Interpreting notation: showing localization commutes with finite product.

Question

I have a simple question.

In Vakil's notes The Rising Sea, there is the following question:

Show that localization commutes with finite products, or equivalently, with finite direct sums. In other words, if $M_1, \ldots, M_n$ are $A$-modules, describe an isomorphism (of $A$-modules, and of $S^{-1}A$-modules) $S^{-1}(M_1 \times \cdots \times M_n) \to S^{-1}M_1 \times \cdots\times S^{-1}M_n$

In this case, $S$ is a multiplicative system-- but of what module? If I naively assume that $S$ is bona-fide the same multiplicative system in each instance it is written, then I believe we get an injection $S^{-1}\prod_{i=1}^n M_i \hookrightarrow \prod_{i=1}^n S^{-1} M_i$, given by the map $(1/s)(m_1,\dots,m_n) \mapsto (m_1/s,\dots,m_n/s)$. Indeed, if $s_1 \neq s_2 \in S$, then in the case $n=2$ this map is not surjective, as nothing maps to $(m_1/s_1,m_2/s_2)$. Even without this "counterexample," it seems to me that it can't be the same system, simply because the modules are different.

So unless I've made a mistake above, I want to presume that we can write the right-hand $S^{-1}$ more accurately as $\prod_{i=1}^n S^{-1}$, and the $S^{-1}A$-module isomorphism is given by: $$\frac{1}{(s_1,\dots,s_n)}(m_1,\dots,m_n) \mapsto (m_1/s_1,\dots,m_n/s_n)$$

Is this the correct interpretation of Vakil's notation? Or am I missing something?

Answer 1

You're missing something: $\frac{1}{s_1s_2}(s_2m_1,s_1m_2)$ maps to $(\frac{m_1}{s_1},\frac{m_2}{s_2})$. This also solves the general problem after a suitable generalization.

There's another misconception to address too, from your second paragraph: multiplicative systems live in rings, not modules. So $S\subset R$, and it wouldn't really make sense to talk about $\prod S$ unless you were looking at modules over $\prod R$.

Answer 2

$S^{-1}M_1 , ... S^{-1} M_n$ are all modules over $S^{-1}R$. $S$ is a multiplicative subset of $R$.

$(m_1 / s_1 , m_2 / s_2) = (\frac{m_1 s_2 }{ s_1 s_2} , \frac{m_2 s_1}{ s_1 s_2}) = \frac{1}{s_1 s_2} (m_1 s_2,m_2 s_1)$ so the map is surjective.

Is this the correct interpretation of Vakil's notation?

No. We don't invert $\prod_{i=1}^n S$. We have a functor $R-mod \rightarrow S^{-1}R -mod$ given by $M \rightarrow S^{-1}M$. We are also given $R$-modules $M_1, ... M_n$

We can take their product first, then apply the $S^{-1}$ functor to get $S^{-1}(M_1 \times ... \times M_n)$. We can also apply the $S^{-1}$ functor first, then take their product in the category of $S^{-1}R$-modules to get $S^{-1} M_1 \times ... \times S^{-1} M_n$.

The exercise is to show that there is an isomorphism between $S^{-1}(M_1 \times ... \times M_n)$ and $S^{-1} M_1 \times ... \times S^{-1} M_n$ that commutes with the projections to $S^{-1} M_i$.