# Localization of finite modules, or: compatibility of ideal norms with localization at a prime number

Let $A$ be an integral domain and $M$ be a finite $A$-module (where "finite" means "finite cardinality"). Let $S\subset A$ be a multiplicative subset. Is the cardinality of $S^{-1}M$ the same as that of $M$?

I'm interested in the case where $A=\mathcal{O}_K$ is the ring of integers of an algebraic number field $K$, $S=\mathbb{Z}\setminus p\mathbb{Z}$ where $p\in \mathbb{Z}$ is a prime number, and $M=\mathcal{O}_K/\mathfrak{a}$, where $(0)\neq \mathfrak{a}\lhd \mathcal{O}_k$.

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Do you mean $S^{-1}M$ instead of $S^{-1}A$? Otherwise, I don't see where $M$ comes in. – froggie Nov 18 '12 at 16:05
@froggie: sorry, I've updated the question. – user46225 Nov 18 '12 at 16:28

I guess you want to compare $M$ with $S^{-1}M$. The answer depends on $S$.

Let $I=\mathrm{Ann}_A(M)$ be the annihilator of $M$.

• If $S\cap I\ne\emptyset$, then $S^{-1}M=0$.

• More generally, we have $S^{-1}M=\{ x/s \mid x\in M, s\in S\}$. Let $F$ be the set of $x\in M$ such that $S\cap \mathrm{Ann}_A(x)=\emptyset$. Then $x/s=0$ if and only if $x\notin F$. So $S^{-1}M=\{ x/s \mid x\in F, s\in S\}$.

Now suppose $S\cap \mathrm{Ann}_A(x)=\emptyset$ for all $x\in M$. Then the canonical map $M\to S^{-1}M$ is injective. Let us show it is an isomorphism.

Let $x/s\in S^{-1}M$. Consider the multiplication-by-$s$ map $M\to M$, $y\mapsto sy$. It is injective hence bijective because $M$ is finite. So there exists $y\in M$ such that $x=sy$. Thus $x/s=y/1$ belongs to the image of $M\to S^{-1}M$ and we are done.

In your case, to have the same cardinality, you need $p\notin \mathfrak m$ for any maximal ideal of $O_K$ containing $\mathfrak a$.

Edit. Under the finiteness hypothesis on $M$, the canonical map $M\to S^{-1}M$ is always surjective.

Proof: Fix an $x/s\in S^{-1}M$. We have an increasing sequence of submodules $$M_n=\{ y\in M \mid s^ny=0\}.$$ As $M$ is finite, this sequence is stationary. Suppose $M_{r}=M_n$ for all $n\ge r$. Consider the (well defined) multiplication-by-$s^r$ map on $M/M_r$. If $s^r\bar{y}=0$, then $s^ry\in M_r$ and $s^r(s^ry)=0$. Hence $y\in M_{2r}=M_r$ and $\bar{y}=0$. Therefore $M/M_r\to M/M_r$ is injective, thus bijective. Let $y\in M$ such that $s^ry=x$ in $M/M_r$. We have $s^ry=x+z$ with $s^rz=0$. So $$s^{r-1}y=x/s+ zs^r/s^{r+1}=x/s.$$

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Thank you for your answer. Please see this follow-up question: math.stackexchange.com/questions/239972/… – user46225 Nov 18 '12 at 17:42

Let $A = \mathbb{Z}$, and take $M = \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/3\mathbb{Z}$. If $S = \mathbb{Z}\smallsetminus 2\mathbb{Z}$, then $$S^{-1}M = S^{-1}(\mathbb{Z}/2\mathbb{Z})\times S^{-1}(\mathbb{Z}/3\mathbb{Z}).$$ But $S^{-1}(\mathbb{Z}/2\mathbb{Z})= \mathbb{Z}/2\mathbb{Z}$ and $S^{-1}(\mathbb{Z}/3\mathbb{Z}) = 0$, so $S^{-1}M = \mathbb{Z}/2\mathbb{Z},$ which does not have the same cardinality as $M$.

This counterexample does fall into the framework of your more specific question. Here $A = \mathcal{O}_\mathbb{Q}$ and $\mathfrak{a} = 6\mathbb{Z}$.

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Thank you for your answer. Please see this follow-up question: math.stackexchange.com/questions/239972/… – user46225 Nov 18 '12 at 17:43