Abstract Algebra – How Tor Functor Commutes with Direct Limits

abstract-algebracommutative-algebrahomological-algebra

Could somebody please provide a sketch of a proof of the fact that the Tor functor commutes with direct limits?

I have been trying to show that the Tor of a module with the direct limit of a family of modules satisfies the required universal property, but it seems too complex.

Best Answer

What is true is that $\rm Tor$ commutes with filtered colimits. We already know that $M\otimes -$ commutes with colimits, now take a right $R$-module $M$ and a system $(N_i,\psi_{ji})$ of left $R$-modules over some filtered set $I$. First, we can obtain a short exact sequence of filtered systems

$$0\to (K_i,\rho_{ji})\to (P_i,\tilde\psi_{ji})\to (N_i,\psi_{ji})\to 0$$

by covering $N_i$ by $P_i=R^{(N_i)}$ by $e_n\to n$ and defining $\tilde \psi_{ji}(e_n)=e_{\psi_{ji}n}$, the condition $\psi_{kj}\psi_{ji}=\psi_{ki}$ is immediately inherited by the $\tilde \psi_{ji}$, similarly for the induced morphisms $\rho_{ji}:K_i\to K_j$. Since $I$ is filtered, $\rm colim$ is exact giving an exact sequence

$$0\to {\rm colim}\; K_i \to {\rm colim}\; P_i \to {\rm colim}\;N_i\to 0$$

Call the terms $K,P,N$ for brevity. Since each $P_i$ is flat and $I$ is filtered, $P$ is flat. We may then use the long exact sequence for $\rm Tor$ and obtain a diagram $\require{AMScd}$ $$\begin{CD} \\ {\rm Tor_1}\;(M,P) @>>> {\rm Tor_1}\;(M,N) @>>> M\otimes K @>>> M\otimes P \\ {}&{}& {} &{}& @VVV @VVV \\ {\rm colim}\;{\rm Tor_1}\;(M,P_i)@>>> {\rm colim}\;{\rm Tor_1}\;(M,N_i)@>>> {\rm colim}\; M\otimes K_i@>>> {\rm colim}\; M\otimes P_i' \\ \end{CD}$$

The first two columns vanish since $P_i,P$ are flat, and the last two columns are connected by natural isomorphisms that give a commutative diagram. You can check that if you have an incomplete commutative diagram with exact rows $$\begin{CD} \\ 0@>>> A @>>> B @>>> C \\ {}&{}& {} &{}& @VVV @VVV \\ 0 @>>> A' @>>> B' @>>> C' \\ \end{CD}$$

you may always complete it, and the morphism introduced is an isomorphism if both vertical arrows are. This gives the isomorphism for $n=1$. The case $n\geqslant 1$ is handled by dimension shifting. Indeed, we get a diagram $$\begin{CD} \\ 0@>>> {\rm Tor}_2(M,{\rm colim}\;N_i) @>\partial >> {\rm Tor}_1(M,{\rm colim}\;K_i) @>>> 0 \\ {}&{}& {} &{}& @VVV \\ 0 @>>> {\rm colim}\;{\rm Tor}_2(M,N_i) @>\partial >> {\rm colim}\;{\rm Tor}_1(M,K_i) @>>> 0 \\ \end{CD}$$

and we get the desired isomorphism by conjugating the vertical isomorphism. One may show the isomorphism induced is natural, much like that of $M\otimes {\rm colim}$ and ${\rm colim}\; M\otimes $, but that is a bit more tortuous. Note this gives naturality of all the upper isomorphisms, since they are obtained by composing the natural connection morphisms and the natural isomorphism in the case $n=1$.