## 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

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

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

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

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

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.
$\square$

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:

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

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

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.
$\square$

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:

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

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

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

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

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.
$\square$

## Comments (2)

Comment #2334 by Noah Olander on

Comment #2405 by Johan on