# $2ix^4 − 10x^3 + (4 − 2i)x + 8 + 6i$ irreducible in $\mathbb Z[i][x]$ and in $\mathbb Q[i][x]$

abstract-algebrairreducible-polynomialspolynomials

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

• $$\alpha=1$$, $$\alpha=(2-i)^2$$ (when $$a=0$$), and
• $$\alpha=-(2-i)$$ (when $$a=2$$)

(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 $$\Bbb{Q}(i)[x]$$.