What kind of structure is $\mathbb{Z}[i]/\langle 2+2i\rangle$?

Question

Consider the factor ring $R=\mathbb{Z}[i]/\langle 2+2i\rangle$, where $\langle 2+2i\rangle$ is the ideal of the Gaussian integers such that for all $z\in\mathbb{Z}[i]$, $z(2+2i)\in\langle 2+2i\rangle$. Since $2+2i+\langle 2+2i\rangle=\langle 2+2i\rangle$, we can say that $2+2i=0$.

From this, we can square both sides to say that $8=0$ in $\langle 2+2i\rangle$, and we can subtract $2$ from both sides to say that $2i=-2$. Note that we cannot say that $i=-1$, because this would imply that $1+i\in\langle 2+2i\rangle$. It is trivial to prove that this is not the case.

With this in mind, we can see that the order of $R$ is 16, and our 16 unique elements are in the set $S=\{0, 1, 2, 3, 4, 5, 6, 7, i, 1+i, 2+i, 3+i, 4+i, 5+i, 6+i,7+i\}$.

However, equipped with addition, multiplication, and the aforementioned identities, what exactly is $R$? It's clearly at least a group under addition. I believe it's also a ring, since distribution appears to work just fine, though there may be something I'm overlooking. I know that there are (to isomorphism) a finite number of distinct groups with a finite order. In particular, there are only 14 distinct groups of order 16. So, which of them is isomorphic to $R$?

A naive answer would be $\mathbb{Z}_8\times\mathbb{Z}_2$, but we can quickly see this is not the case. If it was, and we said $f$ is a homomorphism $f: R\rightarrow S=(a,b)$ where $a=\Re(z),b=\Im(z),$ for $z\in R$. Then, by definition of a homomorphism, we would say $f(z\times w)=f(z)\times f(w)$. We quickly see this is not true - consider the case where $z=1+i$. Here we find that $z^2=2i=-2=-2+8=6$, and as such, $f(z\times z)=f(6+0i)=(6,0)$. However, $f(z)\times f(z)=f(1+i)\times f(1+i)=(1,1)\times(1,1)=(1,1)$. Therefore, $f$ is not a homomorphism. For a similar reason we can also see that $R$ is not isomorphic to $\mathbb{Z}_{16}$, $\mathbb{Z}_4\times\mathbb{Z}_4$, or $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$. What is left? What is it actually isomorphic to?

Answer 1

First, as @LeeMosher commented, and as you suspected, yes, there is a natural ring structure on any quotient $R/I$ where $I$ is an ideal of a ring $R$.

And, as others commented, the cardinality of $\mathbb Z[i]/(1+i)^3$ is $8=2^3$ (not $16=2^4$). A little more generally, for any prime element $\varpi$ in $\mathbb Z[i]$ (note that $1+i$ is prime), the cardinality of $\mathbb Z[i]/\varpi^n$ is $\mathrm{cardinality}(\mathbb Z[i]/\varpi)^n$. There are various ways to argue this, and perhaps the clearest are not the most elementary...

So, yes, what are the possible abelian groups of order $8$? By the structure theorem, there are three, $\mathbb Z/2\oplus \mathbb Z/2\oplus\mathbb Z/2$, $\mathbb Z/2\oplus \mathbb Z/4$, and $\mathbb Z/8$.

For comparison, the quotient ring $\mathbb Z/8$ is, as an additive group, of course, $\mathbb Z/8$. That is, it's cyclic.

$\mathbb Z[i]/(1+i)^3$ is not cyclic, because $4\cdot 1=4=-(1+i)^4=0$, so $4\cdot a=4\cdot (1\cdot a)=(4\cdot 1)\cdot a=0\cdot a=0$. Still, $1+1=2\not=0$. So the additive group you have is $\mathbb Z/2\oplus \mathbb Z/4$.

As a ring, there is also an injection $\mathbb Z/4\to \mathbb Z[i]/(1+i)^4$, which we can follow by the quotient $\mathbb Z[i]/(1+i)^4\to \mathbb Z[i]/(1+i)^3$. Since $\mathbb Z/4$ is generated by $1$, and since its image in $\mathbb Z[i]/(1+i)^3$ is still order $4$, the map is injective. Thus, there's an injection of the ring $\mathbb Z/4$ to $\mathbb Z[i]/(1+i)^3$. But of course the image is not the whole thing.

There are fancier reasons for $\mathbb Z/2^n$ not to be isomorphic to $\mathbb Z[i]/(1+i)^n$, so it's not just a one-off. Namely, all such quotients are images of the $2$-adic completion $\mathbb Z_2$ of $\mathbb Z$, and of the $(1+i)$-adic completion $\mathbb Z[i]_{1+i}$ of $\mathbb Z[i]$. If there were a $\sqrt{-1}$ in $\mathbb Q_2$ (which there is not, as we can see already mod $4$), then these would be isomorphic. In contrast, for example, $\mathbb Z/5^n$ is isomorphic to $\mathbb Z[i]/(2+i)^n$, and in two different ways, because there is a $5$-adic $\sqrt{-1}$. :)

EDIT: as asked in a comment... while $\mathbb Z[i]/(1+i)^3\approx \mathbb Z/2\oplus \mathbb Z/4$ as abelian groups, can we determine whether or not these two things are isomorphic as rings? Methodology-wise, if we could hit upon a ring isomorphism by experiment, that'd be good; however, after failing at this for a while, we'd want a way to prove that they're not isomorphic as rings (without listing all the group isomorphisms and seeing that they fail to be ring isomorphisms). So, how to characterize these two rings? The $\mathbb Z[i]/(1+i)^3$ is a quotient of a commutative ring by a power of a prime ideal, so every zero divisor $x$ (meaning $x\not=0$, and there is $y\not=0$ such that $xy=0$) is actually nilpotent (meaning $x^n=0$ for some $n$). In contrast, $\mathbb Z/2\oplus \mathbb Z/4$ has non-nilpotent zero-divisors, such as $x=1\oplus 0$ and $y=0\oplus 1$: their product is $0$, but $x^2=x$ and $y^2=y$, etc.