathfrak{m}$ is irreducible in $A/\mathfrak{m}$, then $f$ is irreducible in $A$.

If I reduce the polinomal in $\mathbb{Z}_2$ ($p=2$, prime) to $a(x)=x^4+1$, then $a(1)=0$ in $\mathbb{Z}_2$. If I choose $p=3$, all is fine.

I don't understand and I can't find a lot of informations about polynomial reduction mod $n$... How should I choose $p$?

Thanks!

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    $\begingroup$ You are out of luck. Here you will find many proofs for the fact that $x^4+1$ is reducible modulo every prime. So there is no such $p$. You need a different tool for this task. Eisenstein or some such. $\endgroup$
    – Jyrki Lahtonen
    Commented Jun 20 at 10:41
  • $\begingroup$ @JyrkiLahtonen Thank you! It seems that I wrongly understand a thing... Does it suffice to find one prime $p$ which produces $a(x)\not =0$ for any $x \in \mathbb{Z}_p$ to be able to prove that a polynomial is irreducible? If so, this would allow me to prove that $f(x)$ is irreducible by finding a $p$, and there could be a separated discussion for proving that the reduction of $f$ is irreducible on $\mathbb{F}_p$ for any prime $p$. Am I right? Thanks! $\endgroup$
    – ITChristian
    Commented Jun 20 at 11:13
  • $\begingroup$ Of course not! Absence of zeros is equivalent to irreducibility only for polynomials of degree at most three. The standard example is $(x^2+1)(x^2+4)$ that has no zeros in $\Bbb{Q}$ but is obviously reducible. You simply cannot user reduction modulo $p$ with $x^4+1$. $\endgroup$
    – Jyrki Lahtonen
    Commented Jun 20 at 11:35

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