I am revising for exams and have got stuck on the question in my revision,

Decide if $2ix^4 − 10x^3 + (4 − 2i)x + 8 + 6i$ is irreducible in $\mathbb Z[i][x]$ and in $\mathbb Q[i][x]$ and justify your claim.

Usually, I would check if the polynomial was primitive, show by Eisenstein it is irreducible in $\mathbb Z[i][x]$ and then use Gauss to show it is also irreducible in $\mathbb Q[i][x]$ but I don't think this is primitive and 2 can also be factored out.

Does this mean it isn't irreducible in either? Surely not if the question is worth 12 marks?

## Best Answer

Over $\Bbb{Z}[i]$ do whatever your book says about non-unit constant factors (i.e. non-primitive polynomials). I'm used to the convention that irreducibility in $\Bbb{Z}[i][x]$ means

not a product of lower degree polynomials, but other knowledgeable commenters disagree.Anyway, the familiar techniques from $\Bbb{Z}[x]$ have their analogues in $\Bbb{Z}[i][x]$ as $\Bbb{Z}[i]$ is a UFD.

After cancelling the factor $2i$ you are left with the monic (hence primitive) polynomial $$ f(x)=x^4+5ix^3+(-1-2i)x+3-4i. $$ Eisenstein's criterion does not work, but we can reduce modulo a prime, and see what we learn. Let's try the other prime factor $p=2+i$ of $5$. We have $\Bbb{Z}[i]/\langle p\rangle\simeq\Bbb{Z}_5$ and $i\equiv3\pmod p.$

Reduction modulo $p$ gives thus $$ \overline{f}(x)=x^4-2x+1\in\Bbb{Z}_5[x]. $$ We see that $\overline{f}(1)=0$, so it has a factor $x-1$. Division leaves us with $$ \overline{f}(x)=(x-1)(x^3+x^2+x-1).\tag{1} $$ It is quick to check that the cubic factor has no zeros in $\Bbb{Z}_5$. So, by virtue of being cubic, it is irreducible over $\Bbb{Z}_5$.

For us this has the important corollary that

the only way $f(x)$ can factor over $\Bbb{Z}[i]$ is as a product of a linear and a cubic factor.In other words, unless $f(x)$ has a zero in $\Bbb{Q}(i)$, it is irreducible. The analogue of the rational root theorem then kicks in. That putative root, call it $\alpha$, must actually be an element of $\Bbb{Z}[i]$, and also a factor of $3-4i=i(2-i)^2$. In other words $\alpha=i^a(2-i)^b$ with $a\in\{0,1,2,3\}$, $b\in\{0,1,2\}$. Furthermore, equation $(1)$ implies that $\alpha\equiv1\pmod p$. As $$ 1\equiv i^a(2-i)^b\equiv3^a(2-3)^b=3^a(-1)^b\pmod p $$ we can deduce that $a$ must be even and the parity of $b$ is determined. The list of candidates is thus(see professor Lubin's answer for an explanation as to why only the last candidate is actually viable). It is easy to check that none of those three candidates is a zero of $f$. Hence $f$ is irreducible in $\Bbb{Z}[i][x]$ and, by Gauss' Lemma and friends, also in $\Bbb{Q}(i)[x]$.