Let $R=\mathbb{Z}[\sqrt{-5}]$ and let $\mathfrak{i}=(2,1+\sqrt{-5})$ the ideal generated in $R$ by $2$ and $1+\sqrt{-5}$. I want to prove that $\mathfrak{i}$ is prime. So i considered the surjective morphism of rings given by $$\phi:\frac{R}{2R}\longrightarrow \frac{R}{\mathfrak{i}}$$ sending $r+2R$ to $r +\mathfrak{i}$, for every $r\in R$.
As an abelian group, $R$ is isomorphic to $\mathbb{Z}\oplus\mathbb{Z}$ hence the quotient $\frac{R}{2R}$ is isomorphic to $\frac{\mathbb{Z}}{2\mathbb{Z}}\oplus\frac{\mathbb{Z}}{2\mathbb{Z}}$ and so has cardinality $=4$. By surjectivity of $\phi$, the cardinality of $\frac{R}{\mathfrak{i}}$ should be a divisor of $4$, so $1,2$ or $4$. Suppose it is $1$: then $R=\mathfrak{i}$, but this is impossible since $R^2=R$ while $\mathfrak{i}^2=2R$.
Suppose $4$: then $\phi$ is an isomorphism, hence $2R=\mathfrak{i}$, so $4R=2R$, which is falce since $4R$ is properly contained in $2R$.
Hence the only possibility is for cardinality $=2$, which means that $\frac{R}{\mathfrak{i}}$ is a field and $\mathfrak{i}$ certainly prime.
My question is: do you think what i said is right? In particular, i'm not sure of what i said about the case $=4$, where i deduce $2R=\mathfrak{i}$ from the fact that $\phi$ is an iso.