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I have a three-part problem I'm working on, and the first and last part are easy, but the second is giving me trouble.

Let $n\in\mathbb{Z}$ with $\sqrt{n}\notin\mathbb{Z}$, and consider the ring $R=\mathbb{Z}[\sqrt{n}]=\{a+b\sqrt{n}\,|\,a,b\in\mathbb{Z}\}$. Prove:

(a) for every non-zero $z\in R$, there is a non-zero $z'\in R$ such that $zz'\in\mathbb{Z}$.

(b) Conclude from (a) that $R/I$ is finite for every non-zero ideal $I$.

(c) Conclude from (b) that every non-zero prime ideal of $R$ is maximal.

Now, part (a) is very straightforward: take $z=a+b\sqrt{n}$ non-zero in $R$. Then $a$ and $b$ are not both zero. So $z'=a-b\sqrt{n}$ is also non-zero in $R$. And $zz'=a^2-b^2n\in\mathbb{Z}$.

Further, given that we believe (b), part (c) is also easy: let $I$ be a non-zero prime ideal in $R$. Then $R/I$ is a finite integral domain, and thus a field. So in fact $I$ is a maximal ideal.

However, I'm having trouble with (b). What I have so far: Let $I$ be a non-zero ideal of $R$ and suppose $R/I$ is not finite. Take $z\neq 0$ in $I$. Then there is some $z'\neq 0$ in $R$ such that $zz'\in\mathbb{Z}$. But $I$ is an ideal, so in fact $zz'\in I$....

And here's where I've stalled. Any help is appreciated. Thanks!

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The crucial thing to spot is that, if we have some integer m in I, then in fact the whole ideal mR is in I, so [R : I] is at most [R : mR], and you can work out explicitly what this is in this case. –  Billy Aug 22 '11 at 0:15
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3 Answers

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As you say, there is a non-zero integer $m \in I$. We have a surjective homomorphism of abelian groups $$ \mathbf Z \oplus \mathbf Z \to R/I \qquad \text{given by} \qquad (a, b) \mapsto a + b\sqrt{n} \mod{I}. $$ Without doing much work, what can you say about the kernel of this map? It clearly contains $(m, 0)$.

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Thanks for your help! –  Bey Aug 22 '11 at 0:35
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HINT $\rm\ mod\ I:\: \ m\equiv 0\:\ \Rightarrow\:\ A\: +\: B\ \sqrt{n}\:\ \equiv\:\ a\: +\: b\ \sqrt{n}\quad$ for $\rm\ 0\le a,b < m\ \Rightarrow\ |R/I| \le m^2$

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Mostly as a comment, I would like to mention that the result of part b) holds under a more general hypothesis. Namely, let $R$ be any domain whose additive group $(R,+)$ is isomorphic to the additive group $(\mathbb{Z}^n,+)$ for some $n$. (In particular, each of the the given rings has additive group isomorphic to that of $\mathbb{Z}^2$.) Then $R$ is a finite norm domain: for any nonzero ideal $I$ of $R$, $R/I$ is finite. As in the OP's question, this implies that the Krull dimension is at most one, since a finite integral domain is a field.

For a proof see e.g. Proposition 6 of this expository article on factorization in integral domains. The basic idea is the same as above: it is enough to show that any nonzero ideal contains a nonzero integer, and indeed it is enough to show this for principal ideals, since every nonzero ideal contains a nonzero principal ideal.

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A question on terminology: When you say, "Let R be a domain...," what exactly do you mean? I've never heard domain used without some descriptor -- e.g., integral domain, Euclidean domain, etc. –  Bey Aug 22 '11 at 3:54
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@Bey "Integral domain" is frequently contracted to "domain". –  Bill Dubuque Aug 22 '11 at 4:32
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@Bey: By a domain I mean a commutative ring without zero divisors -- so, yes, the same thing as an "integral domain". –  Pete L. Clark Aug 22 '11 at 4:51
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