# Colimits and tensor product ground rings

Is it true that $\varinjlim (M \otimes_{A_i} N) = M \otimes_A N$ where $A = \varinjlim A_i$ and $M$ and $N$ are $A$-modules? Take maps $f : A_j \rightarrow A_k$ and $m : M \otimes_{A_j} N \rightarrow M \otimes_{A_k} N$ with $m \otimes n \mapsto m \otimes n$ is the $A_k$-linear map induced by $f$. Suppose that there is an $A$-module $P$ with maps $h_j: M \otimes_{A_j} N \rightarrow P$ and $h_k : M \otimes_{A_k} N \rightarrow P$ such that the diagram given by the maps above commutes. Since we have maps $A_j \rightarrow A$, we can treat the map $\phi : M \times N \rightarrow M \otimes_A N$ with $(m, n) \mapsto m \otimes n$ as an $A_j$-linear map. Using the map $\phi_j : M \times N \rightarrow M \otimes_{A_j} N$, we create a bilinear map $h_j\phi_j : M \times N \rightarrow P$.

It looks like this induces a linear map $u : M \otimes_A N \rightarrow P$ such that $h_j\phi_j = u\phi$, but I'm having trouble keeping track of which maps are linear or bilinear with respect to what ring while using the universal property of $A$ to induce linear maps $M \otimes_{A_i} N \rightarrow M \otimes N$ and considering bilinear maps $\phi_j : M \times N \rightarrow M \otimes_{A_j} N$. So, I don't know if this map even makes sense as a linear map with respect to a particular ring.

This came up when I tried to show that $(\mathscr{F} \otimes_{\mathcal{O}_X} \mathscr{G})_p = \mathscr{F}_p \otimes_{(\mathcal{O}_X)_p} \mathscr{G}_p$ for $\mathcal{O}_X$-modules $\mathscr{F}$ and $\mathscr{G}$, where $(\mathcal{O}_X, X)$ is a ringed space. Since tensor products (and left adjoints in general) commute with colimits, we have that $(\mathscr{F} \otimes_{\mathcal{O}_X} \mathscr{G})_p = \mathscr{F}_p \otimes_{\mathcal{O}_X(U)} \mathscr{G}_p$ for all open $U \subset X$. If the statement above is true, we are done since colimits commute with colimits.

• You should check that the colimit has the universal property of the tensor product. I suggest you write down in the question body what you have done, in detail. Jul 21 '15 at 23:19

Let $\phi_i : M \times N \rightarrow M \otimes_{A_i} N$ be the maps taking $(m, n) \mapsto (m \otimes n)$ in $M \otimes_{A_i} N$ and $\psi : M \times N \rightarrow P$ be a bilinear map with respect to $A$. Since we have maps $A_i \rightarrow A$, we can treat linear (or bilinear) maps with respect to $A$ as those with respect to $A_i$. By the universal property of the tensor product, we have $A_i$-linear maps $h_i : M \otimes_{A_i} N \rightarrow P$ such that $\psi = h_i\phi_i$.
Now we construct a map $M \times N \rightarrow \varinjlim (M \otimes_{A_i} N)$ which is bilinear with respect to any of the $A_i$. Let $f_i : M \otimes_{A_i} N \rightarrow \varinjlim (M \otimes_{A_i} N)$ be the maps to the colimit as defined in the universal property. We take the maps $m : M \otimes_{A_i} N \rightarrow M \otimes_{A_j} N$ to be those sending $m \otimes n \mapsto m \otimes n$. We clearly have that $\phi_j = m\phi_i$. Combining this with the fact that $f_i = f_jm$, we have that $f_i\phi_i = f_j\phi_j$. Since there is a map similar to $m$ between any two tensor products of the form $M \otimes_{A_i} N$, this means that all of the $f_i\phi_i$ are equal to each other. We take $\phi : M \times N \rightarrow \varinjlim(M \otimes_{A_i} N)$ to be this map. This map is bilinear with respect to any of the $A_i$ since the composition of a bilinear map with a linear map is a bilinear map.
Since $h_i\phi_i = h_i\phi_j = \psi$ and $\phi_i$ is a surjection, we have that $h_jm = h_i$. By the universal property of the colimit, this gives a map $u : \varinjlim(M \otimes_{A_i} N) \rightarrow P$ such that $uf_i = h_i$ for all $i$. We can easily check that $u\phi = \psi$.
Here are my questions: Why is $u$ linear with respect to $A$? Is this because we are considering objects which are $A_i$ modules and we take morphisms between modules over the same ring to be linear maps with respect to that ring? I think this means that $u$ is linear with respect to each $A_i$, but I don't know how this extends to all of $A$.