Irreducible in $(\mathbb{Z}[i])[x]$ Prove that $p(x)=x^3-6x^2+4ix+1+3i$ is irreducible in $(\mathbb{Z}[i])[x]$.
I'm not so clear on irreducibility in specific instances like this one. I know I need to show that if $p(x)=q(x)r(x)$ then W.L.O.G. $q(x)$ is a unit. Since $\mathbb{Z}[i]$ is an integral domain, it follows that $(\mathbb{Z}[i])[x]$ is also an integral domain, so the units are simply $\{1,-1,i,-i\}$. Some back and forth feedback would be preferred to a straightaway proof.
I know that $\deg(p(x))=3$ $\Rightarrow \deg(q(x))+\deg(r(x))=3$. The question becomes how to show $\deg(q(x))=0$.
 A: Hint $ $ If $\,p = \pi\bar \pi$ divides all non-extremal coefs of $\,f(x)\,$ and $\,p\,||\,N(f(0))$ then the Eisenstein irreducibility test applies for prime $\,\pi\,$ or $\,\bar \pi,\,$ since e.g. $\,\pi^2\!\mid f(0)\,\Rightarrow\,p^2\! =N(\pi^2)\mid N(f(0)).$
A: I like the solutions by others, but if you want something different, then you can use the fact that $\Bbb{Z}[i]/\langle 4+i\rangle\cong \Bbb{Z}_{17}$. Reducing the coefficients of your polynomial modulo $\frak{p}=4+i$ (basically replacing $i$ with $-4$ and reducing rational integers modulo $17$) gives the polynomial
$$
x^3-6x^2-16x-11=x^3-6x^2+x+6\in\Bbb{Z}_{17}[x].
$$
Have fun checking that this has no zeros in $\Bbb{Z}_{17}$, and is thus irreducible in
$\Bbb{Z}_{17}[x]$, and consequently also in $\Bbb{Z}[i][x].$
A: Another method, like @BillDubuque’s:
The sextic $f\overline f$ is equal to $x^6 - 12x^5 + 36x^4 + 2x^3 + 4x^2 + 24x + 10$, a $\mathbb Z$-irreducible polynomial. Oughtn’t a nontrivial $\mathbb Z[i]$-factoring of the original $f$ induce a factoring of $f\overline f\,$?
A: If a third degree plynomial is not irreducible, there has to be a first degree divisor. So if $x^3-6x^2-4ix+1+3i$ is not irreducible, it has a root in $\Bbb Z[i]$ and this root must divide $1+3i$. So you only have to check among the divisors of $1+3i$.
