
## 15.71 Splitting complexes

In this section we discuss conditions which imply an object of the derived category of a ring is a direct sum of its truncations. Our method is to use the following lemma (under suitable hypotheses) to split the canonical distinguished triangles

$\tau _{\leq i}K^\bullet \to K^\bullet \to \tau _{\geq i + 1}K^\bullet \to (\tau _{\leq i}K^\bullet )[1]$

in $D(R)$, see Derived Categories, Remark 13.12.4.

Lemma 15.71.1. Let $R$ be a ring. Let $K^\bullet$ and $L^\bullet$ be complexes of $R$-modules such that $L^\bullet$ is perfect of tor-amplitude in $[a, b]$.

1. If $H^ i(K^\bullet ) = 0$ for $i \geq a$, then $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(L^\bullet , K^\bullet ) = 0$.

2. If $H^ i(K^\bullet ) = 0$ for $i \geq a + 1$, then given any distinguished triangle $K^\bullet \to M^\bullet \to L^\bullet \to K^\bullet [1]$ there is an isomorphism $M^\bullet \cong K^\bullet \oplus L^\bullet$ in $D(R)$ compatible with the maps in the distinguished triangle.

3. If $H^ i(K^\bullet ) = 0$ for $i \geq a$, then the isomorphism in (2) exists and is unique.

Proof. We may assume $L^\bullet$ is a finite complex of finite free $R$-modules with $L^ i = 0$ for $i \not\in [a, b]$, see Lemma 15.69.2. If $H^ i(K^\bullet ) = 0$ for $i \geq a$, then $K^\bullet$ is quasi-isomorphic to $\tau _{\leq a - 1}K^\bullet$, hence we may assume that $K^ i = 0$ for $i \geq a$. Then we obtain

$\mathop{\mathrm{Hom}}\nolimits _{D(R)}(L^\bullet , K^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(R)}(L^\bullet , K^\bullet ) = 0$

by Derived Categories, Lemma 13.19.8. This proves (1). Under the hypotheses of (2) we see that $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(L^\bullet , K^\bullet [1]) = 0$ by (1), hence the distinguished triangle is split by Derived Categories, Lemma 13.4.10. The uniqueness of (3) follows from (1). $\square$

Lemma 15.71.2. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime ideal. Let $K^\bullet$ be a pseudo-coherent complex of $R$-modules. Assume that for some $i \in \mathbf{Z}$ the map

$H^ i(K^\bullet ) \otimes _ R \kappa (\mathfrak p) \longrightarrow H^ i(K^\bullet \otimes _ R^{\mathbf{L}} \kappa (\mathfrak p))$

is surjective. Then there exists an $f \in R$, $f \not\in \mathfrak p$ such that $\tau _{\geq i + 1}(K^\bullet \otimes _ R R_ f)$ is a perfect object of $D(R_ f)$ with tor amplitude in $[i + 1, \infty ]$. Thus a canonical isomorphism

$K^\bullet \otimes _ R R_ f \cong \tau _{\leq i}(K^\bullet \otimes _ R R_ f) \oplus \tau _{\geq i + 1}(K^\bullet \otimes _ R R_ f)$

in $D(R_ f)$.

Proof. In this proof all tensor products are over $R$ and we write $\kappa = \kappa (\mathfrak p)$. We may assume that $K^\bullet$ is a bounded above complex of finite free $R$-modules. Let us inspect what is happening in degree $i$:

$\ldots \to K^{i - 1} \xrightarrow {d^{i - 1}} K^ i \xrightarrow {d^ i} K^{i + 1} \to \ldots$

Let $0 \subset V \subset W \subset K^ i \otimes \kappa$ be defined by the formulas

$V = \mathop{\mathrm{Im}}\left( K^{i - 1} \otimes \kappa \to K^ i \otimes \kappa \right) \quad \text{and}\quad W = \mathop{\mathrm{Ker}}\left( K^ i \otimes \kappa \to K^{i + 1} \otimes \kappa \right)$

Set $\dim (V) = r$, $\dim (W/V) = s$, and $\dim (K^ i \otimes \kappa /W) = t$. We can pick $x_1, \ldots , x_ r \in K^{i - 1}$ which map by $d^{i - 1}$ to a basis of $V$. By our assumption we can pick $y_1, \ldots , y_ s \in \mathop{\mathrm{Ker}}(d^ i)$ mapping to a basis of $W/V$. Finally, choose $z_1, \ldots , z_ t \in K^ i$ mapping to a basis of $K^ i \otimes \kappa /W$. Then we see that the elements $d^ i(z_1), \ldots , d^ i(z_ t) \in K^{i + 1}$ are linearly independent in $K^{i + 1} \otimes _ R \kappa$. By Algebra, Lemma 10.78.3 we may after replacing $R$ by $R_ f$ for some $f \in R$, $f \not\in \mathfrak p$ assume that

1. $d^ i(x_ a), y_ b, z_ c$ is an $R$-basis of $K^ i$,

2. $d^ i(z_1), \ldots , d^ i(z_ t)$ are $R$-linearly independent in $K^{i + 1}$, and

3. the quotient $E^{i + 1} = K^{i + 1}/\sum Rd^ i(z_ c)$ is finite projective.

Since $d^ i$ annihilates $d^{i - 1}(x_ a)$ and $y_ b$, we deduce from condition (2) that $E^{i + 1} = \mathop{\mathrm{Coker}}(d^ i : K^ i \to K^{i + 1})$. Thus we see that

$\tau _{\geq i + 1}K^\bullet = (\ldots \to 0 \to E^{i + 1} \to K^{i + 2} \to \ldots )$

is a bounded complex of finite projective modules sitting in degrees $[i + 1, b]$ for some $b$. Thus $\tau _{\geq i + 1}K^\bullet$ is perfect of amplitude $[i + 1, b]$. Since $\tau _{\leq i}K^\bullet$ has no cohomology in degrees $> i$, we may apply Lemma 15.71.1 to the distinguished triangle

$\tau _{\leq i}K^\bullet \to K^\bullet \to \tau _{\geq i + 1}K^\bullet \to (\tau _{\leq i}K^\bullet )[1]$

(Derived Categories, Remark 13.12.4) to conclude. $\square$

Lemma 15.71.3. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime ideal. Let $K^\bullet$ be a pseudo-coherent complex of $R$-modules. Assume that for some $i \in \mathbf{Z}$ the maps

$H^ i(K^\bullet ) \otimes _ R \kappa (\mathfrak p) \longrightarrow H^ i(K^\bullet \otimes _ R^{\mathbf{L}} \kappa (\mathfrak p)) \quad \text{and}\quad H^{i - 1}(K^\bullet ) \otimes _ R \kappa (\mathfrak p) \longrightarrow H^{i - 1}(K^\bullet \otimes _ R^{\mathbf{L}} \kappa (\mathfrak p))$

are surjective. Then there exists an $f \in R$, $f \not\in \mathfrak p$ such that

1. $\tau _{\geq i + 1}(K^\bullet \otimes _ R R_ f)$ is a perfect object of $D(R_ f)$ with tor amplitude in $[i + 1, \infty ]$,

2. $H^ i(K^\bullet )_ f$ is a finite free $R_ f$-module, and

3. there is a canonical direct sum decomposition

$K^\bullet \otimes _ R R_ f \cong \tau _{\leq i - 1}(K^\bullet \otimes _ R R_ f) \oplus H^ i(K^\bullet )_ f \oplus \tau _{\geq i + 1}(K^\bullet \otimes _ R R_ f)$

in $D(R_ f)$.

Proof. We get (1) from Lemma 15.71.2 as well as a splitting $K^\bullet \otimes _ R R_ f = \tau _{\leq i}K^\bullet \otimes _ R R_ f \oplus \tau _{\geq i + 1}K^\bullet \otimes _ R R_ f$ in $D(R_ f)$. Applying Lemma 15.71.2 once more to $\tau _{\leq i}K^\bullet \otimes _ R R_ f$ we obtain (after suitably choosing $f$) a splitting $\tau _{\leq i}K^\bullet \otimes _ R R_ f = \tau _{\leq i - 1}K^\bullet \otimes _ R R_ f \oplus H^ i(K^\bullet )_ f$ in $D(R_ f)$ as well as the conclusion that $H^ i(K)_ f$ is a flat perfect module, i.e., finite projective. $\square$

Lemma 15.71.4. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime ideal. Let $i \in \mathbf{Z}$. Let $K^\bullet$ be a pseudo-coherent complex of $R$-modules such that $H^ i(K^\bullet \otimes _ R^{\mathbf{L}} \kappa (\mathfrak p)) = 0$. Then there exists an $f \in R$, $f \not\in \mathfrak p$ and a canonical direct sum decomposition

$K^\bullet \otimes _ R R_ f = \tau _{\geq i + 1}(K^\bullet \otimes _ R R_ f) \oplus \tau _{\leq i - 1}(K^\bullet \otimes _ R R_ f)$

in $D(R_ f)$ with $\tau _{\geq i + 1}(K^\bullet \otimes _ R R_ f)$ a perfect complex with tor-amplitude in $[i + 1, \infty ]$.

Proof. This is an often used special case of Lemma 15.71.2. A direct proof is as follows. We may assume that $K^\bullet$ is a bounded above complex of finite free $R$-modules. Let us inspect what is happening in degree $i$:

$\ldots \to K^{i - 2} \to R^{\oplus l} \to R^{\oplus m} \to R^{\oplus n} \to K^{i + 2} \to \ldots$

Let $A$ be the $m \times l$ matrix corresponding to $K^{i - 1} \to K^ i$ and let $B$ be the $n \times m$ matrix corresponding to $K^ i \to K^{i + 1}$. The assumption is that $A \bmod \mathfrak p$ has rank $r$ and that $B \bmod \mathfrak p$ has rank $m - r$. In other words, there is some $r \times r$ minor $a$ of $A$ which is not in $\mathfrak p$ and there is some $(m - r) \times (m - r)$-minor $b$ of $B$ which is not in $\mathfrak p$. Set $f = ab$. Then after inverting $f$ we can find direct sum decompositions $K^{i - 1} = R^{\oplus l - r} \oplus R^{\oplus r}$, $K^ i = R^{\oplus r} \oplus R^{\oplus m - r}$, $K^{i + 1} = R^{\oplus m - r} \oplus R^{\oplus n - m + r}$ such that the module map $K^{i - 1} \to K^ i$ kills of $R^{\oplus l - r}$ and induces an isomorphism of $R^{\oplus r}$ onto the corresponding summand of $K^ i$ and such that the module map $K^ i \to K^{i + 1}$ kills of $R^{\oplus r}$ and induces an isomorphism of $R^{\oplus m - r}$ onto the corresponding summand of $K^{i + 1}$. Thus $K^\bullet$ becomes quasi-isomorphic to

$\ldots \to K^{i - 2} \to R^{\oplus l - r} \to 0 \to R^{\oplus n - m + r} \to K^{i + 2} \to \ldots$

and everything is clear. $\square$

Lemma 15.71.5. Let $R$ be a ring and let $\mathfrak p \subset R$ be a prime. Let $K \in D^-(R)$ be pseudo-coherent. Set $d_ i = \dim _{\kappa (\mathfrak p)} H^ i(K \otimes _ R^\mathbf {L} \kappa )$. 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 15.71.4 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 15.70.6. $\square$

Lemma 15.71.6. 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 15.71.4 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 10.22.1. Moreover, Lemma 15.71.4 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 15.69.11 and 15.63.16. $\square$

Lemma 15.71.7. 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 in $D^{-}(R)$, then all conditions are equivalent.

Proof. By Lemma 15.69.9 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 15.69.9 to the map $R_\mathfrak m \to R_\mathfrak p$ to see that (3) implies (2). Applying Lemma 15.69.9 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 15.71.4 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 15.62.3 and 15.62.9). Thus by Nakayama's lemma (Algebra, Lemma 10.19.1) 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 in $D^{-}(R)$. 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 15.69.11 and Algebra, Lemma 10.16.10). After possibly choosing a smaller $a$ we may assume that also $H^ i(K^\bullet \otimes ^\mathbf {L}_ R \kappa ) = 0$ for $i < a$. By Lemma 15.71.4 after replacing $R$ by $R_ f$ for some $f \in R$, $f \not\in \mathfrak m$ we can write $K^\bullet = \tau _{\leq a - 1}K^\bullet \oplus \tau _{\geq a}K^\bullet$ in $D(R)$. Since $H^ i(K^\bullet ) = 0$ for $i < a$ we see that $\tau _{\leq a - 1}K^\bullet = 0$ in $D(R)$ as desired. $\square$

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

Lemma 15.71.8. 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 15.63.19 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 15.71.6. $\square$

The following lemma is a local version of Lemma 15.71.8. It can be used to find perfect complexes over regular local rings.

Lemma 15.71.9. 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 15.60.1 and the fact that $B/\mathfrak m_ AB$ has finite global dimension $d$ (by (2) and Algebra, Proposition 10.109.1) 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 15.71.6. $\square$

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