Isomorphism of quotient rings

Question

In a course on algebraic number theory, the lecturer says

$$\mathcal{O}_K\cong \mathbb Z\left[\frac{1+\sqrt d}{2}\right] \cong\frac{\mathbb Z[x]}{\left( x^2-x-\frac{d-1}{4} \right)}.$$

This means, for a prime $p\in \mathbb Z$, we have

$$\frac{\mathcal{O}_K}{p\mathcal{O}_K}\cong \frac{\mathbb F_p[x]}{\left( x^2-x-\frac{d-1}{4} \right)}.$$

When I tried to justify this claim, I found up a theorem from a course I took on group theory (the third isomorphism theorem) which said that if $G$ is a group with $L, K \unlhd G$, $K\leq L$ then

$$\frac{G}{L} \cong \frac{G/K}{L/K}$$

and thought about what the right generalization of this would be for rings. I think in my case it would be

\begin{align*} \frac{\mathcal{O}_K}{p\mathcal{O}_K} &\cong \left. \frac{\mathbb Z[x]}{\left( x^2-x-\frac{d-1}{4} \right)} \middle/ p\left(\frac{ \mathbb Z[x]}{\left( x^2-x-\frac{d-1}{4} \right)}\right) \right.\\ &\cong \frac{\mathbb Z[x]}{p\mathbb Z[x]} \end{align*}

What am I doing wrong? Does the result for groups not generalize to rings? Have I made a silly mistake?

Answer 1

So the isomorphism theorem you're quoting is true for rings as well as groups, you just assume $I\subseteq J\subseteq R$ and $I,J$ are ideals. Then

$$R/I\bigg/ J/I\cong R/J$$

Now what are the ideals in your case and what's the ring? Well $R=\Bbb Z[x]$, $J=\left(p,x^2-x-{d-1\over 4}\right)$ and $I_1=(p), I_2=\left(x^2-x-{d-1\over 4}\right)$.

Then since we have

$$\Bbb F_p[x]/\left(x^2-x-{d-1\over 4}\right)\cong R/I_1\bigg/ J/I_1\cong R/J\cong R/I_2\bigg/ J/I_2\cong \mathcal{O}_K/p\mathcal{O}_K$$

we can conclude the result you're looking for.