Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$d \in \mathbb{Z}$ 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.

share|cite|improve this question
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
up vote 3 down vote accepted

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.

share|cite|improve this answer
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
thanks guys!!!! – PandaMan Dec 19 '13 at 5:48
You have a sign wrong at the definition of $n$. – Mariano Suárez-Alvarez Jan 5 '14 at 0:36
@MarianoSuárez-Alvarez Right, thank you! – benh Jan 5 '14 at 2:44

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 $e$.

share|cite|improve this answer
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-Alvarez Dec 18 '13 at 19:17
Great, this makes sense to me. – BlakeM Dec 18 '13 at 19:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.