The Stacks project

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 $\mu _ i \otimes 1 = \psi \circ \lambda _ i$.

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 (7)

Comment #394 by Fan on

I heard of an argument of deducing this Lemma from Lemma 10.12.8 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.

Comment #5989 by Shota Inoue on

In the second proof, should probably be , and seems to be undefined.

Comment #7380 by T.C. on

In the second proof, shouldn't it be instead of ?

There are also:

  • 8 comment(s) on Section 10.12: Tensor products

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00DD. Beware of the difference between the letter 'O' and the digit '0'.