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It is a well-known fact that taking direct limits commutes with tensor products in the following sense:

Let $I$ be a directed set and suppose for every $i\in I$ we have modules $M_i,N_i$ over a ring $A_i$. Furthermore, for every $i,j\in I, i\leq j$ we have $M_i\to N_j, N_i\to N_j, A_i\to A_j$ such that all the compatibility conditions hold (also for the action of the A_i's). Then $\varinjlim M_i \otimes_{A_i} N_i = \varinjlim M_i\otimes_{\varinjlim A_i} \varinjlim N_i$.

A specific case is given by: $(\varinjlim M_i) \otimes_A N = \varinjlim (M_i\otimes_A N)$ where $A$ and $N$ are fixed. Here one can also say that $- \otimes N$ is a left-adjoint functor from the category of $A$-modules to itself (hence it preserves all colimits).

I wondered if there a similar category theoretical argument for the general case? I have seen direct proofs of the general statement (in fact this has already been asked on this website) but I am not looking that. Thanks in advance.

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  • $\begingroup$ You could probably use the fact that filtered categories are sifted. $\endgroup$ – Zhen Lin Nov 25 '14 at 8:27
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"I have seen direct proofs of the general statement [...] but I am not looking that.": Since you haven't specified what a direct proof or a category theoretic proof is for you, I am not sure if the following proof will satisfy your requirements. If not, please tell me why and we will try to find something else.

Proof 1. Recall that in general $M \otimes_R N$ is the coequalizer of the two maps $M \otimes R \otimes N \to M \otimes N$ given by $m \otimes r \otimes n \mapsto mr \otimes n$ resp. $m \otimes rn$. Hence, $\varinjlim_i M_i \otimes_{\varinjlim_i A_i} \varinjlim_i N_i$ is the coequalizer of $\varinjlim_i M_i \otimes \varinjlim_i A_i \otimes \varinjlim_i N_i \rightrightarrows \varinjlim_i M_i \otimes \varinjlim_i N_i$. But we have $$\varinjlim_i M_i \otimes \varinjlim_i N_i = \varinjlim_i \varinjlim_j M_i \otimes N_j = \varinjlim_i M_i \otimes N_i.$$ The last equality uses that $I$ is directed, which implies that the diagonal $I \to I \times I$ is cofinal. Hence, $\varinjlim_i M_i \otimes_{\varinjlim_i A_i} \varinjlim_i N_i$ is the coequalizer of $\varinjlim_i (M_i \otimes A_i \otimes N_i) \rightrightarrows \varinjlim_i (M_i \otimes N_i)$. Since colimits commute with colimits, this is $\varinjlim_i$ of the coequalizer of $M_i \otimes A_i \otimes N_i \rightrightarrows M_i \otimes N_i$, i.e. $\varinjlim_i M_i \otimes_{A_i} N_i$.

Proof 2. Both sides are $\varinjlim_i A_i$-modules. So let us apply the Yoneda Lemma. Let $T$ be some $\varinjlim_i A_i$-module.

$\hom(\varinjlim_i M_i \otimes_{\varinjlim_i A_i} \varinjlim_i N_i,T)\\\cong \hom(\varinjlim_i M_i,\underline{\hom}(\varinjlim_i N_i,T))\\\cong \varprojlim_i \hom(M_i,\underline{\hom}(\varinjlim_i N_i,T)|_{A_i})\\\cong \varprojlim_i \hom(M_i,\varprojlim_{j \geq i} \underline{\hom}(N_j,T|_{A_j})|_{A_i})\\\cong \varprojlim_i ~\varprojlim_{j \geq i} \hom(M_i, \underline{\hom}(N_j,T|_{A_j})|_{A_i})\\\cong \varprojlim_i ~\varprojlim_{j \geq i} \hom(M_i \otimes_{A_i} A_j, \underline{\hom}(N_j,T|_{A_j}))\\\cong \varprojlim_i ~\varprojlim_{j \geq i} \hom((M_i \otimes_{A_i} A_j) \otimes_{A_j} N_j, T|_{A_j})\\\cong \varprojlim_i ~\varprojlim_{j \geq i} \hom(M_i \otimes_{A_i} N_j, T|_{A_j})\\\cong \varprojlim_i \hom(M_i \otimes_{A_i} N_i,T|_{A_i})\\\cong \hom(\varinjlim_i (M_i \otimes_{A_i} N_i),T)$

Hence, $\varinjlim_i M_i \otimes_{\varinjlim_i A_i} \varinjlim_i N_i \cong \varinjlim_i (M_i \otimes_{A_i} N_i)$.

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  • $\begingroup$ Very nice, thanks a lot! I was indeed looking for some kind of colimit functor with respect to a suitable diagram. But your second solution is also nice. $\endgroup$ – AYK Nov 25 '14 at 15:29

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