The Stacks project

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10.108 Rings of finite global dimension

The following lemma is often used to compare different projective resolutions of a given module.

Lemma 10.108.1 (Schanuel's lemma). Let $R$ be a ring. Let $M$ be an $R$-module. Suppose that

\[ 0 \to K \xrightarrow {c_1} P_1 \xrightarrow {p_1} M \to 0 \quad \text{and}\quad 0 \to L \xrightarrow {c_2} P_2 \xrightarrow {p_2} M \to 0 \]

are two short exact sequences, with $P_ i$ projective. Then $K \oplus P_2 \cong L \oplus P_1$. More precisely, there exist a commutative diagram

\[ \xymatrix{ 0 \ar[r] & K \oplus P_2 \ar[r]_{(c_1, \text{id})} \ar[d] & P_1 \oplus P_2 \ar[r]_{(0, p_2)} \ar[d] & M \ar[r] \ar@{=}[d] & 0 \\ 0 \ar[r] & P_1 \oplus L \ar[r]^{(\text{id}, c_2)} & P_1 \oplus P_2 \ar[r]^{(p_1, 0)} & M \ar[r] & 0 } \]

whose vertical arrows are isomorphisms.

Proof. Consider the module $N$ defined by the short exact sequence $0 \to N \to P_1 \oplus P_2 \to M \to 0$, where the last map is the sum of the two maps $P_ i \to M$. It is easy to see that the projection $N \to P_1$ is surjective with kernel $L$, and that $N \to P_2$ is surjective with kernel $K$. Since $P_ i$ are projective we have $N \cong K \oplus P_2 \cong L \oplus P_1$. This proves the first statement.

To prove the second statement (and to reprove the first), choose $a : P_1 \to P_2$ and $b : P_2 \to P_1$ such that $p_1 = p_2 \circ a$ and $p_2 = p_1 \circ b$. This is possible because $P_1$ and $P_2$ are projective. Then we get a commutative diagram

\[ \xymatrix{ 0 \ar[r] & K \oplus P_2 \ar[r]_{(c_1, \text{id})} & P_1 \oplus P_2 \ar[r]_{(0, p_2)} & M \ar[r] & 0 \\ 0 \ar[r] & N \ar[r] \ar[d] \ar[u] & P_1 \oplus P_2 \ar[r]_{(p_1, p_2)} \ar[d]_ T \ar[u]^ S & M \ar[r] \ar@{=}[d] \ar@{=}[u] & 0 \\ 0 \ar[r] & P_1 \oplus L \ar[r]^{(\text{id}, c_2)} & P_1 \oplus P_2 \ar[r]^{(p_1, 0)} & M \ar[r] & 0 } \]

with $T$ and $S$ given by the matrices

\[ S = \left( \begin{matrix} \text{id} & 0 \\ a & \text{id} \end{matrix} \right) \quad \text{and}\quad T = \left( \begin{matrix} \text{id} & b \\ 0 & \text{id} \end{matrix} \right) \]

Then $S$, $T$ and the maps $N \to P_1 \oplus L$ and $N \to K \oplus P_2$ are isomorphisms as desired. $\square$

Definition 10.108.2. Let $R$ be a ring. Let $M$ be an $R$-module. We say $M$ has finite projective dimension if it has a finite length resolution by projective $R$-modules. The minimal length of such a resolution is called the projective dimension of $M$.

It is clear that the projective dimension of $M$ is $0$ if and only if $M$ is a projective module. The following lemma explains to what extent the projective dimension is independent of the choice of a projective resolution.

Lemma 10.108.3. Let $R$ be a ring. Suppose that $M$ is an $R$-module of projective dimension $d$. Suppose that $F_ e \to F_{e-1} \to \ldots \to F_0 \to M \to 0$ is exact with $F_ i$ projective and $e \geq d - 1$. Then the kernel of $F_ e \to F_{e-1}$ is projective (or the kernel of $F_0 \to M$ is projective in case $e = 0$).

Proof. We prove this by induction on $d$. If $d = 0$, then $M$ is projective. In this case there is a splitting $F_0 = \mathop{\mathrm{Ker}}(F_0 \to M) \oplus M$, and hence $\mathop{\mathrm{Ker}}(F_0 \to M)$ is projective. This finishes the proof if $e = 0$, and if $e > 0$, then replacing $M$ by $\mathop{\mathrm{Ker}}(F_0 \to M)$ we decrease $e$.

Next assume $d > 0$. Let $0 \to P_ d \to P_{d-1} \to \ldots \to P_0 \to M \to 0$ be a minimal length finite resolution with $P_ i$ projective. According to Schanuel's Lemma 10.108.1 we have $P_0 \oplus \mathop{\mathrm{Ker}}(F_0 \to M) \cong F_0 \oplus \mathop{\mathrm{Ker}}(P_0 \to M)$. This proves the case $d = 1$, $e = 0$, because then the right hand side is $F_0 \oplus P_1$ which is projective. Hence now we may assume $e > 0$. The module $F_0 \oplus \mathop{\mathrm{Ker}}(P_0 \to M)$ has the finite projective resolution

\[ 0 \to P_ d \to P_{d-1} \to \ldots \to P_2 \to P_1 \oplus F_0 \to \mathop{\mathrm{Ker}}(P_0 \to M) \oplus F_0 \to 0 \]

of length $d - 1$. By induction applied to the exact sequence

\[ F_ e \to F_{e-1} \to \ldots \to F_2 \to P_0 \oplus F_1 \to P_0 \oplus \mathop{\mathrm{Ker}}(F_0 \to M) \to 0 \]

of length $e - 1$ we conclude $\mathop{\mathrm{Ker}}(F_ e \to F_{e - 1})$ is projective (if $e \geq 2$) or that $\mathop{\mathrm{Ker}}(F_1 \oplus P_0 \to F_0 \oplus P_0)$ is projective. This implies the lemma. $\square$

Lemma 10.108.4. Let $R$ be a ring. Let $M$ be an $R$-module. Let $d \geq 0$. The following are equivalent

  1. $M$ has projective dimension $\leq d$,

  2. there exists a resolution $0 \to P_ d \to P_{d - 1} \to \ldots \to P_0 \to M \to 0$ with $P_ i$ projective,

  3. for some resolution $\ldots \to P_2 \to P_1 \to P_0 \to M \to 0$ with $P_ i$ projective we have $\mathop{\mathrm{Ker}}(P_{d - 1} \to P_{d - 2})$ is projective if $d \geq 2$, or $\mathop{\mathrm{Ker}}(P_0 \to M)$ is projective if $d = 1$, or $M$ is projective if $d = 0$,

  4. for any resolution $\ldots \to P_2 \to P_1 \to P_0 \to M \to 0$ with $P_ i$ projective we have $\mathop{\mathrm{Ker}}(P_{d - 1} \to P_{d - 2})$ is projective if $d \geq 2$, or $\mathop{\mathrm{Ker}}(P_0 \to M)$ is projective if $d = 1$, or $M$ is projective if $d = 0$.

Proof. The equivalence of (1) and (2) is the definition of projective dimension, see Definition 10.108.2. We have (2) $\Rightarrow $ (4) by Lemma 10.108.3. The implications (4) $\Rightarrow $ (3) and (3) $\Rightarrow $ (2) are immediate. $\square$

Lemma 10.108.5. Let $R$ be a local ring. Let $M$ be an $R$-module. Let $d \geq 0$. The equivalent conditions (1) – (4) of Lemma 10.108.4 are also equivalent to

  1. there exists a resolution $0 \to P_ d \to P_{d - 1} \to \ldots \to P_0 \to M \to 0$ with $P_ i$ free.

Lemma 10.108.6. Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. Let $d \geq 0$. The equivalent conditions (1) – (4) of Lemma 10.108.4 are also equivalent to

  1. there exists a resolution $0 \to P_ d \to P_{d - 1} \to \ldots \to P_0 \to M \to 0$ with $P_ i$ finite projective.

Proof. Choose a resolution $\ldots \to F_2 \to F_1 \to F_0 \to M \to 0$ with $F_ i$ finite free (Lemma 10.70.1). By Lemma 10.108.4 we see that $P_ d = \mathop{\mathrm{Ker}}(F_{d - 1} \to F_{d - 2})$ is projective at least if $d \geq 2$. Then $P_ d$ is a finite $R$-module as $R$ is Noetherian and $P_ d \subset F_{d - 1}$ which is finite free. Whence $0 \to P_ d \to F_{d - 1} \to \ldots \to F_1 \to F_0 \to M \to 0$ is the desired resolution. $\square$

Lemma 10.108.7. Let $R$ be a local Noetherian ring. Let $M$ be a finite $R$-module. Let $d \geq 0$. The equivalent conditions (1) – (4) of Lemma 10.108.4, condition (5) of Lemma 10.108.5, and condition (6) of Lemma 10.108.6 are also equivalent to

  1. there exists a resolution $0 \to F_ d \to F_{d - 1} \to \ldots \to F_0 \to M \to 0$ with $F_ i$ finite free.

Proof. This follows from Lemmas 10.108.4, 10.108.5, and 10.108.6 and because a finite projective module over a local ring is finite free, see Lemma 10.77.2. $\square$

Lemma 10.108.8. Let $R$ be a ring. Let $M$ be an $R$-module. Let $n \geq 0$. The following are equivalent

  1. $M$ has projective dimension $\leq n$,

  2. $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(M, N) = 0$ for all $R$-modules $N$ and all $i \geq n + 1$, and

  3. $\mathop{\mathrm{Ext}}\nolimits ^{n + 1}_ R(M, N) = 0$ for all $R$-modules $N$.

Proof. Assume (1). Choose a free resolution $F_\bullet \to M$ of $M$. Denote $d_ e : F_ e \to F_{e - 1}$. By Lemma 10.108.3 we see that $P_ e = \mathop{\mathrm{Ker}}(d_ e)$ is projective for $e \geq n - 1$. This implies that $F_ e \cong P_ e \oplus P_{e - 1}$ for $e \geq n$ where $d_ e$ maps the summand $P_{e - 1}$ isomorphically to $P_{e - 1}$ in $F_{e - 1}$. Hence, for any $R$-module $N$ the complex $\mathop{\mathrm{Hom}}\nolimits _ R(F_\bullet , N)$ is split exact in degrees $\geq n + 1$. Whence (2) holds. The implication (2) $\Rightarrow $ (3) is trivial.

Assume (3) holds. If $n = 0$ then $M$ is projective by Lemma 10.76.2 and we see that (1) holds. If $n > 0$ choose a free $R$-module $F$ and a surjection $F \to M$ with kernel $K$. By Lemma 10.70.7 and the vanishing of $\mathop{\mathrm{Ext}}\nolimits _ R^ i(F, N)$ for all $i > 0$ by part (1) we see that $\mathop{\mathrm{Ext}}\nolimits _ R^ n(K, N) = 0$ for all $R$-modules $N$. Hence by induction we see that $K$ has projective dimension $\leq n - 1$. Then $M$ has projective dimension $\leq n$ as any finite projective resolution of $K$ gives a projective resolution of length one more for $M$ by adding $F$ to the front. $\square$

Lemma 10.108.9. Let $R$ be a ring. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of $R$-modules.

  1. If $M$ has projective dimension $\leq n$ and $M''$ has projective dimension $\leq n + 1$, then $M'$ has projective dimension $\leq n$.

  2. If $M'$ and $M''$ have projective dimension $\leq n$ then $M$ has projective dimension $\leq n$.

  3. If $M'$ has projective dimension $\leq n$ and $M$ has projective dimension $\leq n + 1$ then $M''$ has projective dimension $\leq n + 1$.

Proof. Combine the characterization of projective dimension in Lemma 10.108.8 with the long exact sequence of ext groups in Lemma 10.70.7. $\square$

Definition 10.108.10. Let $R$ be a ring. The ring $R$ is said to have finite global dimension if there exists an integer $n$ such that every $R$-module has a resolution by projective $R$-modules of length at most $n$. The minimal such $n$ is then called the global dimension of $R$.

The argument in the proof of the following lemma can be found in the paper [Auslander] by Auslander.

Lemma 10.108.11. Let $R$ be a ring. Suppose we have a module $M = \bigcup _{e \in E} M_ e$ where the $M_ e$ are submodules well-ordered by inclusion. Assume the quotients $M_ e/\bigcup \nolimits _{e' < e} M_{e'}$ have projective dimension $\leq n$. Then $M$ has projective dimension $\leq n$.

Proof. We will prove this by induction on $n$.

Base case: $n = 0$. Then $P_ e = M_ e/\bigcup _{e' < e} M_{e'}$ is projective. Thus we may choose a section $P_ e \to M_ e$ of the projection $M_ e \to P_ e$. We claim that the induced map $\psi : \bigoplus _{e \in E} P_ e \to M$ is an isomorphism. Namely, if $x = \sum x_ e \in \bigoplus P_ e$ is nonzero, then we let $e_{max}$ be maximal such that $x_{e_{max}}$ is nonzero and we conclude that $y = \psi (x) = \psi (\sum x_ e)$ is nonzero because $y \in M_{e_{max}}$ has nonzero image $x_{e_{max}}$ in $P_{e_{max}}$. On the other hand, let $y \in M$. Then $y \in M_ e$ for some $e$. We show that $y \in \mathop{\mathrm{Im}}(\psi )$ by transfinite induction on $e$. Let $x_ e \in P_ e$ be the image of $y$. Then $y - \psi (x_ e) \in \bigcup _{e' < e} M_{e'}$. By induction hypothesis we conclude that $y - \psi (x_ e) \in \mathop{\mathrm{Im}}(\psi )$ hence $y \in \mathop{\mathrm{Im}}(\psi )$. Thus the claim is true and $\psi $ is an isomorphism. We conclude that $M$ is projective as a direct sum of projectives, see Lemma 10.76.3.

If $n > 0$, then for $e \in E$ we denote $F_ e$ the free $R$-module on the set of elements of $M_ e$. Then we have a system of short exact sequences

\[ 0 \to K_ e \to F_ e \to M_ e \to 0 \]

over the well-ordered set $E$. Note that the transition maps $F_{e'} \to F_ e$ and $K_{e'} \to K_ e$ are injective too. Set $F = \bigcup F_ e$ and $K = \bigcup K_ e$. Then

\[ 0 \to K_ e/\bigcup \nolimits _{e' < e} K_{e'} \to F_ e/\bigcup \nolimits _{e' < e} F_{e'} \to M_ e/\bigcup \nolimits _{e' < e} M_{e'} \to 0 \]

is a short exact sequence of $R$-modules too and $F_ e/\bigcup _{e' < e} F_{e'}$ is the free $R$-module on the set of elements in $M_ e$ which are not contained in $\bigcup _{e' < e} M_{e'}$. Hence by Lemma 10.108.9 we see that the projective dimension of $K_ e/\bigcup _{e' < e} K_{e'}$ is at most $n - 1$. By induction we conclude that $K$ has projective dimension at most $n - 1$. Whence $M$ has projective dimension at most $n$ and we win. $\square$

Lemma 10.108.12. Let $R$ be a ring. The following are equivalent

  1. $R$ has finite global dimension $\leq n$,

  2. every finite $R$-module has projective dimension $\leq n$, and

  3. every cyclic $R$-module $R/I$ has projective dimension $\leq n$.

Proof. It is clear that (1) $\Rightarrow $ (2) and (2) $\Rightarrow $ (3). Assume (3). Choose a set $E \subset M$ of generators of $M$. Choose a well ordering on $E$. For $e \in E$ denote $M_ e$ the submodule of $M$ generated by the elements $e' \in E$ with $e' \leq e$. Then $M = \bigcup _{e \in E} M_ e$. Note that for each $e \in E$ the quotient

\[ M_ e/\bigcup \nolimits _{e' < e} M_{e'} \]

is either zero or generated by one element, hence has projective dimension $\leq n$ by (3). By Lemma 10.108.11 this means that $M$ has projective dimension $\leq n$. $\square$

Lemma 10.108.13. Let $R$ be a ring. Let $M$ be an $R$-module. Let $S \subset R$ be a multiplicative subset.

  1. If $M$ has projective dimension $\leq n$, then $S^{-1}M$ has projective dimension $\leq n$ over $S^{-1}R$.

  2. If $R$ has finite global dimension $\leq n$, then $S^{-1}R$ has finite global dimension $\leq n$.

Proof. Let $0 \to P_ n \to P_{n - 1} \to \ldots \to P_0 \to M \to 0$ be a projective resolution. As localization is exact, see Proposition 10.9.12, and as each $S^{-1}P_ i$ is a projective $S^{-1}R$-module, see Lemma 10.93.1, we see that $0 \to S^{-1}P_ n \to \ldots \to S^{-1}P_0 \to S^{-1}M \to 0$ is a projective resolution of $S^{-1}M$. This proves (1). Let $M'$ be an $S^{-1}R$-module. Note that $M' = S^{-1}M'$. Hence we see that (2) follows from (1). $\square$


Comments (1)

Comment #689 by Keenan Kidwell on

In the paragraph following 00O4, "extend" should be "extent."


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