4
$\begingroup$

Let $N,P$ be submodules of an $R$-module $M$ and let $S$ be a multiplicative subset of $R$. I think I proved $S^{-1}(N \cap P) = S^{-1}N \cap S^{-1} P$ but since my proof is not the same as the one given in Atiyah-MacDonald on page 39 I suspect there is something wrong with it. Can you tell me please what's wrong here:

Claim: $S^{-1}(N \cap P) = S^{-1}N \cap S^{-1} P$

Proof:

$\frac{m}{s} \in S^{-1}N \cap S^{-1} P \iff$ $\frac{m}{s} \in S^{-1}N$ and $\frac{m}{s} \in S^{-1}P \iff m \in N$ and $m \in P \iff m \in N \cap P \iff \frac{m}{s} \in S^{-1}(N \cap P)$.

  • 2
    So for example, if you take $\mathbb{Z}$ and its module $\mathbb{Z}$ with submodule $N=6\mathbb{Z}$, then localize at the powers of 2, then $\frac{6}{2}=3$ is in the localization of $N$, but $3$ is not in $N$.2012-06-13
  • 1
    @rschwieb Same example! I wish I were able to think of something more geometric; there's probably something with faithful flatness that I'm forgetting.2012-06-13
  • 0
    @DylanMoreland bah why did you delete your comment? The whole reason I put up the concrete example is to illustrate your comment :P2012-06-13
  • 0
    @rschwieb Ah, sorry! I'm trying to be better about cleaning up comments. The only reason it was ever a comment was because I hadn't thought of an example and wanted to see whether the proof still went through.2012-06-13
  • 0
    Looks like they were posted approximately simultaneously.2012-06-13

1 Answers 1

4

In your notation, it doesn't have to be the case that $m \in N$. There just needs to be an $s' \in S$ such that $s'm \in N$. For example, take $R = M = \mathbf Z$, $N = 6\mathbf Z$, and $S = \{1, 2, 2^2, \ldots\}$. But having made this change, I think you can complete your proof.

  • 0
    Yes.. and then everything matches! Gonna clear that out. Thanks!2012-06-13
  • 0
    Thank you! In particular for the example. Now: $$ \frac{m}{s} \in S^{-1}(N \cap P) \iff \exists s^\prime: s^\prime m \in N \cap P \iff \exists s^\prime: s^\prime m \in N \land s^\prime m \in P$$ $$ \iff \frac{m}{s} \in S^{-1}N \cap S^{-1}P$$2012-06-13
  • 3
    @MattN. In the last $\Leftarrow$, is it so clear that you can choose the same $s'$ for both $N$ and $P$? I think there's something to say.2012-06-13
  • 0
    If $sm \in N$ and $s^\prime m \in P$ then $ss^\prime m \in N \cap P$ since $N,P$ are submodules and hence closed under the action of $R$. : ) I'm glad I asked this question. Thank you!2012-06-13