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I need to factor $x^{14}+8x^{13}+3$ over the rationals, and there is a hint to use reduction mod 3. The reduction is $x^{14}+2x^{13}=x^{13}(x+2)$, but I know it has no rational roots (they would have to be $\pm 3$ by the rational roots theorem), so no linear terms in the factoring and all this tells me is that if g(x) is a factor then it reduces to a power of $x$ or to $x+2$ mod3, but that still leaves many possibilities. Any ideas?

user26857
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Psy
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  • The polynomial is irreducible over the $\mathbb Q[x]$. – Lucian Mar 15 '15 at 02:20
  • How did you get to that conclusion? – Psy Mar 15 '15 at 02:27
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    @user28111 It having no rational roots does not mean it is irreducible – Psy Mar 15 '15 at 02:58
  • @Panja.S. The rational root theorem implies it doesn't. – hjhjhj57 Mar 15 '15 at 03:22
  • I can't get the answer, but if $x$ is a root $x+2$ can't equal zero, by the rational root theorem. So the root of the polynomial is of the form $x=-(3k)^{1/13}$ for some natural $k$. Another fact you can use is that since this must be irrational, its negative must also be a root. I couldn't get anywhere from there, but it's a start. – hjhjhj57 Mar 15 '15 at 03:44
  • @Psy, quite right, I completely misread what you wrote. – Noah Schweber Mar 15 '15 at 06:24

1 Answers1

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By Gauss's lemma, $p(x) = x^{14}+8x^{13}+3$ is irreducible in $\mathbb{Q}[x]$ iff it is so in $\mathbb{Z}[x]$. Suppose $p(x) = f(x)g(x)$, where $f(x),g(x)\in \mathbb{Z}[x]$. As $f(x)g(x) = x^{13}(x+2)$ in $\mathbb{Z}_3[x]$ and $\mathbb{Z}_3[x]$ is UFD, we may assume that $f(x)=x^r$ and $g(x)=x^{13-r}(x+2)$ in $\mathbb{Z}_3[x]$. Notice that one of the constant terms of $f, g$ is not a multiple of $3$, and so $r = 0$ or $13$.

Suppose $r=13$, i.e., $f(x)=x^{13}, g(x)=x+2$ in $\mathbb{Z}_3[x]$. Since $\deg f(x)g(x) = 14$, this is possible only if $\deg f=13$ and $\deg g=1$. In other words, $p$ has a linear factor $g$ in $\mathbb{Z}[x]$, which is not the case.

Therefore $r=0$, i.e., $f(x)=1, g(x)=x^{14}+2x^{13}$ in $\mathbb{Z}_3[x]$. The same degree argument shows that $\deg f = 0$ and $\deg g=14$. So in this case, $f=\pm 1$ and $g=\pm p$, and so $p$ is irreducible.

Zilin J.
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