The gaussian integers modulo a pair of complex numbers 
Taking a pair of Gaussian integers $u,v$ satisfying
There exist $a,b \in \mathbb Z$

*

*$v^2 = av+bu$

*$u^2 = av+bu$

*$uv = av+bu$
(Diffrent $a,b$)
define the equivalence relation $\sim$ by $z\sim w \iff \exists a,b \in \mathbb Z, z-w=au+bv$
Let $\mathbb Z[i]/\sim$ be the gaussian integers modulo $u,v$

For example, if $u=2i,v=2$ we get a four element ring with $0,1,i,1+i$ with the multiplication table:
$$\begin{matrix} \cdot && 0 && 1 && i && 1+i \\ 0 && 0 && 0 && 0 && 0 \\ 1 && 0 && 1 && i && 1+i\\ i&& 0 && i && 1 && 1+i \\ 1+i && 0 && 1+i && 1+i && 0\end{matrix}$$
Are these objects studied?
Do any of you have a reference to an article about them?
if not, I have some questions:

*

*is there a way to use $u,v$ to find properties of the ring? for example, in the example there was a nilpotent element, $1+i$, how could I have predicted it using $2,2i$?

*Do there exist pairs for which this is a field? if so, is there a criterion for these?

*Is there a criterion for when this is the zero(trivial) ring?

*Can this be expressed as some "multiplication" of normal cyclic rings? For example, the ring I mentioned is isomorphic to $\mathbb Z_4$($0 \mapsto 0, 1 \mapsto 1, 2 \mapsto i+1, 3\mapsto i$), I've been trying to do it on $2-2i,2+2i$ and have not managed it.

 A: Another way to state your assumption about $u$ and $v$ is that the additive subgroup generated by $u$ and $v$ is closed under multiplication. Thus, this additive subgroup is close to being an ideal, but in general it need not be!
As a result, multiplication on $\mathbb{Z}[i]/\sim$ is not always well-defined. For example, let $u = 2, v = 3$. Then $u^2 = 4 = 2u + 0v$, $v^2 = 9 = 0u + 3v$, and $uv = 6 = 3u + 0v$. Note that $1 - 0 = (-1)u + 1v$, so $[1] = [0]$ in $\mathbb{Z}[i]/\sim$ (where $[\bullet]$ means "the equivalence class of $\bullet$"). If multiplication were well-defined, we would get
$$[m] = [m][1] = [m][0] = [0]$$
for all $m \in \mathbb{Z}[i]$, but this is not the case, for example $[i] \neq [0]$.
In fact, $([x],[y]) \mapsto [xy]$ is a well-defined ring multiplication operation on $\mathbb{Z}[i]/\sim$ if and only if the additive subgroup generated by $u$ and $v$ is an ideal. This turns out to be equivalent to the following:

*

*There exist integers $a$ and $b$ such that $iu = au + bv$

*There exist integers $a$ and $b$ such that $iv = au + bv$
These two conditions actually imply the three conditions you gave! Note that your example of $u = 2i$, $v = 2$ does satisfy these conditions, which is why you got an actual ring out in the end.
For the rest of this answer, I'll suppose we've imposed the two conditions above instead of the three conditions you gave, so that these quotients always make sense as rings. Let $I = \{au + bv : a,b \in \mathbb{Z}\}$, which we now know is an ideal of $\mathbb{Z}[i]$, and let $R = \mathbb{Z}[i]/I = \mathbb{Z}[i]/\sim$. Let me try to answer your questions:

Are these objects studied?

Yes! It turns out that these objects are just arbitrary quotient rings of $\mathbb{Z}[i]$. Quotient rings are a basic construction in ring theory, and $\mathbb{Z}[i]$ is very well-studied.
Why are these things arbitrary quotients of $\mathbb{Z}[i]$? This is the same as saying that every ideal of $\mathbb{Z}[i]$ is generated, as an additive group, by two elements. This is true because every ideal of $\mathbb{Z}[i]$ is principal ($\mathbb{Z}[i]$ is a Euclidean domain), and the principal ideal $(x)$ is generated by $x$ and $ix$ as an additive group.

Do any of you have a reference to an article about them?

No, this would be like finding an article about cyclic groups -- these objects are too basic to be the focus of an article. To learn more about them, I'd recommend reading some introductory abstract algebra books (ring theory in particular). To dive a little deeper, you could read some introductory algebraic number theory, where $\mathbb{Z}[i]$ is an important classical example.

Is there a way to use $u,v$ to find properties of the ring? for example, in the example there was a nilpotent element, $1+i$, how could I have predicted it using $2,2i$?

Yes, there is a way to use $u$ and $v$ to find properties of the ring. Since $u$ and $v$ generate $I$ as an additive subgroup, they also generate $I$ as an ideal. Now $I$ is generated (as an ideal) by $x = \gcd(u,v)$. Note: the choice of $\gcd$ is non-unique, but any $\gcd$ will generate $I$, and you can compute a $\gcd$ using the Euclidean algorithm. In your specific example, we can take $x = 2 = \gcd(2,2i)$.
Now $R = \mathbb{Z}[i]/I = \mathbb{Z}[i]/(x)$ will have nilpotent elements if and only if $(x)$ is not a radical ideal (this is just by definition of "radical"). $(x)$ is radical if and only if $x$ is square-free (standard argument about PID's). An element of a ring is said to be square-free if $y^2 \mid x \implies y$ is a unit, and in $\mathbb{Z}[i]$ the units are $1, -1, i, -i$. In your specific example, $x = 2$ is not square-free, because $2 = i(1-i)^2$, so $R$ has a nilpotent element. In this case, we can choose the nilpotent element to be $[1-i]$ (which is equal to $[1+i]$), because $(1-i)^2 = (-i)2 \in I$. In general, to check whether or not $R$ has nilpotent elements, we do the following:

*

*Compute the prime factorizations $u$ and $v$ in $\mathbb{Z}[i]$ (there is an algorithm for doing this).

*Use these prime factorizations to compute $x = \gcd(u,v)$ and its prime factorization.

*If $x$ has any repeated prime factor, so $x = p^2 y$ for some gaussian prime $p$, then $[py]$ will be nilpotent in $R$ (in particular, its square will be zero). Otherwise, $R$ will not have any nilpotent elements.


Do there exist pairs for which this is a field? if so, is there a criterion for these?

Yes, $R = \mathbb{Z}[i]/I$ will be a field if and only if $I = (x)$ is a maximal ideal, if and only if $x$ is prime. This is actually particularly easy to check: $x = a + bi$ is prime if and only if one of the following holds:

*

*$a = 0$, $\lvert b \rvert$ is prime, and $\lvert b \rvert \equiv 3 \pmod{4}$

*$b = 0$, $\lvert a \rvert$ is prime, and $\lvert a \rvert \equiv 3 \pmod{4}$

*$a \neq 0$, $b \neq 0$, and $a^2 + b^2$ is prime


Is there a criterion for when this is the zero(trivial) ring?

Yes. $R = \mathbb{Z}[i]/I$ is the zero ring if and only if $I = \mathbb{Z}[i]$, if and only if $1 \in I = (x)$, if and only if $x$ is a unit, if and only if $u$ and $v$ are coprime. Again, this is computable by just finding the factorizations of $u$ and $v$.

Can this be expressed as some "multiplication" of normal cyclic rings?

This is a little tricker, and the answer is no! See here for a general discussion of how to compute the isomorphism type of the quotient. Important examples are:

*

*If $x = a + bi$ with $a$ and $b$ coprime, then $R \cong \mathbb{Z}_{a^2 + b^2}$

*If $x = 3$, then $R \cong \mathbb{Z}_3[X]/(X^2+1)$ is a field (as discussed earlier), and we have $\lvert R \rvert = 9$ since this field is $2$-dimensional over $\mathbb{Z}_3$. Thus, $R \cong \mathbb{F}_9$ is not a product of quotients of $\mathbb{Z}$.

All of this is (more or less) elementary abstract algebra, so I'd recommend learning more abstract algebra if you want to understand these things more deeply.
