## Adjunctions and derivation

Suppose we are given two additive functors between abelian categories $F: \mathcal A \to \mathcal B$ and $G: \mathcal B \to \mathcal A$ which form an adjoint $F \vdash G$.

These immediately yield an adjoint pair at the level of the homotopy categories \(F: K(\mathcal A) \leftrightarrows K(\mathcal B) : G\)

**Lemma**. Assume that $\mathcal A$ has enough projectives and countable sums while $\mathcal B$ has enough injectives and countable products. Then we have adjoint pair $\mathbf{L}F \vdash \mathbf{R}G$.

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

Let $P \to X$ be a quasi-isomorphism with $P$ K-projective and let $Y \to I$ be a quasi-isomorphism with $I$ K-injective. We have a chain of natural isomorphisms $$ \begin{aligned} \operatorname{Hom}_{D(\mathcal B)}(\mathbf{L}F X, Y ) & \cong \operatorname{Hom}_{D(\mathcal B)}(F P, Y) \\ & \cong \operatorname{Hom}_{K(\mathcal B)}(FP,I) \\ & \cong \operatorname{Hom}_{K(\mathcal A)}(P,GI) \\ & \cong \operatorname{Hom}_{D(\mathcal A)}(X,GI) \\ & \cong \operatorname{Hom}_{D(\mathcal A)}(X,\mathbf{R}GY) \end{aligned} $$ ■

## Composition and derivation

Now let’s assume that we have additive functors between abelian categories $F_1: \mathcal A \to \mathcal B$ and $F_2: \mathcal B \to \mathcal C$.

In general, \(\mathbf{R}F_1 \circ \mathbf{R}F_2 \not \cong \mathbf{R}(F_1 \circ F_2)\) and similarly for left derived functors.

**Example**. Let’s consider $k$ as a bimodule over $k[x]$ and look at the functor \(F := - \otimes_{k[x]} k : \operatorname{Mod} k[x] \to \operatorname{Mod} k[x]\)

If $\mathbf{L}F \circ \mathbf{L}F \cong \mathbf{L} F^2$, we would have \(M \overset{\mathbf{L}}{\otimes}_{k[x]} \overset{\mathbf{L}}{\otimes}_{k[x]} k \cong M \overset{\mathbf{L}}{\otimes}_{k[x]} k \otimes_{k[x]} k\) In particular, if we took $M = R$, then we would need \(k \overset{\mathbf{L}}{\otimes}_{k[x]} k \cong k \otimes_{k[x]} k\) which is problematic since \(H^{-1}( k \overset{\mathbf{L}}{\otimes}_{k[x]} k ) = k.\)

However, there are some situations where the composition of derived functors is equal to the derived functor of the composition.

**Proposition**. Assume that $\mathcal A$ has enough injectives $I$ for which $F_1I$ is adapted to $F_2$. Then the natural map \(\mathbf{R}F_1 \circ \mathbf{R}F_2 \to \mathbf{R}(F_1 \circ F_2)\) is a quasi-isomorphism.

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

We can compute $\mathbf{R}F_2$ applied to $\mathbf{R}F_1 X$ using $\mathbf{R}F_1 X$ itself since applying $F_2$ to complexes of adapted objects preserves quasi-isomorphisms. ■

## Morita equivalence

We now turn to studying the following question. To shorten notation we write \(D(R) := D(\operatorname{Mod} R)\) for $R$ a ring.

**Question**. Assume \(D(R) \cong D(S).\) How do $R$ and $S$ compare?

The naive guess is that $R$ and $S$ are isomorphic. This is not true but also not terribly far off.

Before we head to an equivalence of derived categories, we should note that \(\operatorname{Mod} R \cong \operatorname{Mod} S\) does not imply $R \cong S$ in general.

**Definition**. Two rings $R$ and $S$ are called *Morita equivalent* if there is an equivalence of categories of module categories. \(\operatorname{Mod} R \cong \operatorname{Mod} S\)

The name comes a paper of Kiiti Morita which gives a characterization such equivalences.

**Theorem**. $R$ and $S$ are Morita equivalent if and only if there exists an $R$-$S$ bimodule $U$ for which \(- \otimes_R U : \operatorname{Mod} R \to \operatorname{Mod} S\) is an equivalence.

You can extract properties which are equivalent to $U$ furnishing an equivalence but we will not elaborate here as we will be focused on derived Morita equivalence next.

What we will provide the following counter-example to the naive expectation. From reading the proof, you can tease out some of these condition, though.

**Proposition**. Let $R$ be a commutative ring and let \(M_n(R) := \operatorname{End}_R(R^{\oplus n})\) for some $n > 0$. Then, $R$ and $M_n(R)$ are Morita equivalent.

Let $U := R^{\oplus n}$. Consider the functor \(- \otimes_R U : \operatorname{Mod} R \to \operatorname{Mod} M_n(R)\)

This comes with a right adjoint \(\operatorname{Hom}_{M_n(R)}(U,-): \operatorname{Mod} M_n(R) \to \operatorname{Mod} R\)

We consider the full subcategories $\mathcal C$ where the unit \(\eta : M \to \operatorname{Hom}_{M_n(R)}(U,M \otimes_R U)\) is an isomorphism and $\mathcal D$ where the counit \(\epsilon : \operatorname{Hom}_{M_n(R)}(U,N) \otimes_R U \to N\) is an isomorphism.

We will prove the proposition be establishing a few properties of $\mathcal C$ and $\mathcal D$ which will then imply \(\begin{aligned} \mathcal C & \cong \operatorname{Mod} R \\ \mathcal D & \cong \operatorname{Mod} M_n(R) \end{aligned}\)

**Lemma**. The homomorphism \(\begin{aligned} R & \to \operatorname{Hom}_{M_n(R)}(U,U) \\ r & \mapsto r \cdot 1_U \end{aligned}\) is an isomorphism.

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

Let $\phi: U \to U$ be a $M_n(R)$-module morphism. Then we have a commutative diagram

**Lemma**. $R \in \mathcal C$ and $M_n(R) \in \mathcal D$.

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

We need to check that the map $$ \begin{aligned} R & \to \operatorname{Hom}_{M_n(R)}(U, R \otimes_R U) \\ r & \mapsto r \cdot 1_U \end{aligned} $$ is an isomorphism as is $$ \begin{aligned} \operatorname{Hom}_{M_n(R)}(U,U) \otimes_R U & \to U \\ \phi \otimes u & \mapsto \phi(u) \end{aligned} $$ For the first map, via the natural isomorphism $R \otimes_R U \cong U$, we reduce to the previous lemma. The previous lemma also gives the second. ■

**Lemma**. The natural map \(\bigoplus_{i \in I} \operatorname{Hom}_{M_n(R)}(U,N_i) \to \operatorname{Hom}_{M_n(R)}(U, \bigoplus_{i \in I} N_i)\) is an isomorphism for any $N_a, a \in A$.

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

Note that $$ M_n(R) \cong U^{\oplus n} $$ in $\operatorname{Mod} M_n(R)$. Thus, $U$ is a retract of $M_n(R)$ and the natural map $$ \bigoplus_{i \in I} \operatorname{Hom}_{M_n(R)}(M_n(R),N_i) \to \operatorname{Hom}_{M_n(R)}(M_n(R), \bigoplus_{i \in I} N_i) $$ is an isomorphism. Thus, it is also for $U$. ■

**Lemma**. Assume that $M_i \in \mathcal C$ for $i \in I$ then $\bigoplus_{i \in I} M_i \in \mathcal C$.

Similarly, if $N_i \in \mathcal D$ for $i \in I$, then $\bigoplus_{i \in I} N_i \in \mathcal D$.

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

We have to check that $$ \bigoplus_{i \in I} M_i \to \operatorname{Hom}_{M_n(R)}(U, \bigoplus_{i \in I} M_i \otimes_R U) $$ is an isomorphism. For each each $i$ we have isomorphisms $$ M_i \to \operatorname{Hom}_{M_n(R)}(U, M_i \otimes_R U) $$ Taking direct sums gives an isomorphism $$ \bigoplus_{i \in I} M_i \to \bigoplus_{i \in I} \operatorname{Hom}_{M_n(R)}(U, M_i \otimes_R U) $$ The original map is the composition of this with the map $$ \bigoplus_{i \in I} \operatorname{Hom}_{M_n(R)}(U, M_i \otimes_R U) \to \operatorname{Hom}_{M_n(R)}(U, \bigoplus_{i \in I} M_i \otimes_R U) $$ The final map is an isomorphism from the previous lemma. Checking the other statement is analogous. ■

One final property.

**Lemma**. Assume we have an exact sequence \(F^{-1} \to F^0 \to M \to 0\) with $F^i \in \mathcal C$. Then $M \in \mathcal C$.

Similarly, if we have an exact sequence \(G^{-1} \to G^0 \to N \to 0\) with $G^i \in \mathcal D$, then $N \in \mathcal D$.

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

Since $U$ is projective both as an $R$-module and as an $M_n(R)$-module, we know that $- \otimes_R$ and $\operatorname{Hom}(U,-)$ are exact. Thus we have a commutative diagram

We can now give the proof the proposition.

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

The full subcategory $\mathcal C \subseteq \operatorname{Mod} R$ - contains $R$ - is closed under all coproducts and - is closed under cokernels. Since any module admits a presentation $$ R^{\oplus I_1} \to R^{\oplus I_0} \to M \to 0 $$ we see that all modules are in $\mathcal C$. The same argument shows that $\mathcal D = \operatorname{Mod} M_n(R)$. ■