$\mathbf{Z}[\sqrt{-3}]$ and its ideals $(2)$ and $(2,1+\sqrt{-3})$

Question

Consider $\mathbf{Z}[\sqrt{-3}]$, that is obviously not the ring of integers of $\mathbf{Q}[\sqrt{-3}]$, and its ideal $P=(2,1+\sqrt{-3})$. I know that $P^2=(2)P$ and that $P\neq (2)$. I think that $P$ is a prime ideal since its norm $N(P)=N(2)=2^2$ is a power of a prime. I want to prove that $P$ is the unique prime ideal containing $(2)$ and so $(2)$ is not product of prime ideal in $\mathbf{Z}[\sqrt{-3}]$. I think that if $(2)=PQ$, with $Q$ prime ideal, then $N(2)=N(P)N(Q)$ and so $N(Q)=1$, then $Q$ is trivial. But I can't see why $P$ is the unique prime ideal containing $(2)$. Thanks in advance!

Answer 1

My approach would be as follows: Consider the surjective ring homomorphism $$\phi: \mathbb{Z}[\sqrt{-3}] \rightarrow \mathbb{F}_2$$ defined by $$\phi(a+b\sqrt{-3}) = (a+b) \mod 2$$Then the kernel of $\phi$ is your ideal $P$ and thus, according to one of the ring isomorphism theorems, $$\mathbb{Z}[\sqrt{-3}]/P \cong \mathbb{F}_2$$ which is a field, so $P$ is maximal and hence also prime.