Unique prime ideal containing $(2)$ in $\mathbb{Z}[\sqrt{-3}]$

Question

I'm having trouble with an algebraic number theory problem. Let $R = \mathbb{Z}[\sqrt{-3}]$. The problem is to show that $(2, 1 + \sqrt{-3})$ is the unique prime ideal containing the ideal $(2)$, and to conclude that $(2)$ can't be written as a product of prime ideals. Aside from actually solving the problem I don't even understand how proving this implies that $(2)$ can't be factored. Wouldn't we want to show that $(2)$ contains no prime ideals instead? Thanks.

Answer 1

Assume $\mathfrak p$ is a prime ideal of $\mathbb Z[\sqrt{-3}]$ that contains $(2)$. Then $$ (1+\sqrt{-3})(1-\sqrt{-3})=1+3=4=2\cdot 2\in (2)\subset\mathfrak p, $$ hence $1+\sqrt{-3}\in\mathfrak p$ or $1-\sqrt{-3}\in\mathfrak p$. In fact, we know that both are contained in $\mathfrak p$, since $$ (2,1+\sqrt{-3})=(2,-1-1\sqrt{-3})=(2,1-\sqrt{-3})\subset\mathfrak p. $$ To show that $\mathfrak p=(2,1+\sqrt{-3})$, we show that $(2,1+\sqrt{-3})$ is maximal: $$ \frac{\mathbb Z[\sqrt{-3}]}{(2,1+\sqrt{-3})}\cong\frac{\mathbb Z[X]}{(2,1+X,X^2+3)}\cong\frac{\mathbb F_2[X]}{(1+X,X^2+1)}\cong\frac{\mathbb F_2[X]}{(X+1)}\cong\mathbb F_2, $$ where I used that $X^2+1=(X+1)^2=\mathbb F_2[x]$. Since $\mathbb F_2$ is a field, it follows that $(2,1+\sqrt{-3})$ is maximal.

Note that $\mathbb Z[\sqrt{-3}]$ is a number ring, i.e., a subring of a number field ($\mathbb Q(\sqrt{-3})$ is this case). Since each non-zero ideal in a number ring has finite index, it follows that $\mathbb Z[\sqrt{-3}]$ has Krull dimension $1$ (since it is not a field). We also know that $\mathbb Z[\sqrt{-3}]$ is Noetherian. Therefore we have unique primary decomposition of ideals. Note that by the argument above $(2,1+\sqrt{-3})$ is the only ideal that divides $(2)$, hence if $(2)$ factors are a product of prime ideals, we must have $(2)=(2,1+\sqrt{-3})^n$ for some $n\in\mathbb N$. However, note that $$ (2,1+\sqrt{-3})^2=(2,1+\sqrt{-3})(2,1-\sqrt{-3})=(4,2+2\sqrt{-3})=(2)\cdot (2,1+\sqrt{-3}), $$ which is only possibe if $n=1$ (by uniqueness of primary decomposition), but $n=1$ means $(2)=(2,1+\sqrt{-3})$ which is absurd.