Show that $M=\bigcap_{\mathfrak{p}\in\operatorname{Spec}(R)}M_\mathfrak{p}=\bigcap_{\mathfrak{m}\in\text{Max}(R)}M_\mathfrak{m}$ for certain $M$.

$\newcommand{\Spec}{\operatorname{Spec}}$ $\newcommand{\mSpec}{\operatorname{Max}}$

This is a homework from my algebra course. I am in a situation where I think I have found a solution, though somehow there's a condition in the question that I don't need.

Important: I don't want help on the problem itself, I just want to know what's wrong with my proof!

Exercise: Let $R$ be an integral domain with quotient field $K$. Let $M \subset K$ be an $R$-submodule of $K$. Then for each prime ideal $\mathfrak{p} \subset R$ we can regard $M_\mathfrak{p} \subset K$.

Show that $M = \bigcap_{\mathfrak{p} \in \Spec(R)} M_\mathfrak{p} = \bigcap_{\mathfrak{m} \in \mSpec(R)} M_\mathfrak{m}$

as $R$-submodules of $K$.

My proof: the inclusions from left to right are obvious (since $R$ is an integral domain and $M$ is torsionfree the inclusion $M \rightarrow M_\mathfrak{a}$, $a \mapsto \frac{a}{1}$ is injective, so $M$ can be seen as an $R$-submodule of any of the $M_\mathfrak{p}$. The inclusion $\bigcap_{\mathfrak{p} \in \Spec(R)} M_\mathfrak{p} \subset \bigcap_{\mathfrak{m} \in \mSpec(R)} M_\mathfrak{m}$ is always trivial.

Now for the inclusion $\bigcap_{\mathfrak{m} \in \mSpec(R)} M_\mathfrak{m} \subset M$:

Let $\frac{1}{s} \cdot m \in \bigcap_{\mathfrak{m} \in \mSpec(R)} M_\mathfrak{m} \Rightarrow \forall \mathfrak{m} \in \mSpec(R): \frac{1}{s} \cdot m \in M_\mathfrak{m} \Rightarrow \forall \mathfrak{m} \in \max(R): s \notin \mathfrak{m}$.

Therefore $s$ is a unit in $R$ and $\frac{1}{s} \cdot m = s^{-1} \cdot m \in M$.

However, I don't see where my proof uses the fact that $M$ is a submodule of $K$ (torsionfree and integral domain would be enough). This confuses me a bit, so I am afraid to have made a mistake. It would be nice if someone could check this solution and tell me what I have done wrong, because right now I seem to be blind. Thanks a lot.

Edit: I've just realized that all the $M_\mathfrak{p}$ must be submodules of some module $P$ for the $\bigcap$ to be defined. But is this really all the problem?

• Dear @ BenjaLim, Injectivity of $M\hookrightarrow M_\mathfrak{p}$ is determined by the fact that the set of non-zero elements in $M$ annihilated by some element of $R\setminus\mathfrak{p}$ is $\{0\}$. In this because, because $R$ is a domain and $M$ is torsion-free, and $0\notin R\setminus\mathfrak{p}$, the injectivity holds. But even for a domain $R$, given an arbitrary $R$-module $M$ and a prime $\mathfrak{p}$ of $R$, $M\rightarrow M_\mathfrak{p}$ will have kernel consisting of the aforementioned set of elements. – Keenan Kidwell May 4 '13 at 1:01
• @BenjaLim: it's just the other way around. Being a domain is necessary, but not sufficient. However, you might have noticed that I mentioned the fact that I use $M$ torsionfree and $R$ integral domain in my proof, but I don't use the fact of $M$ being "special" in the way of being a submodule of a quotient field (which in my opinion is quite a special case). – Louis May 4 '13 at 1:25
• Being a domain is not necessary or sufficient for there to exist a module injecting into one of its localizations. To talk about torsion-free modules, $R$ should be a domain, and for torsion-free modules, the injectivity holds. But if $R$ is any non-zero ring at all, and $u$ is a unit, then $R\rightarrow R[u^{-1}]$ is an isomorphism, and in particular injective, whether or not $R$ is a domain. On the other hand, for $R=\mathbf{Z}$ and $M=\mathbf{Z}/6\mathbf{Z}$, if we invert $6$, $M$ localizes to zero, so it doesn't inject into its localization. – Keenan Kidwell May 4 '13 at 1:45
• – Watson Jan 24 '17 at 10:37

The fact that you can take the intersection of the $M_\mathfrak{m}$ uses that $M\subseteq K$ ($K$ provides an $R$-module into which all the $M_\mathfrak{m}$ embeds, so it makes sense to take the intersection).
You write a typical element of $\bigcap_\mathfrak{m}M_\mathfrak{m}$ as $m/s$, and then conclude that $s\notin\mathfrak{m}$ for all $\mathfrak{m}$. This is incorrect though you have the right idea. A random element $x$ of $K$ can be written as $a/b$ with $a,b\in A$, but the $a$ and $b$ are irrelevant in this case. What you need to do is start with an arbitrary element $x\in K$, and suppose that it lies in $\bigcap_\mathfrak{m}M_\mathfrak{m}$. Now what does it mean that $x\in M_\mathfrak{m}$ for a given $\mathfrak{m}$? It means there exist $m_\mathfrak{m},s_\mathfrak{m}\in R$ with $s_\mathfrak{m}\notin\mathfrak{m}$ so that $x=m_\mathfrak{m}/s_\mathfrak{m}$. I won't say anymore since you've specifically asked not to be given a complete answer.
• I don't really see my mistake by writing $m/s$ to be honest. We have defined $S^{-1}M$ as the module of pairs $(m,s) \in M \times S$ with the usual operations and modulo the usual equivalence. So I know, when I pick "some element" in $\bigcap_\mathfrak{m} M_\mathfrak{m}$ I'm making choices, but they should not really matter. $s$ will always be in $S = R\backslash \mathfrak{m}$ as the notion $M_\mathfrak{m}$ means location on $R \backslash \mathfrak{m}$. No? – Louis May 4 '13 at 1:13
• Dear @Louis, The point is, you cannot assume that there is a single $s$ which serves as a denominator for your element in each $M_\mathfrak{m}$. You only know that, for each $\mathfrak{m}$, there is some denominator $s_\mathfrak{m}$, but it may depend on $\mathfrak{m}$. And there is no single set $S$ being used here. You have a bunch of localizations at different multiplicative sets. For $K$, yes, the set $S$ you're localizing at is indeed $R\setminus\{0\}$. – Keenan Kidwell May 4 '13 at 1:42