Tensor product of a free module commutes with arbitrary intersection

Question

I was wondering if the following holds true:

Suppose that we have two modules $M$ and $N$ over a commutative ring $R$. Suppose that $M$ is a finitely generated free $R$ module. Suppose $(N_i)_{i\in I}$ are submodules of $N$, where $I$ is arbitrary. We identify each $M \otimes N_i$ with a submodule of $M \otimes N$, using that $M$ is flat and similarly identify $M \otimes (\bigcap_{i \in I} N_i) $ with a submodule of $M \otimes N$. Then $$ \bigcap_{i \in I} (M \otimes N_i) = M \otimes (\bigcap_{i \in I} N_i) $$ as submodules of $M\otimes N$.

1) A proof of this was provided here in the case where $M$ is merely flat, but at the cost of only allowing $I$ to be finite.

2) In the answer for this question, it is shown that tensoring by $M$ commutes with limits when $M$ is even finitely generated and projective.

To be able to apply point 2) here, I was wondering whether I can regard the intersection of submodules $N_i$ as a limit of some functor $\mathcal{F}$ into $R-\text{Mod}$ (The notion of limit that I use is the one from the textbook of Aluffi in Chapter VIII).

I believe that this should be so, but my lack of background in category theory makes me feel uncertain about this claim. Of course, I understand why this is so if we were looking into the category of sets and $N_i$ were some subsets of a set $N$.

I believe that the intersection of two submodules of a modules can be regarded as a fibered product, and I believe fibered products are examples of limits. But I am not so sure about arbitrary intersections.

Answer 1

let $M\cong R^n$. Then $M\otimes N\cong \oplus_{j=1}^nN$ as $R$-modules. Let's see where the submodules u are looking for go. $M\otimes N_i \cong \oplus_{j=1}^nN_i $ under this identification. So $\cap _iM\otimes N_i \cong \cap_i \oplus_{j=1}^nN_i=\oplus_{j=1}^n\cap_iNi\cong M\otimes \cap_iN_i$. So in the identification the 2 submodules are equal. Thus the 2 submodules are indeed equal.