I am trying to figure out if any of the following rings are isomorphic $$ \frac{\mathbb{Z}[\sqrt{5}]}{(2)},\;\frac{\mathbb{Z}[\sqrt{-2}]}{(2)},\; \frac{\mathbb{Z}[\sqrt{-3}]}{(1+\sqrt{-3})} $$ I believe two of them are and I think I can write each down in a simpler way but am hoping to confirm.
I think $$ \frac{\mathbb{Z}[\sqrt{5}]}{(2)}\cong \{ a+bx:\;a,b\in\mathbb{Z}/2\mathbb{Z}\} $$ since we may equivalently first quotient $\mathbb{Z}[x]$ by $2$ and then by $x^2-1$, killing all quadratic terms and above.
Similarly, I think $$ \frac{\mathbb{Z}[\sqrt{-2}]}{(2)}=\{ a+bx:\;a,b\in\mathbb{Z}/2\mathbb{Z}\} $$ and $$ \frac{\mathbb{Z}[\sqrt{-3}]}{(1+\sqrt{-3})}\cong \mathbb{Z}/4\mathbb{Z} $$ since this is equivalent to first mapping $x$ to $1$ and then $4$ to $0$.
$$\mathbf{Z}[\sqrt{5}]/(2) \cong \mathbf{Z}/2\mathbf{Z}[x]/(x+1)^2 \cong \mathbf{Z}/2\mathbf{Z}[\varepsilon]/(\varepsilon^2)$$
This is the dual numbers over $\mathbf{Z}/2\mathbf{Z}$. The first isomorphism is because $$x^2 - 5 \equiv x^2 - 1 \equiv (x - 1)(x + 1) \equiv (x+1)^2 \pmod 2. $$ The second is the substitution $\varepsilon = x + 1$.
The ring $\mathbf{Z}[\sqrt{-2}]/(2)$ is isomorphic to the same ring. It is not isomorphic (as a ring) to $\mathbf{Z}/2\mathbf{Z} \oplus \mathbf{Z}/2\mathbf{Z}$ since, for example, $\sqrt{-2}$ is a non-zero element whose square is $0$; no element of $\mathbf{Z}/2\mathbf{Z} \oplus \mathbf{Z}/2\mathbf{Z}$ has this property.
The last ring is indeed isomorphic to $\mathbf{Z}/4\mathbf{Z}$ and not isomorphic to the first two because every element of $\mathbf{Z}/2\mathbf{Z}[\varepsilon]/(\varepsilon^2)$ has order 2.