Let $f(x) = x^3 - 4ix^2 + 16x - (1+3i) \in (\mathbb{Z}[i])[x]$. Is $f$ irreducible over $(\mathbb{Z}[i])[x]$ ?

By Gauss Lemma, I think that the irreducibility of $f$ over $(\mathbb{Z}[i])[x]$ is the same as irreducibility over $(\mathbb{Q}[i])[x]$. For question regarding irreducibility, I think I should either apply Eisenstein or Gauss Lemma. But Eisenstein might not helpful in this case. For Gauss Lemma, I think I might need to do preliminary test like applying analogous version of Rational root test over $(\mathbb{Q}[i])[x]$ to $f$ to see if it has linear factor. Anyway, when I try to substitute some factor of $1 + 3i$ to $f(x)$, I find that it is hard to calculate if $f(x) = 0 $ or not.

Can anyone suggest a good way to detect the irreducubility of a strange polynomial like this one ?

  • 1
    $\begingroup$ Eisenstein will work. $\endgroup$ – Wojowu Dec 13 '15 at 12:17
  • $\begingroup$ @Wojowu What irreducible $p \in \mathbb{Z}[i]$ should I use ? I guess $p = 1 + 3i$, But $1 + 3i$ does not divide $4$ over $\mathbb{Z}[i]$ since $\frac{4}{1+3i} = \frac{-4}{8}(1 - 3i) $. ???? Can you point out a bit more which $p$ I should consider ? $\endgroup$ – Both Htob Dec 13 '15 at 12:26
  • $\begingroup$ $1+3i$ is not irreducible. $\endgroup$ – Gerry Myerson Dec 13 '15 at 12:27
  • $\begingroup$ As Gerry said, $1+3i$ is not irreducible, as $1+3i=(1+i)(2+i)$. This suggests you should use one of these factors. $\endgroup$ – Wojowu Dec 13 '15 at 12:43
  • $\begingroup$ @Wojowu okay, I will try those two, If I need further help, I might ask more later. Thank you very much, Wojowu and Gerry Myerson. $\endgroup$ – Both Htob Dec 13 '15 at 13:03

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