## Kernels continued

Given a compact generator $P$ of $D(A)$ and a compact generator $Q$ of $D(B)$, we saw last time that $P \otimes_k Q$ is a compact generator for $D(A \otimes_k B)$.

**Lemma**. If $P$ and $Q$ are compact, there is a quasi-isomorphism \(\mathbf{R} \operatorname{End}(P) \otimes_k \mathbf{R} \operatorname{End}(Q) \to \mathbf{R} \operatorname{End}(P \otimes_k Q)\)

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

We have $$ \mathbf{R} \operatorname{Hom}(P \otimes_k Q, P \otimes_k Q) \cong \mathbf{R} \operatorname{Hom}(P, \mathbf{R} \operatorname{Hom}(Q,P \otimes_k Q)) $$ As a $B$-module, $P \otimes_k Q$ is quasi-isomorphic to a sum of shifts $Q$ since we are still assuming that $k$ is field. We know that $$ \bigoplus \mathbf{R} \operatorname{Hom}(Q,Q) \to \mathbf{R} \operatorname{Hom}(Q,\bigoplus Q) $$ is a quasi-isomorphism. Thus, so is $$ P \otimes_k \mathbf{R} \operatorname{Hom}(Q,Q) \to \mathbf{R} \operatorname{Hom}(Q,P \otimes_k Q) $$ Appealing to the same argument shows that $$ \mathbf{R} \operatorname{Hom}(P,P) \otimes_k \mathbf{R} \operatorname{Hom}(Q,Q) \to \mathbf{R} \operatorname{Hom}(P,P \otimes_k \mathbf{R} \operatorname{Hom}(Q,Q)) $$ is a quasi-isomorphism. ■

Next, we consider the functor \(M \to \Upsilon(M) := \mathbf{R}\operatorname{Hom}(P, M \overset{\mathbf{L}}{\otimes}_B Q)\) where $M$ is a $A$-$B$ bimodule. Let \(Q^\vee := \mathbf{R}\operatorname{Hom}(Q,B)\) as a right $B$-module. Since $Q$ is compact, the map \(Q^\vee \overset{\mathbf{L}}{\otimes}_B Q \to \mathbf{R}\operatorname{Hom}(Q,Q)\) is a quasi-isomorphism. The object $Q^\vee$ is a compact generator for $D(B^{op})$ as \((-)^\vee : (\operatorname{perf} B)^{op} \to \operatorname{perf} B^{op}\) is an equivalence. Thus, if we plug in the $P \otimes_k Q^\vee$ into the functor above we get a quasi-isomorphism between \(\mathbf{R}\operatorname{Hom}(P, P \otimes_k Q^\vee \overset{\mathbf{L}}{\otimes}_B Q) \sim \mathbf{R}\operatorname{Hom}(P,P) \otimes_k \mathbf{R} \operatorname{Hom}(Q,Q)\) So we have \(\Upsilon : D(A \otimes_k B^{op}) \to D(\mathbf{R}\operatorname{Hom}(P,P) \otimes_k \mathbf{R} \operatorname{Hom}(Q,Q)^{\op})\)

Now letâ€™s assume that we have a compact generator $P$ of $D(A)$ and we take

\(B := \mathbf{R}\operatorname{End}(P)\) Then $B \otimes_k P^\vee$ is a compact generator of $D(B \otimes A^{op})$. Applying $\Upsilon$ we get an equivalence \(\Upsilon : D(B \otimes_k A^{op}) \to D(B \otimes_k B^{op})\) Thus, there is some object $U$ for which \(\mathbf{R}\operatorname{Hom}(B, U \overset{\mathbf{L}}{\otimes}_B P) \sim \Delta_B\)

Since \(\mathbf{R}\operatorname{Hom}(B, U \overset{\mathbf{L}}{\otimes}_B P) \cong U \overset{\mathbf{L}}{\otimes}_B P\) we see that \(U \overset{\mathbf{L}}{\otimes}_B P \cong B\) as $B$-modules. Thus, \(\Phi_U : D(A) \to D(B)\) takes the compact generator $P$ to the compact generator $B$. Moreover, it is fully-faithful on $P[l]$. Applying the proposition from the previous class, we have an equivalence.

**Theorem**. For two dg-algebras $A$-$B$, if $D(A) \cong D(B)$, then there exists a $U \in D(B \otimes_k A^{op})$ such that \(\Phi_U : D(A) \to D(B)\) is equivalence.

## Smoothness and properness

We can use bimodules to give a notation of smoothness.

**Definition**. A dg-algebra $A$ is called *homologically smooth* if $\Delta_A$ is a compact $A$-$A$ dg-bimodule.

Recall the *global dimension* of a ring is the maximal $n$ with $\operatorname{Ext}^n_R(M,N) \neq 0$ for some $M$ and $N$.

**Lemma**. Let $R$ be a ring. If $R$ is homologically smooth, then $R$ has finite global dimension.

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

Since $R$ is a compact $R$-$R$ bimodule, then it admits a finite length projective resolution, $P^\bullet \to \Delta_R$. Then, $P^i \otimes_R M$ is a projective module over $R$ and we know that $P^\bullet \otimes_R M$ has no cohomology except in degree $0$ where it is $M$. This means we have a projective resolution of $M$ of length at most the length of the projective resolution of $\Delta_R$. ■

**Example**. Consider $R = k[x]/(x^2)$. We try to resolve $\Delta_R$ as a bimodule. To start, we have \(k[y,z]/(y^2,z^2) \to k[x]/(x^2) \to 0\) where $y,z \mapsto x$. The kernel is generated by $(y-z)$. Since \(y^2 - z^2 = (y+z)(y-z)\) We see that \((y-z) \cong k[y,z]/(y+z,y^2,z^2) \cong k[x]/(x^2)\) This immediately implies that \(\operatorname{Ext}^l_{k[y,z]/(y^2,z^2)}(k[x]/(x^2),k[x]/(x^2)) \neq 0\) for all $l \geq 0$. In particular, $\Delta_R$ cannot be perfect, hence compact. So $R$ is not homologically smooth.

**Definition**. A dg-algebra $A$ over a commutative ring $k$ is *proper* over $k$ if $A$ is perfect as a $k$ dg-module (chain complex).

In particular, we need that $H^i(A) = 0$ for all but finitely many $i$.