How can I show $\mathbb Z[\sqrt{-3}]$ is one dimensional?

Question

I have to show that the ring $A=\mathbb Z[\sqrt{-3}]$ is one dimensional. I know this means that it contains no chains $P\subset P'\subset P''$ of prime ideals, but I really don't see where to begin. The only thing I can think of is Krull's height theorem, but applying it here would mean showing $A$ is principal, and I don't think it is. Or is it?

Answer 1

Interpreting $A$ as $\mathbb Z[x]/(x^2+3)$, you are looking at a quotient of the two dimensional ring $\mathbb Z[x]$ by a prime ideal.