Processing math: 100%

The Stacks project

Lemma 10.75.5. Let R be a ring. For any i \geq 0 the functors \text{Mod}_ R \times \text{Mod}_ R \to \text{Mod}_ R, (M, N) \mapsto \text{Tor}_ i^ R(M, N) and (M, N) \mapsto \text{Tor}_ i^ R(N, M) are canonically isomorphic.

Proof. Let F_\bullet be a free resolution of the module M and let G_\bullet be a free resolution of the module N. Consider the double complex (A_{i, j}, d, \delta ) defined as follows:

  1. set A_{i, j} = F_ i \otimes _ R G_ j,

  2. set d_{i, j} : F_ i \otimes _ R G_ j \to F_{i-1} \otimes G_ j equal to d_{F, i} \otimes \text{id}, and

  3. set \delta _{i, j} : F_ i \otimes _ R G_ j \to F_ i \otimes G_{j-1} equal to \text{id} \otimes d_{G, j}.

This double complex is usually simply denoted F_\bullet \otimes _ R G_\bullet .

Since each G_ j is free, and hence flat we see that each row of the double complex is exact except in homological degree 0. Since each F_ i is free and hence flat we see that each column of the double complex is exact except in homological degree 0. Hence the double complex satisfies the conditions of Lemma 10.75.3.

To see what the lemma says we compute R(A)_\bullet and U(A)_\bullet . Namely,

\begin{eqnarray*} R(A)_ i & = & \mathop{\mathrm{Coker}}(A_{1, i} \to A_{0, i}) \\ & = & \mathop{\mathrm{Coker}}(F_1 \otimes _ R G_ i \to F_0 \otimes _ R G_ i) \\ & = & \mathop{\mathrm{Coker}}(F_1 \to F_0) \otimes _ R G_ i \\ & = & M \otimes _ R G_ i \end{eqnarray*}

In fact these isomorphisms are compatible with the differentials \delta and we see that R(A)_\bullet = M \otimes _ R G_\bullet as homological complexes. In exactly the same way we see that U(A)_\bullet = F_\bullet \otimes _ R N. We get

\begin{eqnarray*} \text{Tor}_ i^ R(M, N) & = & H_ i(F_\bullet \otimes _ R N) \\ & = & H_ i(U(A)_\bullet ) \\ & = & H_ i(R(A)_\bullet ) \\ & = & H_ i(M \otimes _ R G_\bullet ) \\ & = & H_ i(G_\bullet \otimes _ R M) \\ & = & \text{Tor}_ i^ R(N, M) \end{eqnarray*}

Here the third equality is Lemma 10.75.3, and the fifth equality uses the isomorphism V \otimes W = W \otimes V of the tensor product.

Functoriality. Suppose that we have R-modules M_\nu , N_\nu , \nu = 1, 2. Let \varphi : M_1 \to M_2 and \psi : N_1 \to N_2 be morphisms of R-modules. Suppose that we have free resolutions F_{\nu , \bullet } for M_\nu and free resolutions G_{\nu , \bullet } for N_\nu . By Lemma 10.71.4 we may choose maps of complexes \alpha : F_{1, \bullet } \to F_{2, \bullet } and \beta : G_{1, \bullet } \to G_{2, \bullet } compatible with \varphi and \psi . We claim that the pair (\alpha , \beta ) induces a morphism of double complexes

\alpha \otimes \beta : F_{1, \bullet } \otimes _ R G_{1, \bullet } \longrightarrow F_{2, \bullet } \otimes _ R G_{2, \bullet }

This is really a very straightforward check using the rule that F_{1, i} \otimes _ R G_{1, j} \to F_{2, i} \otimes _ R G_{2, j} is given by \alpha _ i \otimes \beta _ j where \alpha _ i, resp. \beta _ j is the degree i, resp. j component of \alpha , resp. \beta . The reader also readily verifies that the induced maps R(F_{1, \bullet } \otimes _ R G_{1, \bullet })_\bullet \to R(F_{2, \bullet } \otimes _ R G_{2, \bullet })_\bullet agrees with the map M_1 \otimes _ R G_{1, \bullet } \to M_2 \otimes _ R G_{2, \bullet } induced by \varphi \otimes \beta . Similarly for the map induced on the U(-)_\bullet complexes. Thus the statement on functoriality follows from the statement on functoriality in Lemma 10.75.3. \square


Comments (0)

There are also:

  • 2 comment(s) on Section 10.75: Tor groups and flatness

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.