## 15.76 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

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

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

Lemma 15.76.1. Let $R$ be a ring. Let $K$ and $L$ be objects of $D(R)$. Assume $L$ has projective-amplitude in $[a, b]$, for example if $L$ is perfect of tor-amplitude in $[a, b]$.

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

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

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

**Proof.**
The assumption that $L$ has projective-amplitude in $[a, b]$ means we can represent $L$ by a complex $L^\bullet $ of projective $R$-modules with $L^ i = 0$ for $i \not\in [a, b]$, see Definition 15.68.1. If $L$ is perfect of tor-amplitude in $[a, b]$, then we can represent $L$ by a complex $L^\bullet $ of finite projective $R$-modules with $L^ i = 0$ for $i \not\in [a, b]$, see Lemma 15.74.2. If $H^ i(K) = 0$ for $i \geq a$, then $K$ is quasi-isomorphic to $\tau _{\leq a - 1}K$. Hence we can represent $K$ by a complex $K^\bullet $ of $R$-modules with $K^ i = 0$ for $i \geq a$. Then we obtain

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

by Derived Categories, Lemma 13.19.8. This proves (1). Under the hypotheses of (2) we see that $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(L, K[1]) = 0$ by (1), hence the distinguished triangle is split by Derived Categories, Lemma 13.4.11. The uniqueness of (3) follows from (1).
$\square$

Lemma 15.76.2. 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 $[i + 1, \infty ]$ and 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 10.79.4 we may after replacing $R$ by $R_ f$ for some $f \in R$, $f \not\in \mathfrak p$ assume that

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

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

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 15.76.1 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 15.76.3. 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

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

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

there is a canonical direct sum decomposition

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

in $D(R_ f)$.

**Proof.**
We get (1) from Lemma 15.76.2 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 15.76.2 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 15.76.4. 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 15.76.2. 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 15.76.5. Let $R$ be a ring. Let $K \in D^-(R)$. Let $a \in \mathbf{Z}$. Assume that for any injective $R$-module map $M \to M'$ the map $\mathop{\mathrm{Ext}}\nolimits ^{-a}_ R(K, M) \to \mathop{\mathrm{Ext}}\nolimits ^{-a}_ R(K, M')$ is injective. Then there is a unique direct sum decomposition $K \cong \tau _{\leq a}K \oplus \tau _{\geq a + 1}K$ and $\tau _{\geq a + 1}K$ has projective-amplitude in $[a + 1, b]$ for some $b$.

**Proof.**
Consider the distinguished triangle

\[ \tau _{\leq a}K \to K \to \tau _{\geq a + 1}K \to (\tau _{\leq a}K)[1] \]

in $D(R)$, see Derived Categories, Remark 13.12.4. Observe that $\mathop{\mathrm{Ext}}\nolimits ^{-a}_ R(\tau _{\leq a}K, M) = \mathop{\mathrm{Hom}}\nolimits _ R(H^ a(K), M)$ and $\mathop{\mathrm{Ext}}\nolimits ^{-a - 1}_ R(\tau _{\leq a}K, M) = 0$, see Derived Categories, Lemma 13.27.3. Thus the long exact sequence of $\mathop{\mathrm{Ext}}\nolimits $ gives an exact sequence

\[ 0 \to \mathop{\mathrm{Ext}}\nolimits ^{-a}_ R(\tau _{\geq a + 1}K, M) \to \mathop{\mathrm{Ext}}\nolimits ^{-a}_ R(K, M) \to \mathop{\mathrm{Hom}}\nolimits _ R(H^ a(K), M) \]

functorial in the $R$-module $M$. Now if $I$ is an injective $R$-module, then $\mathop{\mathrm{Ext}}\nolimits ^{-a}_ R(\tau _{\geq a + 1}K, I) = 0$ for example by Derived Categories, Lemma 13.27.2. Since every module injects into an injective module, we conclude that $\mathop{\mathrm{Ext}}\nolimits ^{-a}_ R(\tau _{\geq a + 1}K, M) = 0$ for every $R$-module $M$. By Lemma 15.68.2 we conclude that $\tau _{\geq a + 1}K$ has projective-amplitude in $[a + 1, b]$ for some $b$ (this is where we use that $K$ is bounded above). We obtain the splitting by Lemma 15.76.1.
$\square$

Lemma 15.76.6. Let $R$ be a ring. Let $K \in D^-(R)$. Let $a \in \mathbf{Z}$. Assume $\mathop{\mathrm{Ext}}\nolimits ^{-a}_ R(K, M) = 0$ for any $R$-module $M$. Then there is a unique direct sum decomposition $K \cong \tau _{\leq a - 1}K \oplus \tau _{\geq a + 1}K$ and $\tau _{\geq a + 1}K$ has projective-amplitude in $[a + 1, b]$ for some $b$.

**Proof.**
By Lemma 15.76.5 we have a direct sum decomposition $K \cong \tau _{\leq a}K \oplus \tau _{\geq a + 1}K$ and $\tau _{\geq a + 1}K$ has projective-amplitude in $[a + 1, b]$ for some $b$. Clearly, we must have $H^ a(K) = 0$ and we conclude that $\tau _{\leq a}K = \tau _{\leq a - 1}K$ in $D(R)$.
$\square$

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