
Lemma 10.11.9 (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 10.11.8. Thus $M' \mapsto M' \otimes _ R N$ commutes with all colimits, see Categories, Lemma 4.24.5.

Second direct proof. Let $P = \mathop{\mathrm{colim}}\nolimits (M_ i \otimes N)$, $M = \mathop{\mathrm{colim}}\nolimits M_ i$. 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 $\psi : P \to M \otimes N$, with $\lambda _ i : M_ i \otimes N \to P$ given by $\lambda _ i = \pi \circ (\iota _ i \otimes 1)$.

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 $P$. $\square$

Comment #394 by Fan on

I heard of an argument of deducing this Lemma from Lemma 10.11.8 and Lemma 4.24.5. Does that work?

Comment #395 by Fan on

The second Lemma says left adjoint functors are right exact, somewhere in 4.23. I may have messed up with the labels. Also the preview function is not working here.

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