$d \in \mathbb{Z}$ is a square-free integer ($d \ne 1$, and $d$ has no factors of the form $c^2$ except $c = \pm 1$), and let $R=\mathbb{Z}[\sqrt{d}]= \{ a+b\sqrt{d} \mid a,b \in \mathbb{Z} \}$. Prove that every nonzero prime ideal $P \subset R$ is a maximal ideal.
I have a possible outline which I think is good enough to follow.
I think that we need to first prove that every ideal $I \subset R$ is finitely generated.
So if $I$ is non-zero, then $I \cap \mathbb{Z}$ is a non-zero ideal in $\mathbb{Z}$.
Then I need to find $I \cap \mathbb{Z} = \{ xa \mid a \in \mathbb{Z} \}$ for some $x \in \mathbb{Z}$. That way if I let $J$ be the set of all integers $b$ such that $a+b\sqrt{d} \in I$ for some $a\in \mathbb{Z}$, then if there exists a integer $y$ such that $J=\{ yt \mid t\in \mathbb{Z} \}$, then there must exist $s \in \mathbb{Z}$ such that $s+y\sqrt{d} \in I$.
Then all I need to show is that $I = ( x,s+y\sqrt{d} )$.
Now I need to derive that the factor ring $R / P$ is a finite ring without zero divisors, also finite, then since every finite integral domain is a field, every prime ideal $P \subset R$ is a maximal ideal, then I'll be done.
Best Answer
Sounds good. In order to show that $I \cap \Bbb Z$ is a non-zero ideal, it is enough to notice that for $a+b\sqrt{d}\in I$ you have $(a+b\sqrt{d})(a-b\sqrt{d}) = a^2 -db^2 =:n\in \Bbb Z \cap I$. Now by writing $R = \Bbb Z[X]/(X^2-d)$ you can show that $R/P$ is a quotient of the finite ring $(\Bbb Z / n\Bbb Z)[X]/(X^2-d)$ where $X^2-d$ denotes the reduced polynomial in $(\Bbb Z/n \Bbb Z)[X]$. So indeed, $R/P$ is finite without zero divisors.