Lemma 10.12.9 (Tensor products commute with colimits). Let $(M_ i, \mu _{ij})$ be a system over the preordered set $I$. Let $N$ be an $R$-module. Then

\[ \mathop{\mathrm{colim}}\nolimits (M_ i \otimes N) \cong (\mathop{\mathrm{colim}}\nolimits M_ i)\otimes N. \]

Moreover, the isomorphism is induced by the homomorphisms $\mu _ i \otimes 1: M_ i \otimes N \to M \otimes N$ where $M = \mathop{\mathrm{colim}}\nolimits _ i M_ i$ with natural maps $\mu _ i : M_ i \to M$.

**Proof.**
First proof. The functor $M' \mapsto M' \otimes _ R N$ is left adjoint to the functor $N' \mapsto \mathop{\mathrm{Hom}}\nolimits _ R(N, N')$ by Lemma 10.12.8. Thus $M' \mapsto M' \otimes _ R N$ commutes with all colimits, see Categories, Lemma 4.24.5.

Second direct proof. Let $P = \mathop{\mathrm{colim}}\nolimits (M_ i \otimes N)$ with coprojections $\lambda _ i : M_ i \otimes N \to P$. Let $M = \mathop{\mathrm{colim}}\nolimits M_ i$ with coprojections $\mu _ i : M_ i \to M$. Then for all $i\leq j$, the following diagram commutes:

\[ \xymatrix{ M_ i \otimes N \ar[r]_{\mu _ i \otimes 1} \ar[d]_{\mu _{ij} \otimes 1} & M \otimes N \ar[d]^{\text{id}} \\ M_ j \otimes N \ar[r]^{\mu _ j \otimes 1} & M \otimes N } \]

By Lemma 10.8.7 these maps induce a unique homomorphism $\psi : P \to M \otimes N$ such that $\lambda _ i = \psi \circ (\mu _ i \otimes 1)$.

To construct the inverse map, for each $i\in I$, there is the canonical $R$-bilinear mapping $g_ i : M_ i \times N \to M_ i \otimes N$. This induces a unique mapping $\widehat{\phi } : M \times N \to P$ such that $\widehat{\phi } \circ (\mu _ i \times 1) = \lambda _ i \circ g_ i$. It is $R$-bilinear. Thus it induces an $R$-linear mapping $\phi : M \otimes N \to P$. From the commutative diagram below:

\[ \xymatrix{ M_ i \times N \ar[r]^{g_ i} \ar[d]^{\mu _ i \times \text{id}} & M_ i \otimes N\ar[r]_{\text{id}} \ar[d]_{\lambda _ i} & M_ i \otimes N \ar[d]_{\mu _ i \otimes \text{id}} \ar[rd]^{\lambda _ i} \\ M \times N \ar[r]^{\widehat{\phi }} & P \ar[r]^{\psi } & M \otimes N \ar[r]^{\phi } & P } \]

we see that $\psi \circ \widehat{\phi } = g$, the canonical $R$-bilinear mapping $g : M \times N \to M \otimes N$. So $\psi \circ \phi $ is identity on $M \otimes N$. From the right-hand square and triangle, $\phi \circ \psi $ is also identity on $P$.
$\square$

## Comments (5)

Comment #394 by Fan on

Comment #395 by Fan on

Comment #399 by Johan on

Comment #5989 by Shota Inoue on

Comment #6160 by Johan on

There are also: