The Stacks project

15.77 Recognizing perfect complexes

Some lemmas that allow us to prove certain complexes are perfect.

Lemma 15.77.1. Let $R$ be a ring and let $\mathfrak p \subset R$ be a prime. Let $K$ be pseudo-coherent and bounded below. Set $d_ i = \dim _{\kappa (\mathfrak p)} H^ i(K \otimes _ R^\mathbf {L} \kappa (\mathfrak p))$. If there exists an $a \in \mathbf{Z}$ such that $d_ i = 0$ for $i < a$, then there exists an $f \in R$, $f \not\in \mathfrak p$ and a complex

\[ \ldots \to 0 \to R_ f^{\oplus d_ a} \to R_ f^{\oplus d_{a + 1}} \to \ldots \to R_ f^{\oplus d_{b - 1}} \to R_ f^{\oplus d_ b} \to 0 \to \ldots \]

representing $K \otimes _ R^\mathbf {L} R_ f$ in $D(R_ f)$. In particular $K \otimes _ R^\mathbf {L} R_ f$ is perfect.

Proof. After decreasing $a$ we may assume that also $H^ i(K^\bullet ) = 0$ for $i < a$. By Lemma 15.76.4 after replacing $R$ by $R_ f$ for some $f \in R$, $f \not\in \mathfrak p$ we can write $K^\bullet = \tau _{\leq a - 1}K^\bullet \oplus \tau _{\geq a}K^\bullet $ in $D(R)$ with $\tau _{\geq a}K^\bullet $ perfect. Since $H^ i(K^\bullet ) = 0$ for $i < a$ we see that $\tau _{\leq a - 1}K^\bullet = 0$ in $D(R)$. Hence $K^\bullet $ is perfect. Then we can conclude using Lemma 15.75.6. $\square$

Lemma 15.77.2. Let $R$ be a ring. Let $a, b \in \mathbf{Z}$. Let $K^\bullet $ be a pseudo-coherent complex of $R$-modules. The following are equivalent

  1. $K^\bullet $ is perfect with tor amplitude in $[a, b]$,

  2. for every prime $\mathfrak p$ we have $H^ i(K^\bullet \otimes _ R^{\mathbf{L}} \kappa (\mathfrak p)) = 0$ for all $i \not\in [a, b]$, and

  3. for every maximal ideal $\mathfrak m$ we have $H^ i(K^\bullet \otimes _ R^{\mathbf{L}} \kappa (\mathfrak m)) = 0$ for all $i \not\in [a, b]$.

Proof. We omit the proof of the implications (1) $\Rightarrow $ (2) $\Rightarrow $ (3). Assume (3). Let $i \in \mathbf{Z}$ with $i \not\in [a, b]$. By Lemma 15.76.4 we see that the assumption implies that $H^ i(K^\bullet )_{\mathfrak m} = 0$ for all maximal ideals of $R$. Hence $H^ i(K^\bullet ) = 0$, see Algebra, Lemma 10.23.1. Moreover, Lemma 15.76.4 now also implies that for every maximal ideal $\mathfrak m$ there exists an element $f \in R$, $f \not\in \mathfrak m$ such that $K^\bullet \otimes _ R R_ f$ is perfect with tor amplitude in $[a, b]$. Hence we conclude by appealing to Lemmas 15.74.12 and 15.66.16. $\square$

Lemma 15.77.3. Let $R$ be a ring. Let $K^\bullet $ be a pseudo-coherent complex of $R$-modules. Consider the following conditions

  1. $K^\bullet $ is perfect,

  2. for every prime ideal $\mathfrak p$ the complex $K^\bullet \otimes _ R R_{\mathfrak p}$ is perfect,

  3. for every maximal ideal $\mathfrak m$ the complex $K^\bullet \otimes _ R R_{\mathfrak m}$ is perfect,

  4. for every prime $\mathfrak p$ we have $H^ i(K^\bullet \otimes _ R^{\mathbf{L}} \kappa (\mathfrak p)) = 0$ for all $i \ll 0$,

  5. for every maximal ideal $\mathfrak m$ we have $H^ i(K^\bullet \otimes _ R^{\mathbf{L}} \kappa (\mathfrak m)) = 0$ for all $i \ll 0$.

We always have the implications

\[ (1) \Rightarrow (2) \Leftrightarrow (3) \Leftrightarrow (3) \Leftrightarrow (4) \Leftrightarrow (5) \]

If $K^\bullet $ is bounded below, then all conditions are equivalent.

Proof. By Lemma 15.74.9 we see that (1) implies (2). It is immediate that (2) $\Rightarrow $ (3). Since every prime $\mathfrak p$ is contained in a maximal ideal $\mathfrak m$, we can apply Lemma 15.74.9 to the map $R_\mathfrak m \to R_\mathfrak p$ to see that (3) implies (2). Applying Lemma 15.74.9 to the residue maps $R_\mathfrak p \to \kappa (\mathfrak p)$ and $R_\mathfrak m \to \kappa (\mathfrak m)$ we see that (2) implies (4) and (3) implies (5).

Assume $R$ is local with maximal ideal $\mathfrak m$ and residue field $\kappa $. We will show that if $H^ i(K^\bullet \otimes ^\mathbf {L} \kappa ) = 0$ for $i < a$ for some $a$, then $K$ is perfect. This will show that (4) implies (2) and (5) implies (3) whence the first part of the lemma. First we apply Lemma 15.76.4 with $i = a - 1$ to see that $K^\bullet = \tau _{\leq a - 1}K^\bullet \oplus \tau _{\geq a}K^\bullet $ in $D(R)$ with $\tau _{\geq a}K^\bullet $ perfect of tor-amplitude contained in $[a, \infty ]$. To finish we need to show that $\tau _{\leq a - 1}K$ is zero, i.e., that its cohomology groups are zero. If not let $i$ be the largest index such that $M = H^ i(\tau _{\leq a - 1}K)$ is not zero. Then $M$ is a finite $R$-module because $\tau _{\leq a - 1}K^\bullet $ is pseudo-coherent (Lemmas 15.64.3 and 15.64.8). Thus by Nakayama's lemma (Algebra, Lemma 10.20.1) we find that $M \otimes _ R \kappa $ is nonzero. This implies that

\[ H^ i((\tau _{\leq a - 1}K^\bullet ) \otimes _ R^\mathbf {L} \kappa ) = H^ i(K^\bullet \otimes _ R^\mathbf {L} \kappa ) \]

is nonzero which is a contradiction.

Assume the equivalent conditions (2) – (5) hold and that $K^\bullet $ is bounded below. Say $H^ i(K^\bullet ) = 0$ for $i < a$. Pick a maximal ideal $\mathfrak m$ of $R$. It suffices to show there exists an $f \in R$, $f \not\in \mathfrak m$ such that $K^\bullet \otimes _ R^\mathbf {L} R_ f$ is perfect (Lemma 15.74.12 and Algebra, Lemma 10.17.10). This follows from Lemma 15.77.1. $\square$

Lemma 15.77.4. Let $R$ be a ring. Let $K$ be a pseudo-coherent object of $D(R)$. Let $a, b \in \mathbf{Z}$. The following are equivalent

  1. $K$ has projective-amplitude in $[a, b]$,

  2. $K$ is perfect of tor-amplitude in $[a, b]$,

  3. $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(K, N) = 0$ for all finitely presented $R$-modules $N$ and all $i \not\in [-b, -a]$,

  4. $H^ n(K) = 0$ for $n > b$ and $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(K, N) = 0$ for all finitely presented $R$-modules $N$ and all $i > -a$, and

  5. $H^ n(K) = 0$ for $n \not\in [a - 1, b]$ and $\mathop{\mathrm{Ext}}\nolimits ^{-a + 1}_ R(K, N) = 0$ for all finitely presented $R$-modules $N$.

Proof. From the final statement of Lemma 15.74.2 we see that (2) implies (1). If (1) holds, then $K$ can be represented by a complex of projective modules $P^ i$ with $P^ i = 0$ for $i \not\in [a, b]$. Since projective modules are flat (as summands of free modules), we see that $K$ has tor-amplitude in $[a, b]$, see Lemma 15.66.3. Thus by Lemma 15.74.2 we see that (2) holds.

In conditions (3), (4), (5) the assumed vanishing of ext groups $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(K, M)$ for $M$ of finite presentation is equivalent to the vanishing for all $R$-modules $M$ by Lemma 15.65.1 and Algebra, Lemma 10.11.3. Thus the equivalence of (1), (3), (4), and (5) follows from Lemma 15.68.2. $\square$

The following lemma useful in order to find perfect complexes over a polynomial ring $B = A[x_1, \ldots , x_ d]$.

Lemma 15.77.5. Let $A \to B$ be a ring map. Let $a, b \in \mathbf{Z}$. Let $d \geq 0$. Let $K^\bullet $ be a complex of $B$-modules. Assume

  1. the ring map $A \to B$ is flat,

  2. for every prime $\mathfrak p \subset A$ the ring $B \otimes _ A \kappa (\mathfrak p)$ has finite global dimension $\leq d$,

  3. $K^\bullet $ is pseudo-coherent as a complex of $B$-modules, and

  4. $K^\bullet $ has tor amplitude in $[a, b]$ as a complex of $A$-modules.

Then $K^\bullet $ is perfect as a complex of $B$-modules with tor amplitude in $[a - d, b]$.

Proof. We may assume that $K^\bullet $ is a bounded above complex of finite free $B$-modules. In particular, $K^\bullet $ is flat as a complex of $A$-modules and $K^\bullet \otimes _ A M = K^\bullet \otimes _ A^{\mathbf{L}} M$ for any $A$-module $M$. For every prime $\mathfrak p$ of $A$ the complex

\[ K^\bullet \otimes _ A \kappa (\mathfrak p) \]

is a bounded above complex of finite free modules over $B \otimes _ A \kappa (\mathfrak p)$ with vanishing $H^ i$ except for $i \in [a, b]$. As $B \otimes _ A \kappa (\mathfrak p)$ has global dimension $d$ we see from Lemma 15.66.19 that $K^\bullet \otimes _ A \kappa (\mathfrak p)$ has tor amplitude in $[a - d, b]$. Let $\mathfrak q$ be a prime of $B$ lying over $\mathfrak p$. Since $K^\bullet \otimes _ A \kappa (\mathfrak p)$ is a bounded above complex of free $B \otimes _ A \kappa (\mathfrak p)$-modules we see that

\begin{align*} K^\bullet \otimes _ B^{\mathbf{L}} \kappa (\mathfrak q) & = K^\bullet \otimes _ B \kappa (\mathfrak q) \\ & = (K^\bullet \otimes _ A \kappa (\mathfrak p)) \otimes _{B \otimes _ A \kappa (\mathfrak p)} \kappa (\mathfrak q) \\ & = (K^\bullet \otimes _ A \kappa (\mathfrak p)) \otimes ^{\mathbf{L}}_{B \otimes _ A \kappa (\mathfrak p)} \kappa (\mathfrak q) \end{align*}

Hence the arguments above imply that $H^ i(K^\bullet \otimes _ B^{\mathbf{L}} \kappa (\mathfrak q)) = 0$ for $i \not\in [a - d, b]$. We conclude by Lemma 15.77.2. $\square$

The following lemma is a local version of Lemma 15.77.5. It can be used to find perfect complexes over regular local rings.

Lemma 15.77.6. Let $A \to B$ be a local ring homomorphism. Let $a, b \in \mathbf{Z}$. Let $d \geq 0$. Let $K^\bullet $ be a complex of $B$-modules. Assume

  1. the ring map $A \to B$ is flat,

  2. the ring $B/\mathfrak m_ AB$ is regular of dimension $d$,

  3. $K^\bullet $ is pseudo-coherent as a complex of $B$-modules, and

  4. $K^\bullet $ has tor amplitude in $[a, b]$ as a complex of $A$-modules, in fact it suffices if $H^ i(K^\bullet \otimes _ A^\mathbf {L} \kappa (\mathfrak m_ A))$ is nonzero only for $i \in [a, b]$.

Then $K^\bullet $ is perfect as a complex of $B$-modules with tor amplitude in $[a - d, b]$.

Proof. By (3) we may assume that $K^\bullet $ is a bounded above complex of finite free $B$-modules. We compute

\begin{align*} K^\bullet \otimes _ B^{\mathbf{L}} \kappa (\mathfrak m_ B) & = K^\bullet \otimes _ B \kappa (\mathfrak m_ B) \\ & = (K^\bullet \otimes _ A \kappa (\mathfrak m_ A)) \otimes _{B/\mathfrak m_ A B} \kappa (\mathfrak m_ B) \\ & = (K^\bullet \otimes _ A \kappa (\mathfrak m_ A)) \otimes ^{\mathbf{L}}_{B/\mathfrak m_ A B} \kappa (\mathfrak m_ B) \end{align*}

The first equality because $K^\bullet $ is a bounded above complex of flat $B$-modules. The second equality follows from basic properties of the tensor product. The third equality holds because $K^\bullet \otimes _ A \kappa (\mathfrak m_ A) = K^\bullet / \mathfrak m_ A K^\bullet $ is a bounded above complex of flat $B/\mathfrak m_ A B$-modules. Since $K^\bullet $ is a bounded above complex of flat $A$-modules by (1), the cohomology modules $H^ i$ of the complex $K^\bullet \otimes _ A \kappa (\mathfrak m_ A)$ are nonzero only for $i \in [a, b]$ by assumption (4). Thus the spectral sequence of Example 15.62.1 and the fact that $B/\mathfrak m_ AB$ has finite global dimension $d$ (by (2) and Algebra, Proposition 10.110.1) shows that $H^ j(K^\bullet \otimes _ B^{\mathbf{L}} \kappa (\mathfrak m_ B))$ is zero for $j \not\in [a - d, b]$. This finishes the proof by Lemma 15.77.2. $\square$

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