## Strong generation in examples

We saw last time that being strongly generated resulted in representability of many (co)homological functors.

Today, we look at when familiar examples of triangulated categories are strongly generated.

**Question**. Let $R$ be a Noetherian commutative ring. When is $D^b(\operatorname{mod} R)$ strongly generated? When is $\operatorname{perf} R$?

**Definition**. We say that $G$ is a *generator* of a triangulated category $\mathcal T$ for every $E \in \mathcal T$ there is some $n$ with \(E \in \overline{\langle G \rangle}_n\)

Note there is no a priori uniform, in $E$, bound on $n$.

**Lemma**. If $\mathcal T$ is strongly generated, then any generator of $\mathcal T$ is a strong generator.

## **Proof**. (Expand to view)

Assume that $$ \overline{\langle G \rangle}_n = \mathcal T $$ and $H$ is a generator. Then there is some $m$ with $$ G \in \overline{\langle H \rangle}_m $$ But then $$ \mathcal T = \overline{\langle G \rangle}_n \subseteq \overline{\langle \overline{\langle H \rangle}_m \rangle}_n = \overline{\langle H \rangle}_{nm} \subset \mathcal T $$ ■

**Example**. $R$ is a always a generator for $\operatorname{perf} R$. So if $\operatorname{perf} R$ is strongly generated then $R$ must also be a strong generator.

**Lemma**. Let $F: \mathcal T \to \mathcal S$ be an exact, essentially surjective functor. If $\mathcal T$ is strongly generated, then so is $\mathcal S$.

## **Proof**. (Expand to view)

We have $$ F\left( \overline{\langle X \rangle}_m \right) \subset \overline{\langle FX \rangle}_m $$ So if $F\mathcal T = \mathcal S$ and $\overline{\langle X \rangle}_m = \mathcal T$, we get $$ \overline{\langle FX \rangle}_m = \mathcal S $$ ■

**Lemma**. Let $\mathcal T = \langle \mathcal A, \mathcal B \rangle$ be a semi-orthogonal decomposition. Then $\mathcal T$ is strongly generated if and only if $\mathcal A$ and $\mathcal B$ both are.

## **Proof**. (Expand to view)

In a SOD, we have essentially surjective functors $$ \mathcal T \to \mathcal A \\ \mathcal T \to \mathcal B $$ so if $\mathcal T$ is strongly generated then so are both $\mathcal A$ and $\mathcal B$. Assume that $G$ is a strong generator of $\mathcal A$ and $H$ is strong generator of $\mathcal B$. In an SOD, any object sits in a triangle $$ B \to X \to A \to B[1] $$ with $A \in \mathcal A$ and $B \in \mathcal B$. Since $$ A \in \overline{\langle G \rangle}_m $$ and $$ B \in \overline{\langle H \rangle}_n $$ we have $$ X \in \overline{\langle G \oplus H \rangle}_{mn} $$ ■

Assume that Noetherian commutative $R$ is strong generator for $\operatorname{perf} R$. We know that the ideal of $R$ is uniformly nilpotent and this ideal is exactly the maps in $D(R)$ which induce the zero map on cohomology.

Let $t$ be a natural number. Let $M$ be a finitely generated module and $N$ an $R$-module. Then, taking a truncation of projective resolution of $M$ we see there is perfect complex $P$ with a map $P \to M$ such that \(\operatorname{Hom}(P,N[i]) \to \operatorname{Hom}(M,N[i])\) is an isomorphism for $0 \leq i \leq t$.

Now take $N = R$. If for $t > 1$ \(\operatorname{Hom}(M,R[t]) = 0\) then we see that $M$ is a retract of a perfect complex of length $t$.

Since any $n$-fold extension can be factored into a composition of extensions, the uniform bound on the nilpotence tell us that every finitely has a projective resolution of length at most the minimal number $t$ with \(\operatorname{perf} R = \overline{\langle R \rangle}_t\)

As a consequence, \(D^b(\operatorname{mod} R) = \operatorname{perf} R\)

Recall that we can test for regularity of local ring $(R,\mathfrak m)$ by asking whether the projective dimension of $k(\mathfrak{m})$ is finite.

**Corollary**. $R$ is a strong generator for $\operatorname{perf} R$ if and only if $R$ is regular, ie all local rings are regular.

So strong generation of $\operatorname{perf} R$ imposes strong conditions on the ring $R$. It turns out that is not the case for $D^b(\operatorname{mod} R)$.

The following result is due to Rouquier.

**Theorem**. If $R$ is a finitely-generated $k$-algebra with $k$ a perfect field, then $D^b(\operatorname{mod} R)$ is strongly generated.

This result has since been improved.