How to compute quotient rings? I am discovering class numbers and ideal class groups. Going through Wikipedia, the example $\mathbb{Z}[\sqrt{-5}]$ is considered along with the ideal $J=(1+\sqrt{-5}, 2)$. Two facts are troubling me (I might haven't understood that well ideals and quotients):


*

*$\mathbb{Z}[\sqrt{-5}]/(1+\sqrt{-5}) \simeq \mathbb{Z}/6\mathbb{Z}$

*$\mathbb{Z}[\sqrt{-5}]/J \simeq \mathbb{Z}/2\mathbb{Z}$


For the first one, I have no idea of how such computations are handled. For the second, I would say that the quotient of $Z/6$ by $Z/2$ is $Z/3$, isn't it?
This is an instance of the more general question I have in mind: how do one generally compute a quotient ring? Is there any standard manner to do so? (at least for well presented rings or groups)
Thanks for any comment and help.
 A: For the first question: $(1 + \sqrt{-5}) (1-\sqrt{-5}) = 6$ and $(1 + \sqrt{-5})\sqrt{-5} = \sqrt{-5}-5$. In particular, in the quotient all the classes are represented by integers (because $[\sqrt{-5}]=[5]$) included between $0$ and $5$ (because $[6]=[0]$). Your ring could still be isomorphic to a proper quotient of $\mathbb{Z}/6\mathbb{Z}$, but somehow you can check that the positive integers smaller than $6$ are not sent to $0$ (because they are not multiples of $1+\sqrt{-5}$). The second case is similar, but also $2$ is sent to $0$, hence you get a quotient of $\mathbb{Z} / 2 \mathbb{Z}$ and again you can somehow check that at least $1$ is not sent to $0$ (i.e., it is not a linear combination of $1+\sqrt{-5}$ and $2$) (Wikipedia proves that $J$ is not principal, so in particular it is not $(1)$).
The intuition you mentioned is wrong. We are applying the Third isomorphism theorem for rings, which says that
$$ (\mathbb{Z} / 6\mathbb{Z}) / ( 2\mathbb{Z} / 6\mathbb{Z} ) \cong \mathbb{Z} / 2\mathbb{Z}.$$
A: First of all,
Claim: $\;a+b\sqrt{-5}\in\langle 1+\sqrt{-5}\rangle\iff 6\,\mid\,(a-b)\;$ in $\;\Bbb Z\;$
Proof: We have that $\;a+b\sqrt{-5}\in\langle1+\sqrt{-5}\rangle\;$ iff
$$a+b\sqrt{-5}=(r+s\sqrt{-5})(1+\sqrt{-5})=r-5s+(r+s)\sqrt{-4}\iff$$
$$\begin{cases}a=r-5s\\{}\\b=r+s\end{cases}\iff6s=b-a\;$$
since this already gives $\;6\,\mid\,b-a\;$ , but then also $\;5b+a=6b-(b-a)=0\pmod6\;$ , thus
$$a+5b=r-5s+5r+5s=6r\implies 6\,\mid\,(a+5b)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\square$$$${}$$
For example, $\;-5+7\sqrt{-5}\in\langle 1+\sqrt{5}\rangle\;$ because $\;7-(-5)=12=0\pmod6\;$ , and indeed:
$$-5+7\sqrt{-5}=(5+2\sqrt{-5})(1+\sqrt{-5})$$
or
$$2+8\sqrt{-5}=(7+\sqrt{-5})(1+\sqrt{-5})$$
Then, we can define
$$\phi:\Bbb Z[\sqrt{-5}]\to\Bbb Z/6\Bbb Z\;,\;\;\phi(a+b\sqrt{-5}):=(a-b)\pmod6$$
Prove this is a ring epimorphism (the multiplicative part is easy but not trivial. One has to be careful there) whose kernel is precisely $\;\langle 1+\sqrt{-5}\rangle\;$ , and now apply the first isomorphism theorem.
The second part follows, as already mentioned in the other answer, from the third (or second or something) isomorphism theorem, since
$$\Bbb Z[\sqrt{-5}]/J=\Bbb Z[\sqrt{-5}]/\langle\,1+\sqrt{-5},\,2\,\rangle \cong$$
$$\cong\left[\Bbb Z[\sqrt{-5}]/\langle1+\sqrt{-5}\rangle\right]\big/\left[\langle1+\sqrt{-5},\,2\rangle/\langle1+\sqrt{-5}\rangle\right]\cong\left(\Bbb Z/6\Bbb Z\right)\big/\left(2\Bbb Z/6\Bbb Z\right)\cong\Bbb Z/2\Bbb Z$$
because we in fact have (by the second isomorphism theorem:$\;\left(M+N\right)/M\cong N/(M\cap N)\;$
$$\langle1+\sqrt{-5},\,2\rangle/\langle1+\sqrt{-5}\rangle\cong\langle2\rangle/\left(\langle 2\rangle\cap\langle1+\sqrt{-5}\rangle\right)$$
But $\;\langle 2\rangle\cap\langle1+\sqrt{-5}\rangle=\left\{\,a+b\sqrt{-5}\;/\;a-b=0\pmod 3\,\right\}\cong\Bbb Z/3\Bbb Z\;$ and etc.
