## Compute a quotient ring

Let $K = \mathbb{Q}(\sqrt{-37})$ with ring of integers $\mathcal{O} = \mathbb{Z}[\sqrt{-37}]$. Consider the ideal $A = (2,1+\sqrt{-37})$. Show that $\mathcal{O}/A \cong \mathbb{Z}/(2)$.

Let $a+b\sqrt{-37} \in \mathcal{O}$, and suppose $a-b \equiv k$ mod 2 where $k \in \{0,1\}$. Now $a+b\sqrt{-37} \equiv k$ mod $A$. In particular, $\mathcal{O}/A = \{\overline{0}, \overline{1}\}$.

Suppose $1 \in A$. Then we have $1 = 2(a+b\sqrt{-37}) + (1+\sqrt{-37})(h+k\sqrt{-37})$ for some integers $a,b,h,k$. Comparing coefficients, we have $2a+h-37k = 1$ and $2b + h + k = 0$. Mod 2, we have $h+k \equiv 1$ and $h+k \equiv 0$, a contradiction. So $1 \not\equiv 0$ mod $A$.

So $\mathcal{O}/A \cong \mathbb{Z}/(2)$.