# Prime ideals of $\mathbb{Z}[\sqrt{d}]$

How could I prove that if $I$ is a prime ideal of $\mathbb{Z}[\sqrt{d}]$, where $d$ is a squarefree integer, is such that $R/I$ is finite?

I know that $I$ prime implies $R/I$ is an integral domain and if it is finite it is a field, then $R/I$ is a field and therefore $I$ is maximal.

I also know that $\mathbb{Z}[\sqrt{d}]$ is a Princial Ideal Domain since this is an Euclidean Domain and I could use this to prove what I need (every prime is maximal).

But I would like to understand why is $R/I$ finite in that case? I have difficulties to work with quotient rings, that is why I would like to understand. Can anybody help me?

## 1 Answer

All nonzero ideals in $R=\Bbb Z[\sqrt d]$ have the property that $R/I$ is finite. Let $a+b\sqrt d\in I$, $(a,b)\ne(0,0)$. Then $N=a^2-b^2d\in I$ and $N\ne0$. Then $I\supseteq NR$, and $|R/NR|=N^2$. So $|R/I|$ is also finite.

• How do you know the order of $R/NR$ is equal to $N^2$? – Algebear Dec 5 '18 at 23:19
• Also, don't you mean $I\subset NR$, instead of $I\supset NR$? Otherwise, I think the finiteness argument doesn't hold. – Algebear Dec 7 '18 at 10:02
• @GuusPalmer As $\mathbb{Z}$-modules, $\mathbb{Z}[\sqrt{d}] \cong \mathbb{Z} \times \mathbb{Z}$, so $\mathbb{Z}[\sqrt{d}]/(N) \cong \mathbb{Z}/N\mathbb{Z} \times \mathbb{Z}/N\mathbb{Z}$. No, the direction of the containment is correct and moreover, it's what's needed: if modding out by the smaller ideal $NR$ results in a finite quotient, then surely modding out by the larger ideal $I$ also results in a finite quotient. – André 3000 Dec 7 '18 at 17:32
• @André3000 I'm confused. Set theoratically, if you say $A\subset B$ and $A$ is finite, then $B$ can still be infinite. – Algebear Dec 8 '18 at 15:46
• @André3000 Okay, that makes it definitely more clear! This surjectivity is what I missed. – Algebear Dec 8 '18 at 18:55