The Stacks project

15.79 Strong generators and regular rings

Let $R$ be a ring. Denote $D(R)_ c$ the saturated full triangulated subcategory of $D(R)$. We already know that

\[ \langle R \rangle = D_{perf}(R) = D(R)_ c \]

See Lemma 15.78.1 and Proposition 15.78.3. It turns out that if $R$ is regular, then $R$ is a strong generator (Derived Categories, Definition 13.36.3).


Lemma 15.79.1. Let $R$ be a ring. Let $n \geq 1$. Let $K \in \langle R \rangle _ n$ with notation as in Derived Categories, Section 13.36. Consider maps

\[ K \xrightarrow {f_1} K_1 \xrightarrow {f_2} K_2 \xrightarrow {f_3} \ldots \xrightarrow {f_ n} K_ n \]

in $D(R)$. If $H^ i(f_ j) = 0$ for all $i, j$, then $f_ n \circ \ldots \circ f_1 = 0$.

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

\[ P' \xrightarrow {i} P \xrightarrow {p} P'' \to P'[1] \]

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

\[ P \xrightarrow {f_1} K_1 \xrightarrow {f_2} K_2 \xrightarrow {f_3} \ldots \xrightarrow {f_ n} K_ n \]

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. $\square$

Lemma 15.79.2. Let $R$ be a Noetherian ring. If $R$ is a strong generator for $D_{perf}(R)$, then $R$ is regular of finite dimension.

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

\[ P = ( P^{-n - 1} \xrightarrow {d^{-n - 1}} P^{-n} \xrightarrow {d^{-n}} P^{-n + 1} \xrightarrow {d^1} \ldots \xrightarrow {d^{-1}} P^0) \]

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

\[ \xi \in \mathop{\mathrm{Ext}}\nolimits ^ n_ R(\mathop{\mathrm{Coker}}(d^{-1}), \mathop{\mathrm{Coker}}(d^{-n - 1})) = \mathop{\mathrm{Ext}}\nolimits ^ n_ R(M, \mathop{\mathrm{Coker}}(d^{-n - 1})) \]

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

\[ K_ j = (\mathop{\mathrm{Coker}}(d^{-n - 1}) \to P^{-n + 1} \to \ldots \to P^{-j}) \]

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

\[ P \to K_1 \to K_2 \to \ldots \to K_ n \]

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. $\square$

Lemma 15.79.3. Let $R$ be a Noetherian regular ring of dimension $d < \infty $. Let $K, L \in D^-(R)$. Assume there exists an $k$ such that $H^ i(K) = 0$ for $i \leq k$ and $H^ i(L) = 0$ for $i \geq k - d + 1$. Then $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(K, L) = 0$.

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

\[ K^ n[-n] \to K^\bullet \to \sigma _{\leq n - 1}K^\bullet \]

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. $\square$

Lemma 15.79.4. Let $R$ be a Noetherian regular ring of dimension $1 \leq d < \infty $. Let $K \in D(R)$ be perfect and let $k \in \mathbf{Z}$ such that $H^ i(K) = 0$ for $i = k - d + 2, \ldots , k$ (empty condition if $d = 1$). Then $K = \tau _{\leq k - d + 1}K \oplus \tau _{\geq k + 1}K$.

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

\[ \tau _{\leq k - d + 1}K \to K \to \tau _{\geq k + 1}K \to (\tau _{\leq k - d + 1}K)[1] \]

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. $\square$

Lemma 15.79.5. Let $R$ be a Noetherian regular ring of finite dimension. Then $R$ is a strong generator for the full subcategory $D_{perf}(R) \subset D(R)$ of perfect objects.

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

\[ K' \to \bigoplus \nolimits _{a \leq n \leq b} F^ n[-n] \to K \to K'[1] \]

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

\[ 0 \to H^ n(K') \to F^ n \to H^ n(K) \to 0 \]

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. $\square$

Proposition 15.79.6. Let $R$ be a Noetherian ring. The following are equivalent

  1. $R$ is regular of finite dimension,

  2. $D_{perf}(R)$ has a strong generator, and

  3. $R$ is a strong generator for $D_{perf}(R)$.

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