I'm trying to find all the maximal ideal of the ring $\mathbb{Z}[\sqrt{3}] = \{a+b\sqrt{3} : a,b\in \mathbb{Z}\}$.
I have found one, that is $A = \{3a + b\sqrt{3} : a,b \in \mathbb{Z} \} $, and I know a result that $\mathbb{Z}[\sqrt{3}]$ is a Euclidean domain, so it is a PID.
My question is: how can I find other maximal ideals?
Any help would be appreciated. Thanks
Best Answer
Since $\Bbb Z[\sqrt{3}]\cong \Bbb Z[x]/(x^2-3)$, you are looking for the maximal ideals of $\Bbb Z[x]$ which contain $(x^2-3)$. The maximal ideals of $\Bbb Z[x]$ appear among the prime ideals described here, and are exactly the ones generated by two elements.
There are three possibilities that each prime $p$ can fall under:
$x^2-3\pmod{p}$ is irreducible, as in the case when $p=5$. In that case, $(p,x^2-3)$ is maximal. This corresponds to the case when $3$ is not a quadratic residue in $\Bbb F_p$.
$x^2-3\equiv (x-\alpha)(x-\beta) \pmod{p}$, where $\alpha\neq\beta$, as in the case for $p=11$. We have two distinct maximal ideals $(p,(x-\alpha))$ and $(p,(x-\beta))$. This corresponds to the case when $3$ is a quadratic residue in $\Bbb F_p$.
$x^2-3\equiv (x-\alpha)^2 \pmod{p}$, as in the case $p=2$ and $p=3$. (Actually, you should show that these are the only two primes for this case.) Then $(p,(x-\alpha))$ is maximal.