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Specifically, I'm trying to solve this problem:

Prove that $p(x)=x^4+x^3+2x+15$ is an irreducible polynomial in $\mathbb{Q}[x]$ by considering $p(x)$ mod $3$ and showing that $p(x)$ has no rational roots.

I'm able to show this is irreducible by applying the rational root theorem to eliminate the possibility of a linear factor and then brute force eliminating the possible quadratic factors, but I don't see how to do this in the way the problem states. Taking $p(x)$ mod $3$, we have $$x^4+x^3+2x+15\equiv x(x^3+x^2+2)\bmod 3.$$ Then, this cubic term is irreducible mod $3$, but how does this help me derive the desired conclusion?

Atsina
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    If $p(x)$ is not irreducible, then Gauss' Lemma and friends show that it must factor as $p(x)=g(x)h(x)$ with $g(x),h(x)\in\Bbb{Z}[x]$, and obviously also monic. Reducing modulo three shows that $\overline{p}(x)=\overline{g}(x)\overline{h}(x)$. Your discovery then implies that one of $g$ or $h$ must be linear. This, in turn, implies the existence of a rational root. Contradiction. – Jyrki Lahtonen Oct 08 '19 at 03:53
  • This has been explained earlier, but I don't have the time to search for the best match now. I have used the technique myself at least here and here. – Jyrki Lahtonen Oct 08 '19 at 03:53
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    @JyrkiLahtonen Oh, duh. Reduce modulo $3$, conclude that if it factors, there must be a linear factor, show there can't be by rational root theorem. Got it. – Atsina Oct 08 '19 at 03:57
  • That's it. Good job. – Jyrki Lahtonen Oct 08 '19 at 03:57
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    This can be also proven by looking at complex roots of $p(x-1)=x^4+3x^3+3x^2+x+13$, since all lie outside the unit circle and also that the constant coeff is a prime. – Sil Nov 17 '19 at 10:02

2 Answers2

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If I recall my algebra correctly, there's a theorem that says that if $p(x) \in \mathbb{Z}[x]$ is irreducible over $\mathbb{Z}[x],$ then it's irreducible over $\mathbb{Q}[x].$ Therefore, if $p(x)$ has no rational roots but is reducible over $\mathbb{Z}[x],$ then it'll be the product of two quadratics with integer coefficients, but then it should still be the product of two quadratics when going mod $3.$

Brian Moehring
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  • The theorem requires that the coefficients are integers and that the polynomial is monic. But that's fulfilled in this case. – Arthur Oct 08 '19 at 05:38
  • @Arthur Unless I'm misreading the wiki for Gauss' lemma (found due to Jyrki's reference in the comment), it seems that the direction I'm using doesn't require $p(x)$ monic. Am I mistaken? – Brian Moehring Oct 08 '19 at 06:01
  • You're right. Primitive is enough. Monic is just a common special case. – Arthur Oct 08 '19 at 06:02
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I believe it is simpler to check that $p(x)$ is irreducible over $\mathbb{F}_2$, thus it is irreducible over $\mathbb{Q}$. Indeed $$ p(x) \equiv x^4+x^3+1 $$ clearly has no root in $\mathbb{F}_2$, but the only quadratic irreducible polynomial over $\mathbb{F}_2$ is $x^2+x+1$. Since $p(x)\neq (x^2+x+1)^2=x^4+x^2+1$, $p(x)$ has no quadratic factor and it is irreducible.

Jack D'Aurizio
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