# Building quotient rings

The quotient rings are following:

$\mathbb{Z}[i]/(1+i)$,

$\mathbb{Z}[i]/(1+2i)$,

$\mathbb{Z}[\sqrt{-2}]/(2)$,

$\mathbb{Z}[\sqrt{-2}]/(1+ \sqrt{-2})$.

I know that the two first are likely to be isomorphic to $\mathbb{Z}_n$ for some $n$, polynomial residues should be used somehow.

• What is the question? – Jorge António Jun 26 '14 at 9:30
• @JorgeAntónio to describe the quotient rings, they should be $\mathbb{Z}_n$ for some $n$, or some cartesian products of $\mathbb{Z}_n$ rings, or something like that. – Andrey Moiseev Jun 26 '14 at 9:33
• @o2genum I'm not sure this is a concrete question - you already have described the rings! If you want to describe them in another form, you need to be quite specific about what that form should be. Asking whether the first two are isomorphic to $\mathbb{Z}_n$ is a perfectly good question, but "what ring is $\mathbb{Z}[\sqrt{-2}]/(2)$ isomorphic to?" is too open-ended (unless there's a "standard" ring it's isomorphic to that I don't immediately see). – mdp Jun 26 '14 at 9:41
• I am very partial towards applying the isomorphism theorems in cases like this. For example, in the first, we can consider $\phi:\mathbb{Z}[i] \rightarrow \mathbb{Z}_2$ defined such that $(a+bi) \mapsto (a+b)_2$. This might not get you the whole way through your problem, but it can help with at least a couple! – Kaj Hansen Jun 26 '14 at 9:51

I would say that you should look first at the structure of abelian group of each of the quotients you mentioned. You can use the Classification of Finitely Generated Modules over PIDs to classify each quotient as an abelian group. (see for example link)

I will illustrate the method for the first two quotients.

i) Pick the $\mathbb{Z}$-basis $1,1+1$ for $\mathbb{Z}[i]$. By passing to quotient we see that the obtained abelian group is generated by the image of $1$, therefore we get just a copy of $\mathbb{Z}$.

ii) Pick the $\mathbb{Z}$-basis $1+2i,i$ for $\mathbb{Z}[i].$ And again, when passing to the quotient you will get a copy of $\mathbb{Z}$ (where the quotient is generated by the image of $i$).

To understand the ring structure on each quotient you should do some calculations. For instance:

i) The ring $\mathbb{Z}[i]/(1+i)$ is generated as an abelian group by the image of $1$, that we will denote by $\bar{1}$, on the quotient. Therefore given two elements $x=n \bar{1}$ and $y=m \bar{1}$, for given integers $n$ and $m$, what you will get when multiplying them is just $xy= nm \bar{1}^2=nm\bar{1}$. Also $\bar{1}$ is the identity of the ring. From this we conclude that $\mathbb{Z}[i]/(1+i)$ is isomorphic, as a ring, to $\mathbb{Z}$.

ii) $\mathbb{Z}[i]/(1+2i)$ is generated as an abelian group by the image of $i$ on the quotient, that we will denote by $\bar{i}$. In this case we see that $\bar{i}\bar{i}=\bar{i}^2=\bar{i^2}=\bar{-1}=2 \bar{i}$. Where the last equality follows because $-1=-(1+2i)+2i$, in $\mathbb{Z}[i].$ Therefore, given two elements $x=n\bar{i}$ and $y=m\bar{i}$, where both $n$ and $m$ are integers, their product is given by $xy=2nm \bar{i}$. The identity of this ring is $-2 \bar{i}$ as you should check.

I often like to write

$$\mathbf{Z}[\mathbf{i}] / (a + b \mathbf{i}) \cong \mathbf{Z}[x] / (x^2 + 1, a + b x)$$

and simplify. (or alternatively, compute by modding out by $a+bx$ first)