## 10.75 Tor groups and flatness

In this section we use some of the homological algebra developed in the previous section to explain what Tor groups are. Namely, suppose that $R$ is a ring and that $M$, $N$ are two $R$-modules. Choose a resolution $F_\bullet$ of $M$ by free $R$-modules. See Lemma 10.71.1. Consider the homological complex

$F_\bullet \otimes _ R N : \ldots \to F_2 \otimes _ R N \to F_1 \otimes _ R N \to F_0 \otimes _ R N$

We define $\text{Tor}^ R_ i(M, N)$ to be the $i$th homology group of this complex. The following lemma explains in what sense this is well defined.

Lemma 10.75.1. Let $R$ be a ring. Let $M_1, M_2, N$ be $R$-modules. Suppose that $F_\bullet$ is a free resolution of the module $M_1$ and that $G_\bullet$ is a free resolution of the module $M_2$. Let $\varphi : M_1 \to M_2$ be a module map. Let $\alpha : F_\bullet \to G_\bullet$ be a map of complexes inducing $\varphi$ on $M_1 = \mathop{\mathrm{Coker}}(d_{F, 1}) \to M_2 = \mathop{\mathrm{Coker}}(d_{G, 1})$, see Lemma 10.71.4. Then the induced maps

$H_ i(\alpha ) : H_ i(F_\bullet \otimes _ R N) \longrightarrow H_ i(G_\bullet \otimes _ R N)$

are independent of the choice of $\alpha$. If $\varphi$ is an isomorphism, so are all the maps $H_ i(\alpha )$. If $M_1 = M_2$, $F_\bullet = G_\bullet$, and $\varphi$ is the identity, so are all the maps $H_ i(\alpha )$.

Proof. The proof of this lemma is identical to the proof of Lemma 10.71.5. $\square$

Not only does this lemma imply that the Tor modules are well defined, but it also provides for the functoriality of the constructions $(M, N) \mapsto \text{Tor}_ i^ R(M, N)$ in the first variable. Of course the functoriality in the second variable is evident. We leave it to the reader to see that each of the $\text{Tor}_ i^ R$ is in fact a functor

$\text{Mod}_ R \times \text{Mod}_ R \to \text{Mod}_ R.$

Here $\text{Mod}_ R$ denotes the category of $R$-modules, and for the definition of the product category see Categories, Definition 4.2.20. Namely, given morphisms of $R$-modules $M_1 \to M_2$ and $N_1 \to N_2$ we get a commutative diagram

$\xymatrix{ \text{Tor}_ i^ R(M_1, N_1) \ar[r] \ar[d] & \text{Tor}_ i^ R(M_1, N_2) \ar[d] \\ \text{Tor}_ i^ R(M_2, N_1) \ar[r] & \text{Tor}_ i^ R(M_2, N_2) \\ }$

Lemma 10.75.2. Let $R$ be a ring and let $M$ be an $R$-module. Suppose that $0 \to N' \to N \to N'' \to 0$ is a short exact sequence of $R$-modules. There exists a long exact sequence

$\text{Tor}_1^ R(M, N') \to \text{Tor}_1^ R(M, N) \to \text{Tor}_1^ R(M, N'') \to M \otimes _ R N' \to M \otimes _ R N \to M \otimes _ R N'' \to 0$

Proof. The proof of this is the same as the proof of Lemma 10.71.6. $\square$

Consider a homological double complex of $R$-modules

$\xymatrix{ \ldots \ar[r]^ d & A_{2, 0} \ar[r]^ d & A_{1, 0} \ar[r]^ d & A_{0, 0} \\ \ldots \ar[r]^ d & A_{2, 1} \ar[r]^ d \ar[u]^\delta & A_{1, 1} \ar[r]^ d \ar[u]^\delta & A_{0, 1} \ar[u]^\delta \\ \ldots \ar[r]^ d & A_{2, 2} \ar[r]^ d \ar[u]^\delta & A_{1, 2} \ar[r]^ d \ar[u]^\delta & A_{0, 2} \ar[u]^\delta \\ & \ldots \ar[u]^\delta & \ldots \ar[u]^\delta & \ldots \ar[u]^\delta \\ }$

This means that $d_{i, j} : A_{i, j} \to A_{i-1, j}$ and $\delta _{i, j} : A_{i, j} \to A_{i, j-1}$ have the following properties

1. Any composition of two $d_{i, j}$ is zero. In other words the rows of the double complex are complexes.

2. Any composition of two $\delta _{i, j}$ is zero. In other words the columns of the double complex are complexes.

3. For any pair $(i, j)$ we have $\delta _{i-1, j} \circ d_{i, j} = d_{i, j-1} \circ \delta _{i, j}$. In other words, all the squares commute.

The correct thing to do is to associate a spectral sequence to any such double complex. However, for the moment we can get away with doing something slightly easier.

Namely, for the purposes of this section only, given a double complex $(A_{\bullet , \bullet }, d, \delta )$ set $R(A)_ j = \mathop{\mathrm{Coker}}(A_{1, j} \to A_{0, j})$ and $U(A)_ i = \mathop{\mathrm{Coker}}(A_{i, 1} \to A_{i, 0})$. (The letters $R$ and $U$ are meant to suggest Right and Up.) We endow $R(A)_\bullet$ with the structure of a complex using the maps $\delta$. Similarly we endow $U(A)_\bullet$ with the structure of a complex using the maps $d$. In other words we obtain the following huge commutative diagram

$\xymatrix{ \ldots \ar[r]^ d & U(A)_2 \ar[r]^ d & U(A)_1 \ar[r]^ d & U(A)_0 & \\ \ldots \ar[r]^ d & A_{2, 0} \ar[r]^ d \ar[u] & A_{1, 0} \ar[r]^ d \ar[u] & A_{0, 0} \ar[r] \ar[u] & R(A)_0 \\ \ldots \ar[r]^ d & A_{2, 1} \ar[r]^ d \ar[u]^\delta & A_{1, 1} \ar[r]^ d \ar[u]^\delta & A_{0, 1} \ar[r] \ar[u]^\delta & R(A)_1 \ar[u]^\delta \\ \ldots \ar[r]^ d & A_{2, 2} \ar[r]^ d \ar[u]^\delta & A_{1, 2} \ar[r]^ d \ar[u]^\delta & A_{0, 2} \ar[r] \ar[u]^\delta & R(A)_2 \ar[u]^\delta \\ & \ldots \ar[u]^\delta & \ldots \ar[u]^\delta & \ldots \ar[u]^\delta & \ldots \ar[u]^\delta \\ }$

(This is no longer a double complex of course.) It is clear what a morphism $\Phi : (A_{\bullet , \bullet }, d, \delta ) \to (B_{\bullet , \bullet }, d, \delta )$ of double complexes is, and it is clear that this induces morphisms of complexes $R(\Phi ) : R(A)_\bullet \to R(B)_\bullet$ and $U(\Phi ) : U(A)_\bullet \to U(B)_\bullet$.

Lemma 10.75.3. Let $(A_{\bullet , \bullet }, d, \delta )$ be a double complex such that

1. Each row $A_{\bullet , j}$ is a resolution of $R(A)_ j$.

2. Each column $A_{i, \bullet }$ is a resolution of $U(A)_ i$.

Then there are canonical isomorphisms

$H_ i(R(A)_\bullet ) \cong H_ i(U(A)_\bullet ).$

The isomorphisms are functorial with respect to morphisms of double complexes with the properties above.

Proof. We will show that $H_ i(R(A)_\bullet ))$ and $H_ i(U(A)_\bullet )$ are canonically isomorphic to a third group. Namely

$\mathbf{H}_ i(A) := \frac{ \{ (a_{i, 0}, a_{i-1, 1}, \ldots , a_{0, i}) \mid d(a_{i, 0}) = \delta (a_{i-1, 1}), \ldots , d(a_{1, i-1}) = \delta (a_{0, i}) \} }{ \{ d(a_{i + 1, 0}) + \delta (a_{i, 1}), d(a_{i, 1}) + \delta (a_{i-1, 2}), \ldots , d(a_{1, i}) + \delta (a_{0, i + 1}) \} }$

Here we use the notational convention that $a_{i, j}$ denotes an element of $A_{i, j}$. In other words, an element of $\mathbf{H}_ i$ is represented by a zig-zag, represented as follows for $i = 2$

$\xymatrix{ a_{2, 0} \ar@{|->}[r] & d(a_{2, 0}) = \delta (a_{1, 1}) & \\ & a_{1, 1} \ar@{|->}[u] \ar@{|->}[r] & d(a_{1, 1}) = \delta (a_{0, 2}) \\ & & a_{0, 2} \ar@{|->}[u] \\ }$

Naturally, we divide out by “trivial” zig-zags, namely the submodule generated by elements of the form $(0, \ldots , 0, -\delta (a_{t + 1, t-i}), d(a_{t + 1, t-i}), 0, \ldots , 0)$. Note that there are canonical homomorphisms

$\mathbf{H}_ i(A) \to H_ i(R(A)_\bullet ), \quad (a_{i, 0}, a_{i-1, 1}, \ldots , a_{0, i}) \mapsto \text{class of image of }a_{0, i}$

and

$\mathbf{H}_ i(A) \to H_ i(U(A)_\bullet ), \quad (a_{i, 0}, a_{i-1, 1}, \ldots , a_{0, i}) \mapsto \text{class of image of }a_{i, 0}$

First we show that these maps are surjective. Suppose that $\overline{r} \in H_ i(R(A)_\bullet )$. Let $r \in R(A)_ i$ be a cocycle representing the class of $\overline{r}$. Let $a_{0, i} \in A_{0, i}$ be an element which maps to $r$. Because $\delta (r) = 0$, we see that $\delta (a_{0, i})$ is in the image of $d$. Hence there exists an element $a_{1, i-1} \in A_{1, i-1}$ such that $d(a_{1, i-1}) = \delta (a_{0, i})$. This in turn implies that $\delta (a_{1, i-1})$ is in the kernel of $d$ (because $d(\delta (a_{1, i-1})) = \delta (d(a_{1, i-1})) = \delta (\delta (a_{0, i})) = 0$. By exactness of the rows we find an element $a_{2, i-2}$ such that $d(a_{2, i-2}) = \delta (a_{1, i-1})$. And so on until a full zig-zag is found. Of course surjectivity of $\mathbf{H}_ i \to H_ i(U(A))$ is shown similarly.

To prove injectivity we argue in exactly the same way. Namely, suppose we are given a zig-zag $(a_{i, 0}, a_{i-1, 1}, \ldots , a_{0, i})$ which maps to zero in $H_ i(R(A)_\bullet )$. This means that $a_{0, i}$ maps to an element of $\mathop{\mathrm{Coker}}(A_{i, 1} \to A_{i, 0})$ which is in the image of $\delta : \mathop{\mathrm{Coker}}(A_{i + 1, 1} \to A_{i + 1, 0}) \to \mathop{\mathrm{Coker}}(A_{i, 1} \to A_{i, 0})$. In other words, $a_{0, i}$ is in the image of $\delta \oplus d : A_{0, i + 1} \oplus A_{1, i} \to A_{0, i}$. From the definition of trivial zig-zags we see that we may modify our zig-zag by a trivial one and assume that $a_{0, i} = 0$. This immediately implies that $d(a_{1, i-1}) = 0$. As the rows are exact this implies that $a_{1, i-1}$ is in the image of $d : A_{2, i-1} \to A_{1, i-1}$. Thus we may modify our zig-zag once again by a trivial zig-zag and assume that our zig-zag looks like $(a_{i, 0}, a_{i-1, 1}, \ldots , a_{2, i-2}, 0, 0)$. Continuing like this we obtain the desired injectivity.

If $\Phi : (A_{\bullet , \bullet }, d, \delta ) \to (B_{\bullet , \bullet }, d, \delta )$ is a morphism of double complexes both of which satisfy the conditions of the lemma, then we clearly obtain a commutative diagram

$\xymatrix{ H_ i(U(A)_\bullet ) \ar[d] & \mathbf{H}_ i(A) \ar[r] \ar[l] \ar[d] & H_ i(R(A)_\bullet ) \ar[d] \\ H_ i(U(B)_\bullet ) & \mathbf{H}_ i(B) \ar[r] \ar[l] & H_ i(R(B)_\bullet ) \\ }$

This proves the functoriality. $\square$

Remark 10.75.4. The isomorphism constructed above is the “correct” one only up to signs. A good part of homological algebra is concerned with choosing signs for various maps and showing commutativity of diagrams with intervention of suitable signs. For the moment we will simply use the isomorphism as given in the proof above, and worry about signs later.

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. $\square$

Remark 10.75.6. An interesting case occurs when $M = N$ in the above. In this case we get a canonical map $\text{Tor}_ i^ R(M, M) \to \text{Tor}_ i^ R(M, M)$. Note that this map is not the identity, because even when $i = 0$ this map is not the identity! For example, if $V$ is a vector space of dimension $n$ over a field, then the switch map $V \otimes _ k V \to V \otimes _ k V$ has $(n^2 + n)/2$ eigenvalues $+1$ and $(n^2-n)/2$ eigenvalues $-1$. In characteristic $2$ it is not even diagonalizable. Note that even changing the sign of the map will not get rid of this.

Lemma 10.75.7. Let $R$ be a Noetherian ring. Let $M$, $N$ be finite $R$-modules. Then $\text{Tor}_ p^ R(M, N)$ is a finite $R$-module for all $p$.

Proof. This holds because $\text{Tor}_ p^ R(M, N)$ is computed as the cohomology groups of a complex $F_\bullet \otimes _ R N$ with each $F_ n$ a finite free $R$-module, see Lemma 10.71.1. $\square$

Lemma 10.75.8. Let $R$ be a ring. Let $M$ be an $R$-module. The following are equivalent:

1. The module $M$ is flat over $R$.

2. For all $i > 0$ the functor $\text{Tor}_ i^ R(M, -)$ is zero.

3. The functor $\text{Tor}_1^ R(M, -)$ is zero.

4. For all ideals $I \subset R$ we have $\text{Tor}_1^ R(M, R/I) = 0$.

5. For all finitely generated ideals $I \subset R$ we have $\text{Tor}_1^ R(M, R/I) = 0$.

Proof. Suppose $M$ is flat. Let $N$ be an $R$-module. Let $F_\bullet$ be a free resolution of $N$. Then $F_\bullet \otimes _ R M$ is a resolution of $N \otimes _ R M$, by flatness of $M$. Hence all higher Tor groups vanish.

It now suffices to show that the last condition implies that $M$ is flat. Let $I \subset R$ be an ideal. Consider the short exact sequence $0 \to I \to R \to R/I \to 0$. Apply Lemma 10.75.2. We get an exact sequence

$\text{Tor}_1^ R(M, R/I) \to M \otimes _ R I \to M \otimes _ R R \to M \otimes _ R R/I \to 0$

Since obviously $M \otimes _ R R = M$ we conclude that the last hypothesis implies that $M \otimes _ R I \to M$ is injective for every finitely generated ideal $I$. Thus $M$ is flat by Lemma 10.39.5. $\square$

Remark 10.75.9. The proof of Lemma 10.75.8 actually shows that

$\text{Tor}_1^ R(M, R/I) = \mathop{\mathrm{Ker}}(I \otimes _ R M \to M).$

Comment #2334 by Noah Olander on

I think the signs are incorrect in the definition of $H_i (A)$ in the beginning of the proof of lemma 10.4.73. I think it would work if all the minus signs in the denominator were switched to plus signs.

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).