$2ix^4 − 10x^3 + (4 − 2i)x + 8 + 6i$ irreducible in $\mathbb Z[i][x]$ and in $\mathbb Q[i][x]$ 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?
 A: 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 $\Bbb{Q}(i)[x]$.
A: The polynomial is reducible over $\mathbb Z[i]$ since it is divisible by $2$, a non-invertible element.
The polynomial is associated over $\mathbb Q(i)$ with the polynomial $$f=x^4+5ix^3-(2i+1)x+(2-i)^2.$$ Let $\pi=2-i$ and let us suppose that $f$ is reducible. (Notice now that $f\bmod\pi=x^4$.)

*

*$f=(x^2+ax+b)(x^2+cx+d)$ with $a,b,c,d\in\mathbb Z[i]$. By identifying the coefficients (or by working $\bmod\pi$) we get that the prime $\pi$ divides $a,b,c,d$, and this contradicts $ad+bc=-i\pi$.

*$f=(x^3+ax^2+bx+c)(x+d)$ with $a,b,c,d\in\mathbb Z[i]$. By identifying the coefficients  (or by working $\bmod\pi$) we get that $\pi$ divides $a,b,c,d$. From $cd=\pi^2$ it follows that $d$ is associated to $\pi$. Now check that $\pm\pi$ and $\pm i\pi$ are not roots of $f$. (In order to don't work too much write $f=x^4+i\pi\overline{\pi}x^3-i\pi x+\pi^2$.)

We proved that $f$ is irreducible over $\mathbb Q(i)$.
A: Another method, much like that of @user26857, but with the much more advanced tool of Newton Polygon to help us:
Working with the polynomial $-(1+2i)^2-(1+2i)x+(2+i)(1+2i)x^3+x^4$, we see that $(1+2i)$-adically, the values of the nonzero coefficients are $2,1,1,0$. Plot the points $(0,2)$, $(1,1)$, $(3,1)$, and $(4,0)$ — the abscissa is the degree of the monomial, the ordinate is the order of divisibility by $1+2i$. Draw the Newton Polygon by erecting a vertical
half-line on each point, and taking the convex hull of the whole. The lower boundary has two segments: one of width $1$ and slope $-1$, one of width $3$ and slope $-1/3$. So, even over $\Bbb Q_5$ (the $(1+2i)$-adic completion of $\Bbb Q(i)$), there can be at most a linear factor, with root $\rho$ singly divisible by $1+2i$.
Over the UFD $\Bbb Z(i)$, this $\rho$ can only be  a divisor of $(1+2i)^2$, and thus $i^n(1+2i)$, for $0\le i<4$. None of these four is a root, as you can check. Thus our polynomial is $\Bbb Q(i)$-irreducible.
A: Reduce to showing that $f(x) = x^4+5ix^3+(-1-2i)x+3-4i$ is irreducible over $\mathbb{Q}(i)$. If it were reducible, then the "norm" $F(x) = f(x) \cdot \bar f(x)= x^8 + 25 x^6 - 2 x^5 - 14 x^4 - 40 x^3 + 5 x^2 + 10 x + 25$ would be reducible over $\mathbb{Q}$, so over $\mathbb{Z}$. Now one checks that in fact $F$ is irreducible $\mod 7$.
