Related to this question, I am trying to solve the aforementioned question from Ravi Vakil's incredible notes which basically asks to prove that:

if $M_1, \ldots, M_n$ are $A$-modules (where $A$ is a commutative ring, and $S$ is a multiplicative subset of $A$), describe an isomorphism (of $A$-modules, and of $S^{-1}A$-modules) $S^{-1}(M_1 \times \cdots \times M_n) \to S^{-1}M_1 \times \cdots\times S^{-1}M_n$.

If you check that paragraph, Vakil is trying to make the notion of universal property clear. Because of that, my approach was to try to solve the exercise by universal product means. Here is my try:

First we will show that there exists a unique map from $S^{-1}(M_1\times\dots\times M_n)\to S^{-1}M_i$. By the universal property of localization, such a map can be found by defining a map $M_1\times\dots\times M_n\to S^{-1}M_i$. How to define such a map?

By the definition of product, there is a map $M_1\times\dots\times M_n\to M_i$ and another one $M_i\to S^{-1}M_i$. Take their composition and the machine starts working, providing us with a unique morphism $$S^{-1}(M_1\times\dots\times M_n)\to S^{-1}M_i$$ Now, by the definition of the product applied on $(S^{-1}M_1)\times\dots\times (S^{-1}M_n)$, there is a unique map $$(S^{-1}M_1)\times\dots\times (S^{-1}M_n)\to S^{-1}(M_1\times\dots\times M_n)$$

On the other hand, we would like to define another unique mapping from $S^{-1}(M_1\times\dots\times M_n)$ to $(S^{-1}M_1)\times\dots\times (S^{-1}M_n)$. In order to do that, we only need a map $M_1\times\dots\times M_n\to (S^{-1}M_1)\times\dots\times (S^{-1}M_n)$. Such a map can be obtained by defining maps $(S^{-1}M_1)\times\dots\times (S^{-1}M_n)\to M_i$ for each $i$. By the product definition, there is a map $(S^{-1}M_1)\times\dots\times (S^{-1}M_n)\to S^{-1}M_i$ for each $i$. So the only thing needed is a last map $S^{-1}M_i\to M_i$.

But if $M$ is not an $S^{-1}A$-module, the definition of localization cannot yield such a map and this is where I am stuck.


1 Answer 1


It looks like you have some things backwards in the universal property of products. If $P$ is a product of objects $X_1,\dots,X_n$ and you have maps $Y\to X_i$ for each $i$, then you get a map $Y\to P$, not a map $P\to Y$. So your maps $$S^{-1}(M_1\times\dots\times M_n)\to S^{-1}M_i$$ in the first part actually give you a map $$S^{-1}(M_1\times\dots\times M_n)\to (S^{-1}M_1)\times\dots\times (S^{-1}M_n),$$ not the other way around.

The harder part is constructing the inverse map $$(S^{-1}M_1)\times\dots\times (S^{-1}M_n)\to S^{-1}(M_1\times\dots\times M_n).$$ This map cannot be constructed by just using the universal properties abstractly; you need to use something special about what they actually look like in this case (in other words, you need to use some actual facts about modules). One way to do it is to just explicitly write down what the first map $$S^{-1}(M_1\times\dots\times M_n)\to (S^{-1}M_1)\times\dots\times (S^{-1}M_n)$$ looks like on elements and check that it is a bijection, and thus it has an inverse. Another way is to use that finite products of modules are also coproducts, and use the universal property of $(S^{-1}M_1)\times\dots\times (S^{-1}M_n)$ as a coproduct.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .