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

for any $A \in M_n(R)$. Fix a basis $e_1,\ldots,e_n$ for $R^{\oplus n}$. We take $E_{ij}$ $$E_{ij}(e_l) := \begin{cases} e_j & l = i \\ 0 & \text{else} \end{cases}$$ Then, $$\phi \circ E_{ij}(e_i) = \phi(e_j) = \sum \phi_{jl} e_l$$ while $$E_{ij} \circ \phi(e_i) = \phi_{ij} e_j$$ Thus, $\phi$ is a represented by a diagonal matrix: $\phi(e_i) = \phi_{ii} e_i$ for each $i$. And $\phi_{jj} e_j = \phi_{ii} e_j$ so $\phi$ is an $R$ multiple of the identity map.

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

with exact rows and where the left two vertical maps are isomorphisms. The final vertical map is then also an isomorphism. Again, the other statement is analogous.

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 $\mathcal D = \operatorname{Mod} M_n(R)$.