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Let $K$ be a number field, i.e. a subfield of $\mathbb{C}$ of finite degree over $\mathbb{Q}$. Let $\mathscr{O}_K$ be the ring of integers of $K$, i.e. algebraic integers which are in $K$. Let $I$ be an ideal of $\mathscr{O}_K$. I read many times that the quotient $\mathscr{O}_K/I$ is obviously/clearly a finite ring, but i've never seen a proof. Could someone suggest me how to see this?

bateman
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1 Answers1

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I guess you know that, as an Abelian group, $O_K\cong\mathbb Z^k$, where $k=[K:\mathbb Q]$. Now if $0\neq a\in J$ then $(a)\subset J$, and as $a$ divides its norm $Na\in\mathbb Z$, also $(Na)\subset(a)$. And $\mathbb Z^k/(Na)\mathbb Z^k$ is certainly finite.

user8268
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    Of course, the operative thing here is that "having finite quotients" is a property of a PID $R$ that extends to any finite ring extension $S$ of $R$. – Alex Youcis Sep 07 '13 at 18:17
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    @user8268 Why do $a$ divide norm of $a$? Is it because the norm of $a$ is the product of Galois automorphisms evaluated at $a$ and one of these is the identity? – bateman Sep 08 '13 at 06:44
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    @bateman: yes, that's a way how to prove it. Maybe I shouldn't have used the norm in the answer; simply use any $n\neq0$ such that $n/a\in O_K$ - $n$ can be $p(0)$ for any monic $p\in\mathbb Z[x]$ such that $p(a)=0$. – user8268 Sep 08 '13 at 07:47