Calculating quotient rings How do I go about calculating what $R = \mathbb Z [\sqrt{-5} ] / \langle 1 + \sqrt{-5} \rangle $ actually is (i.e. how do I find a simpler ring isomorphic to $R$)?
I can see that $\langle \sqrt{-5} \rangle \subsetneq \langle 1 + \sqrt{-5} \rangle$, so $R \subsetneq \mathbb Z[\sqrt{-5}] / \langle \sqrt{-5} \rangle \cong \mathbb Z$. So it's isomorphic to a proper subring of $\mathbb Z$. I can't see how to get any further with this line of reasoning, though.
I can also see that $a + b \sqrt{-5} \equiv a - b \ (\mathrm{mod} \langle 1 + \sqrt{-5} \rangle ) $, and since $4 \in \langle 1 + \sqrt{-5} \rangle $ any element in $R$ is congruent to $0,1,2$ or $3$. How do I proceed?
Thanks
 A: Neither $\sqrt{-5}$ nor $4$ is in the ideal $(1 + \sqrt{-5})$, and you can check this directly: write $a + b\sqrt{-5}$, $a, b \in \mathbf Z$ for a general element of $\mathbf Z[\sqrt{-5}]$ and note that
\[
(1 + \sqrt{-5})(a + b\sqrt{-5}) = (a - 5b) + (a + b)\sqrt{-5} = \sqrt{-5}
\]
would give a solution to the system $a - 5b = 0$ and $a + b = 1$ over the integers. Why can't this happen? Similarly for $4$. Another remark is that there are no proper subrings of $\mathbf Z$, and if you were looking at ideals $\mathfrak a \subset \mathfrak b$ in a ring $A$ then $A/\mathfrak b$ would be a quotient of $A/\mathfrak a$.
One way of finding the quotient is to view $\mathbf Z[\sqrt{-5}]$ as $\mathbf Z[X]/(X^2 + 5)$. Then $\mathbf Z[\sqrt{-5}]/(1 + \sqrt{-5})$ becomes
\[
(\mathbf Z[X]/(X^2 + 5))/(1 + X) \cong (\mathbf Z[X]/(1 + X))/(X^2 + 5) \cong \mathbf Z/(6).
\]
You could use this idea to find out what $\mathbf Z[\sqrt{-5}]/(\sqrt{-5})$ actually is!
A: Hint  $\rm\: f : \mathbb Z \rightarrow \rm R$ is onto by $\rm\sqrt{-5}\equiv -1$. $\rm\ ker\ f =$ integer multiples $\rm\:n\:$  of $\:1+\sqrt{-5}\:$ in $\rm\: \mathbb Z[\sqrt{-5}].\:$
$$\rm 1+\sqrt{-5}\ |\ n\ \ in\ \ \mathbb Z[\sqrt{-5}]\iff \frac{n}{1+\sqrt{-5}}\: =\: \frac{n(1-\sqrt{-5})}6\in \mathbb Z[\sqrt{-5}] \iff\ 6\ |\ n\ \ in\ \ \mathbb Z$$ 
Thus $\rm\:R\:\cong\:  (im\ f)/(ker\ f)\: \cong\: \mathbb Z/6\:\!\mathbb Z$
A: As noted by Dylan, we have 
$$
A:=\frac{\mathbb Z[\sqrt{-5}]}{\left(1+\sqrt{-5}\right)}\simeq\frac{\mathbb Z[X]}{\mathfrak a}\quad,
$$
where $\mathfrak a$ is the ideal of $\mathbb Z[X]$ generated by $X^2+5$ and $X+1$.
Claim: $\mathfrak a$ is the kernel of the epimorphism 
$$
\phi:\mathbb Z[X]\to\mathbb Z/(6),\quad p(X)\mapsto p(-1)\bmod6.
$$
This will imply $A\simeq\mathbb Z/(6)$. 
The inclusion $\mathfrak a\subset\ker\phi$ is clear. 
The equality 
$$
X^2+5=(X+1)\ (X-1)+6
$$ 
shows that $\mathfrak a$ is generated by $X+1$ and $6$. 
If $p(X)$ is in $\ker\phi$, then $p(-1)=6n$ for some $n$ in $\mathbb Z$, and the equalities
$$
p(X)=(X+1)\ q(X)+p(-1)=(X+1)\ q(X)+6n,
$$
with $q(X)$ in $\mathbb Z[X]$, imply that $p(X)$ is in $\mathfrak a$.
