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.

1. 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.

2. 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.

1. 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.

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

3. 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

1. 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

2. 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 complex1. 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. $\square$

[1] At this point it would perhaps be more appropriate to say “an” in stead of “the” Ext-group.

Comment #682 by Keenan Kidwell on

In the proof of 00LW, isn't the argument sort of set up to induct on depth, instead of dimension (not to say one couldn't induct on dimension, but it seems to me to make more sense to induct on depth since the induction hypothesis applies to $M/xM$ to show that $\delta(M)-1=\delta(M/xM)=i(M/xM)=i(M)-1$, so $\delta(M)=i(M)$.

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