# Prove that every nonzero prime ideal is maximal in $\mathbb{Z}[\sqrt{d}]$

$$d \in \mathbb{Z}$$ 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.

• It's not important for $d$ to be squarefree. The result is true for any $d$ that is not a square (including ${\mathbf Z}[\sqrt{12}]$, for instance). – KCd Dec 15 '13 at 3:59

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.

• You omitted a crucial point: $n\ne 0$ because $\ldots$ – Bill Dubuque Dec 18 '13 at 20:12
• Well, yes... but even if $d$ is a square we can choose a,b appropriately (e.g. by $a \mapsto a+1$) to get $n \neq 0$. – benh Dec 18 '13 at 23:13

If the ideal is prime, almost by definition the quotient has no zero divisors.

On the other hand, since $R$ is a finite generated abelian group, the quotient $R/P$ is also a finitely generated abelian group, and to show it is finite it is enough to show that $R/P$ has finite exponent.

If $P$ is non-zero, there is a non-zero element $x=a+b\sqrt d$ in $P$, and then $e=a^2-db^2=(a-b\sqrt d)x\in P$; you can check easily that $e\neq0$. It follows that the product of every element of $R/P$ by $e$ is zero, and therefore the exponent of the abelian group $R/P$ divides $e$.

• I'm still unsure as to why the factor ring has no zero divisors. I know it may seem obvious but I can't seem to make the connection. – BlakeM Dec 18 '13 at 19:15
• Suppose $a$ and $b$ are elements of $R$ such that the product in $R/P$ of their classes is zero. This means that $ab$ is an element of $P$ and, since the ideeal is prime, that one of $a$ or $b$ is in $P$. – Mariano Suárez-Álvarez Dec 18 '13 at 19:17
• Great, this makes sense to me. – BlakeM Dec 18 '13 at 19:25