How to show that all prime ideals of $\mathbb Z[\sqrt {-3}]$ are maximal?
My attempt:
$\mathbb Z[\sqrt {-3}]\cong \mathbb Z[x]/(x^2+3)$
Let p be prime ideal of $\mathbb Z[\sqrt {-3}]$
SO $\mathbb Z[\sqrt {-3}]/(p(x))$ is integral domain
I could not prove but I think it will be finite.
So it will be field so p become maximal ideal .
Please Help me show above claim
Any Help will be appreciated
Best Answer
You need to exclude the zero ideal: it is prime but not maximal.
If $I$ is a nonzero ideal of $R=\Bbb Z[\sqrt{-3}]$, then $I$ has finite index in $R$, so $R/I$ is a finite ring.
An ideal $I$ in a commutative ring $R$ is maximal iff $R/I$ is a field and is prime iff $R/I$ is an integral domain.
A finite integral domain is a field, by a well-known theorem, so in our example, if $I$ is a non-zero prime ideal, then $R/I$ is an integral domain, so $R/I$ is a field, and $I$ must be maximal.
This argument also works when $R$ any order in an algebraic number field.