The Stacks project

Proof. If $n = 1$, then $K$ is a direct summand in $D(R)$ of a bounded complex $P^\bullet$ whose terms are finite free $R$-modules and whose differentials are zero. Thus it suffices to show any morphism $f : P^\bullet \to K_1$ in $D(R)$ with $H^ i(f) = 0$ for all $i$ is zero. Since $P^\bullet$ is a finite direct sum $P^\bullet = \bigoplus R[m_ j]$ it suffices to show any morphism $g : R[m] \to K_1$ with $H^{-m}(g) = 0$ in $D(R)$ is zero. This follows from the fact that $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(R[-m], K) = H^ m(K)$.

For $n > 1$ we proceed by induction on $n$. Namely, we know that $K$ is a summand in $D(R)$ of an object $P$ which sits in a distinguished triangle

with $P' \in \langle R \rangle _1$ and $P'' \in \langle R \rangle _{n - 1}$. As above we may replace $K$ by $P$ and assume that we have

in $D(R)$ with $f_ j$ zero on cohomology. By the case $n = 1$ the composition $f_1 \circ i$ is zero. Hence by Derived Categories, Lemma 13.4.2 we can find a morphism $h : P'' \to K_1$ such that $f_1 = h \circ p$. Observe that $f_2 \circ h$ is zero on cohomology. Hence by induction we find that $f_ n \circ \ldots \circ f_2 \circ h = 0$ which implies $f_ n \circ \ldots \circ f_1 = f_ n \circ \ldots \circ f_2 \circ h \circ p = 0$ as desired. $\square$

Proof. Assume $D_{perf}(R) = \langle R \rangle _ n$ for some $n \geq 1$. For any finite $R$-module $M$ we can choose a complex

of finite free $R$-modules with $H^ i(P) = 0$ for $i = -n, \ldots , - 1$ and $M \cong \mathop{\mathrm{Coker}}(d^{-1})$. Note that $P$ is in $D_{perf}(R)$. For any $R$-module $N$ we can compute $\mathop{\mathrm{Ext}}\nolimits ^ n_ R(M, N)$ the finite free resolution $P$ of $M$, see Algebra, Section 10.71 and compare with Derived Categories, Section 13.27. In particular, the sequence above defines an element

and for any element $\overline{\xi }$ in $\mathop{\mathrm{Ext}}\nolimits ^ n_ R(M, N)$ there is a $R$-module map $\varphi : \mathop{\mathrm{Coker}}(d^{-n - 1}) \to N$ such that $\varphi$ maps $\xi$ to $\overline{\xi }$. For $j = 1, \ldots , n - 1$ consider the complexes

with $\mathop{\mathrm{Coker}}(d^{-n - 1})$ in degree $-n$ and $P^ t$ in degree $t$. We also set $K_ n = \mathop{\mathrm{Coker}}(d^{-n - 1})[n]$. Then we have maps

which induce vanishing maps on cohomology. By Lemma 15.79.1 since $P \in D_{perf}(R) = \langle R \rangle _ n$ we find that the composition of this maps is zero in $D(R)$. Since $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(P, K_ n) = \mathop{\mathrm{Hom}}\nolimits _{K(R)}(P, K_ n)$ by Derived Categories, Lemma 13.19.8 we conclude $\xi = 0$. Hence $\mathop{\mathrm{Ext}}\nolimits ^ n_ R(M, N) = 0$ for all $R$-modules $N$, see discussion above. It follows that $M$ has projective dimension $\leq n - 1$ by Algebra, Lemma 10.109.8. Since this holds for all finite $R$-modules $M$ we conclude that $R$ has finite global dimension, see Algebra, Lemma 10.109.12. We finally conclude by Algebra, Lemma 10.110.8. $\square$

Proof. Let $K^\bullet$ be a bounded above complex representing $K$, say $K^ i = 0$ for $i \geq n + 1$. After replacing $K^\bullet$ by $\tau _{\geq k + 1}K^\bullet$ we may assume $K^ i = 0$ for $i \leq k$. Then we may use the distinguished triangle

to see it suffices to prove the lemma for $K^ n[-n]$ and $\sigma _{\leq n - 1}K^\bullet$. By induction on $n$, we conclude that it suffices to prove the lemma in case $K$ is represented by the complex $M[-m]$ for some $R$-module $M$ and some $m \geq k + 1$. Since $R$ has global dimension $d$ by Algebra, Lemma 10.110.8 we see that $M$ has a projective resolution $0 \to P_ d \to \ldots \to P_0 \to M \to 0$. Then the complex $P^\bullet$ having $P_ i$ in degree $m - i$ is a bounded complex of projectives representing $M[-m]$. On the other hand, we can choose a complex $L^\bullet$ representing $L$ with $L^ i = 0$ for $i \geq k - d + 1$. Hence any map of complexes $P^\bullet \to L^\bullet$ is zero. This implies the lemma by Derived Categories, Lemma 13.19.8. $\square$

Proof. The vanishing of cohomology shows that we have a distinguished triangle

By Derived Categories, Lemma 13.4.11 it suffices to show that the third arrow is zero. Thus it suffices to show that $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(\tau _{\geq k + 1}K, (\tau _{\leq k - d + 1}K)[1]) = 0$ which follows from Lemma 15.79.3. $\square$

Proof. We will use that an object $K$ of $D(R)$ is perfect if and only if $K$ is bounded and has finite cohomology modules, see Lemma 15.74.14. Strong generators of triangulated categories are defined in Derived Categories, Definition 13.36.3. Let $d = \dim (R)$.

Let $K \in D_{perf}(R)$. We will show $K \in \langle R \rangle _{d + 1}$. By Algebra, Lemma 10.110.8 every finite $R$-module has projective dimension $\leq d$. We will show by induction on $0 \leq i \leq d$ that if $H^ n(K)$ has projective dimension $\leq i$ for all $n \in \mathbf{Z}$, then $K$ is in $\langle R \rangle _{i + 1}$.

Base case $i = 0$. In this case $H^ n(K)$ is a finite $R$-module of projective dimension $0$. In other words, each cohomology is a projective $R$-module. Thus $\mathop{\mathrm{Ext}}\nolimits ^ i_ R(H^ n(K), H^ m(K)) = 0$ for all $i > 0$ and $m, n \in \mathbf{Z}$. By Derived Categories, Lemma 13.27.9 we find that $K$ is isomorphic to the direct sum of the shifts of its cohomology modules. Since each cohomology module is a finite projective $R$-module, it is a direct summand of a direct sum of copies of $R$. Hence by definition we see that $K$ is contained in $\langle R \rangle _1$.

Induction step. Assume the claim holds for $i < d$ and let $K \in D_{perf}(R)$ have the property that $H^ n(K)$ has projective dimension $\leq i + 1$ for all $n \in \mathbf{Z}$. Choose $a \leq b$ such that $H^ n(K)$ is zero for $n \not\in [a, b]$. For each $n \in [a, b]$ choose a surjection $F^ n \to H^ n(K)$ where $F^ n$ is a finite free $R$-module. Since $F^ n$ is projective, we can lift $F^ n \to H^ n(K)$ to a map $F^ n[-n] \to K$ in $D(R)$ (small detail omitted). Thus we obtain a morphism $\bigoplus _{a \leq n \leq b} F^ n[-n] \to K$ which is surjective on cohomology modules. Choose a distinguished triangle

in $D(R)$. Of course, the object $K'$ is bounded and has finite cohomology modules. The long exact sequence of cohomology breaks into short exact sequences

by the choices we made. By Algebra, Lemma 10.109.9 we see that the projective dimension of $H^ n(K')$ is $\leq \max (0, i)$. Thus $K' \in \langle R \rangle _{i + 1}$. By definition this means that $K$ is in $\langle R \rangle _{i + 1 + 1}$ as desired. $\square$

Proof. This is a formal consequence of Lemmas 15.78.1, 15.79.2, and 15.79.5 as well as Derived Categories, Lemma 13.36.6. $\square$