## 10.71 Ext groups

In this section we do a tiny bit of homological algebra, in order to establish some fundamental properties of depth over Noetherian local rings.

Lemma 10.71.1. Let $R$ be a ring. Let $M$ be an $R$-module.

There exists an exact complex

\[ \ldots \to F_2 \to F_1 \to F_0 \to M \to 0. \]

with $F_ i$ free $R$-modules.

If $R$ is Noetherian and $M$ finite over $R$, then we can choose the complex such that $F_ i$ is finite free. In other words, we can find an exact complex

\[ \ldots \to R^{\oplus n_2} \to R^{\oplus n_1} \to R^{\oplus n_0} \to M \to 0. \]

**Proof.**
Let us explain only the Noetherian case. As a first step choose a surjection $R^{n_0} \to M$. Then having constructed an exact complex of length $e$ we simply choose a surjection $R^{n_{e + 1}} \to \mathop{\mathrm{Ker}}(R^{n_ e} \to R^{n_{e-1}})$ which is possible because $R$ is Noetherian.
$\square$

Definition 10.71.2. Let $R$ be a ring. Let $M$ be an $R$-module.

A (left) *resolution* $F_\bullet \to M$ of $M$ is an exact complex

\[ \ldots \to F_2 \to F_1 \to F_0 \to M \to 0 \]

of $R$-modules.

A *resolution of $M$ by free $R$-modules* is a resolution $F_\bullet \to M$ where each $F_ i$ is a free $R$-module.

A *resolution of $M$ by finite free $R$-modules* is a resolution $F_\bullet \to M$ where each $F_ i$ is a finite free $R$-module.

We often use the notation $F_{\bullet }$ to denote a complex of $R$-modules

\[ \ldots \to F_ i \to F_{i-1} \to \ldots \]

In this case we often use $d_ i$ or $d_{F, i}$ to denote the map $F_ i \to F_{i-1}$. In this section we are always going to assume that $F_0$ is the last nonzero term in the complex. The *$i$th homology group of the complex* $F_{\bullet }$ is the group $H_ i = \mathop{\mathrm{Ker}}(d_{F, i})/\mathop{\mathrm{Im}}(d_{F, i + 1})$. A *map of complexes $\alpha : F_{\bullet } \to G_{\bullet }$* is given by maps $\alpha _ i : F_ i \to G_ i$ such that $\alpha _{i-1} \circ d_{F, i} = d_{G, i-1} \circ \alpha _ i$. Such a map induces a map on homology $H_ i(\alpha ) : H_ i(F_{\bullet }) \to H_ i(G_{\bullet })$. If $\alpha , \beta : F_{\bullet } \to G_{\bullet }$ are maps of complexes, then a *homotopy* between $\alpha $ and $\beta $ is given by a collection of maps $h_ i : F_ i \to G_{i + 1}$ such that $\alpha _ i - \beta _ i = d_{G, i + 1} \circ h_ i + h_{i-1} \circ d_{F, i}$. Two maps $\alpha , \beta : F_{\bullet } \to G_{\bullet }$ are said to be *homotopic* if a homotopy between $\alpha $ and $\beta $ exists.

We will use a very similar notation regarding complexes of the form $F^{\bullet }$ which look like

\[ \ldots \to F^ i \xrightarrow {d^ i} F^{i + 1} \to \ldots \]

There are maps of complexes, homotopies, etc. In this case we set $H^ i(F^{\bullet }) = \mathop{\mathrm{Ker}}(d^ i)/\mathop{\mathrm{Im}}(d^{i - 1})$ and we call it the *$i$th cohomology group*.

Lemma 10.71.3. Any two homotopic maps of complexes induce the same maps on (co)homology groups.

**Proof.**
Omitted.
$\square$

Lemma 10.71.4. Let $R$ be a ring. Let $M \to N$ be a map of $R$-modules. Let $N_\bullet \to N$ be an arbitrary resolution. Let

\[ \ldots \to F_2 \to F_1 \to F_0 \to M \]

be a complex of $R$-modules where each $F_ i$ is a free $R$-module. Then

there exists a map of complexes $F_\bullet \to N_\bullet $ such that

\[ \xymatrix{ F_0 \ar[r] \ar[d] & M \ar[d] \\ N_0 \ar[r] & N } \]

is commutative, and

any two maps $\alpha , \beta : F_\bullet \to N_\bullet $ as in (1) are homotopic.

**Proof.**
Proof of (1). Because $F_0$ is free we can find a map $F_0 \to N_0$ lifting the map $F_0 \to M \to N$. We obtain an induced map $F_1 \to F_0 \to N_0$ which ends up in the image of $N_1 \to N_0$. Since $F_1$ is free we may lift this to a map $F_1 \to N_1$. This in turn induces a map $F_2 \to F_1 \to N_1$ which maps to zero into $N_0$. Since $N_\bullet $ is exact we see that the image of this map is contained in the image of $N_2 \to N_1$. Hence we may lift to get a map $F_2 \to N_2$. Repeat.

Proof of (2). To show that $\alpha , \beta $ are homotopic it suffices to show the difference $\gamma = \alpha - \beta $ is homotopic to zero. Note that the image of $\gamma _0 : F_0 \to N_0$ is contained in the image of $N_1 \to N_0$. Hence we may lift $\gamma _0$ to a map $h_0 : F_0 \to N_1$. Consider the map $\gamma _1' = \gamma _1 - h_0 \circ d_{F, 1}$. By our choice of $h_0$ we see that the image of $\gamma _1'$ is contained in the kernel of $N_1 \to N_0$. Since $N_\bullet $ is exact we may lift $\gamma _1'$ to a map $h_1 : F_1 \to N_2$. At this point we have $\gamma _1 = h_0 \circ d_{F, 1} + d_{N, 2} \circ h_1$. Repeat.
$\square$

At this point we are ready to define the groups $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N)$. Namely, choose a resolution $F_{\bullet }$ of $M$ by free $R$-modules, see Lemma 10.71.1. Consider the (cohomological) complex

\[ \mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N) : \mathop{\mathrm{Hom}}\nolimits _ R(F_0, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(F_1, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(F_2, N) \to \ldots \]

We define $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N)$ for $i \geq 0$ to be the $i$th cohomology group of this complex^{1}. For $i < 0$ we set $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) = 0$. Before we continue we point out that

\[ \mathop{\mathrm{Ext}}\nolimits ^0_ R(M, N) = \mathop{\mathrm{Ker}}(\mathop{\mathrm{Hom}}\nolimits _ R(F_0, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(F_1, N)) = \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \]

because we can apply part (1) of Lemma 10.10.1 to the exact sequence $F_1 \to F_0 \to M \to 0$. The following lemma explains in what sense this is well defined.

Lemma 10.71.5. 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 $G_{\bullet }$ 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

\[ H^ i(\alpha ) : H^ i(\mathop{\mathrm{Hom}}\nolimits _ R(F_{\bullet }, N)) \longrightarrow H^ i(\mathop{\mathrm{Hom}}\nolimits _ R(G_{\bullet }, N)) \]

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.**
Another map $\beta : F_{\bullet } \to G_{\bullet }$ inducing $\varphi $ is homotopic to $\alpha $ by Lemma 10.71.4. Hence the maps $\mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N) \to \mathop{\mathrm{Hom}}\nolimits _ R(G_\bullet , N)$ are homotopic. Hence the independence result follows from Lemma 10.71.3.

Suppose that $\varphi $ is an isomorphism. Let $\psi : M_2 \to M_1$ be an inverse. Choose $\beta : G_{\bullet } \to F_{\bullet }$ be a map inducing $\psi : M_2 = \mathop{\mathrm{Coker}}(d_{G, 1}) \to M_1 = \mathop{\mathrm{Coker}}(d_{F, 1})$, see Lemma 10.71.4. OK, and now consider the map $H^ i(\alpha ) \circ H^ i(\beta ) = H^ i(\alpha \circ \beta )$. By the above the map $H^ i(\alpha \circ \beta )$ is the *same* as the map $H^ i(\text{id}_{G_{\bullet }}) = \text{id}$. Similarly for the composition $H^ i(\beta ) \circ H^ i(\alpha )$. Hence $H^ i(\alpha )$ and $H^ i(\beta )$ are inverses of each other.
$\square$

Lemma 10.71.6. Let $R$ be a ring. Let $M$ be an $R$-module. Let $0 \to N' \to N \to N'' \to 0$ be a short exact sequence. Then we get a long exact sequence

\[ \begin{matrix} 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N') \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N'')
\\ \phantom{0\ } \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(M, N') \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(M, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(M, N'') \to \ldots
\end{matrix} \]

**Proof.**
Pick a free resolution $F_{\bullet } \to M$. Since each of the $F_ i$ are free we see that we get a short exact sequence of complexes

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(F_{\bullet }, N') \to \mathop{\mathrm{Hom}}\nolimits _ R(F_{\bullet }, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(F_{\bullet }, N'') \to 0 \]

Thus we get the long exact sequence from the snake lemma applied to this.
$\square$

Lemma 10.71.7. Let $R$ be a ring. Let $N$ be an $R$-module. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence. Then we get a long exact sequence

\[ \begin{matrix} 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(M'', N) \to \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(M', N)
\\ \phantom{0\ } \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(M'', N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(M, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ R(M', N) \to \ldots
\end{matrix} \]

**Proof.**
Pick sets of generators $\{ m'_{i'}\} _{i' \in I'}$ and $\{ m''_{i''}\} _{i'' \in I''}$ of $M'$ and $M''$. For each $i'' \in I''$ choose a lift $\tilde m''_{i''} \in M$ of the element $m''_{i''} \in M''$. Set $F' = \bigoplus _{i' \in I'} R$, $F'' = \bigoplus _{i'' \in I''} R$ and $F = F' \oplus F''$. Mapping the generators of these free modules to the corresponding chosen generators gives surjective $R$-module maps $F' \to M'$, $F'' \to M''$, and $F \to M$. We obtain a map of short exact sequences

\[ \begin{matrix} 0
& \to
& M'
& \to
& M
& \to
& M''
& \to
& 0
\\ & & \uparrow
& & \uparrow
& & \uparrow
\\ 0
& \to
& F'
& \to
& F
& \to
& F''
& \to
& 0
\\ \end{matrix} \]

By the snake lemma we see that the sequence of kernels $0 \to K' \to K \to K'' \to 0$ is short exact sequence of $R$-modules. Hence we can continue this process indefinitely. In other words we obtain a short exact sequence of resolutions fitting into the diagram

\[ \begin{matrix} 0
& \to
& M'
& \to
& M
& \to
& M''
& \to
& 0
\\ & & \uparrow
& & \uparrow
& & \uparrow
\\ 0
& \to
& F_\bullet '
& \to
& F_\bullet
& \to
& F_\bullet ''
& \to
& 0
\\ \end{matrix} \]

Because each of the sequences $0 \to F'_ n \to F_ n \to F''_ n \to 0$ is split exact (by construction) we obtain a short exact sequence of complexes

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(F''_{\bullet }, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(F_{\bullet }, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(F'_{\bullet }, N) \to 0 \]

by applying the $\mathop{\mathrm{Hom}}\nolimits _ R(-, N)$ functor. Thus we get the long exact sequence from the snake lemma applied to this.
$\square$

Lemma 10.71.8. Let $R$ be a ring. Let $M$, $N$ be $R$-modules. Any $x\in R$ such that either $xN = 0$, or $xM = 0$ annihilates each of the modules $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N)$.

**Proof.**
Pick a free resolution $F_{\bullet }$ of $M$. Since $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N)$ is defined as the cohomology of the complex $\mathop{\mathrm{Hom}}\nolimits _ R(F_{\bullet }, N)$ the lemma is clear when $xN = 0$. If $xM = 0$, then we see that multiplication by $x$ on $F_{\bullet }$ lifts the zero map on $M$. Hence by Lemma 10.71.5 we see that it induces the same map on Ext groups as the zero map.
$\square$

Lemma 10.71.9. Let $R$ be a Noetherian ring. Let $M$, $N$ be finite $R$-modules. Then $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N)$ is a finite $R$-module for all $i$.

**Proof.**
This holds because $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N)$ is computed as the cohomology groups of a complex $\mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N)$ with each $F_ n$ a finite free $R$-module, see Lemma 10.71.1.
$\square$

## Comments (1)

Comment #682 by Keenan Kidwell on