# Why is this polynomial irreducible over $\mathbb{Z}[i]$?

There is a passage on the crazy project saying $x^3+12x^2+18x+6$ is irreducible over $\mathbb{Z}[i]$.

I'm trying to use Eisenstien's Criterion to figure it out. I know that 3 is irreducible in $\mathbb{Z}[i]$, so since this polynomial is 3-Eisenstein, that would mean it's irreducible over $\mathbb{Q}[i]$, right? Then why is it also irreducible over $\mathbb{Z}[i]$?

• What about an example like $6X$? It's irreducible over $\mathbb{Q}[X]$, but not irreducible over $\mathbb{Z}[X]$ since $6X=2\cdot 3X$, and $2$ is not a unit in $\mathbb{Z}[X]$, although it is in $\mathbb{Q}[X]$. – Ashley Lin Oct 14 '11 at 1:58
• Thanks for the example Ashley. – Sasha Oct 14 '11 at 1:59
• You could also use the Rational Root Test, since $\mathbb{Z}[i]$ is a UFD. You would need to check the divisors of $6 = 3(1+i)(1-i)$, multiplied by units ($\pm 1$, $\pm i$). – Arturo Magidin Oct 14 '11 at 3:30

Primitive + irreducible over $\mathbb{Q}(i)$ implies irreducible over $\mathbb{Z}[i]$ by Gauss's Lemma, because $\mathbb{Z}[i]$ is a UFD with $\mathbb{Q}(i)$ as its field of fractions.
• Thanks! So it's the primitivity that's also important here. Btw, is the $i$ really essential here? Or could it just as well be the rings $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$? Because it seems that Eisenstein holds just as well if 3 is a prime in $\mathbb{Z}$. – Ashley Lin Oct 14 '11 at 2:04
• We could make a similar argument for any primitive polynomial $f\in R[x]$, where $R$ is UFD: if there is a prime ideal $P\subset R$ such that $f$ is $P$-Eisenstein, then $f$ is irreducible in $F[x]$, hence in $R[x]$ by Gauss's Lemma. So it works for $R=\mathbb{Z}$, $R=\mathbb{Z}[i]$, and even $R=\mathbb{Z}[y]$ where $y$ is another variable. – Zev Chonoles Oct 14 '11 at 2:15