The elements of $\mathbb{Z}[\sqrt{-5}]/(2,\sqrt{-5}+1)$ I'm really confused with this one...

How can I determine the elements of the module $\mathbb{Z}[\sqrt{-5}]/(2,\sqrt{-5}+1)$? Or its cardinality?

Does $$\mathbb{Z}[\sqrt{-5}]/(2,\sqrt{-5}+1)=\{\bar{0},\bar{1},\overline{0+\sqrt{-5}}\}?$$ 
 A: Recall that
$$\mathbb{Z}[\sqrt{-5}]=\{a+b\sqrt{-5}\mid a,b\in\mathbb{Z}\}.$$
Therefore, as an abelian group (i.e., forgetting about the multiplicative structure for a second), we can make an isomorphism.
$$\mathbb{Z}[\sqrt{-5}]\cong\mathbb{Z}^2,\quad 1\mapsto (1,0),\quad \sqrt{-5}\mapsto (0,1).$$
Under this isomorphism, the ideal $(2)\subset\mathbb{Z}[\sqrt{-5}]$ corresponds to the subgroup of $\mathbb{Z}^2$ generated by $(2,0)$ and $(0,2)$. Therefore (again, as abelian groups)
$$\mathbb{Z}[\sqrt{-5}]/(2)\cong \mathbb{Z}^2/\langle (2,0),(0,2)\rangle\cong (\mathbb{Z}/2\mathbb{Z})^2,$$
and because representatives for $\mathbb{Z}^2/\langle (2,0),(0,2)\rangle$ are
$$\overline{(0,0)},\quad \overline{(1,0)},\quad \overline{(0,1)},\quad \overline{(1,1)},$$
we can undo our isomorphism and conclude that representatives for $\mathbb{Z}[\sqrt{-5}]/(2)$ are
$$\overline{0},\quad \overline{1},\quad \overline{\sqrt{-5}},\quad \overline{1+\sqrt{-5}}.$$
Now, $\mathbb{Z}[\sqrt{-5}]/(2,1+\sqrt{-5})$ is just $\mathbb{Z}[\sqrt{-5}]/(2)$ further quotiented by $1+\sqrt{-5}$ (or rather $\overline{1+\sqrt{-5}}$).
You could either determine which of
$$\overline{0},\quad \overline{1},\quad \overline{\sqrt{-5}},\quad \overline{1+\sqrt{-5}}$$
differ by a multiple of $1+\sqrt{-5}$ (remember, this means a $\mathbb{Z}[\sqrt{-5}]$-multiple, not a $\mathbb{Z}$-multiple), or determine what the ideal $(2,1+\sqrt{-5})$ corresponds to as a subgroup of $\mathbb{Z}^2$, or compute representatives modulo $1+\sqrt{-5}$ of these four elements. 
A: Almost. Let's call $\alpha:=\sqrt{-5}$, and let $a\equiv b$ denote that $a-b\in\, (2,\,1+\alpha)$. Then, as $1\equiv -1$, we also have
$$1+\alpha\equiv 0 \ \implies\ \alpha\equiv -1\equiv 1$$
So that, the quotient has only $2$ elements.
For $0\not\equiv 1$ observe that $a+b\alpha\in (2,\,1+\alpha)\,\implies\,2\,|\,a+b$.
A: Number theory perspective of this problem is: prime ideal factorization. 
The prime ideal $(2)$ in $\mathbb{Z}$ is totally ramified in $\mathbb{Z}[\sqrt{-5}]$
and the ideal $(2, 1+\sqrt{-5})$ is a prime ideal lying over $(2)$
In $\mathbb{Z}[\sqrt{-5}]$, we can write 
$$
(2)=(2, 1+\sqrt{-5})^2,$$
and this can be checked by direct calculation. 
Thus, the residue field extension 
$$\mathbb{Z}/2\mathbb{Z}\subset \mathbb{Z}[\sqrt{-5}]/(2, 1+\sqrt{-5})$$
is a trivial extension of finite field $\mathbb{Z}/2\mathbb{Z}$. 
i.e. 
$$\mathbb{Z}[\sqrt{-5}]/(2, 1+\sqrt{-5})\simeq \mathbb{Z}/2\mathbb{Z}.$$ 
A: Consider the map $\phi: \mathbb{Z}[\sqrt{-5} ] \to \mathbb{Z}_2 $ given by $\phi( a+b\sqrt{-5})= a+b.$ This is a surjective ring homomorphism; the hardest part to check is that is respects multiplication. 


*

*$\phi\left( (a+b\sqrt{-5})(c+d\sqrt{-5}) \right) = \phi ( (ac-5bd)+ (ad+bc)\sqrt{5} ) = ac - 5bd + ad+bc$

*$\phi( a+ b\sqrt{-5} ) \phi (c+d\sqrt{-5} ) = (a+b)(c+d) = ac + ad+bc+bd$


These are indeed equal in $\mathbb{Z}_2.$ It is clear that $(2,1+\sqrt{-5})\subseteq \ker \phi.$
If $a+b\sqrt{-5} \in \ker \phi $ then $ a=-b \in \mathbb{Z}_2 $ so $a=-b+2k$ for some $k\in \mathbb{Z}.$ Therefore $a+b\sqrt{-5} = 2(k-b) + b(1+\sqrt{-5})\in (2,1+\sqrt{-5})$ so $\ker \phi = (2,1+\sqrt{-5}).$ The first isomorphism theorem gives the result.

As per BenjaLim's deleted answer, we have
$$\begin{eqnarray*} \Bbb{Z}[\sqrt{-5}]/(2,1 + \sqrt{-5}) &\cong& \Bbb{Z}[x]/(x^2 + 5)/(2,1+x)/(x^2 + 5)\\
&\cong& \Bbb{Z}[x]/(2,1+x)\\
&\cong& \Bbb{Z}/2\Bbb{Z}\end{eqnarray*}.$$
OP asked why the last isomorphism holds. It is because $\mathbb{Z}[x]/(1+x) \cong \mathbb{Z}.$ To see this, consider the homomorphism $\varphi: \mathbb{Z}[x] \to \mathbb{Z}$ given by $\varphi( p(x) ) = p(-1).$ It is not hard to verify it is surjective with kernel $(1+x)$ which yields the isomorphism. Then 
$$\begin{eqnarray*} \Bbb{Z}[x]/(2,1 + x) &\cong& \Bbb{Z}[x]/(x+1)/(2,1+x)/(x+1)\\
&\cong& \Bbb{Z}/(2)= \Bbb{Z}/2\Bbb{Z}\end{eqnarray*}.$$
