Proof. For a proof in the case where A is connected, see [Reference Rogalski41, Lemma 2.1.3]. Because of the subtleties involved in extending to the case of a general graded algebra, we sketch a proof below.

Set $J = J(A)$ . The equivalence of (b) and (c) follows by considering the left (respectively, right) module presentation $0 \to J \to A \to S \to 0$ and applying Schanuel’s lemma along with Lemma 2.2(2) to see that S is finitely presented if and only if $J/J^2$ is a finitely generated module over $A/J = S$ . Since S is finite-dimensional, this occurs if and only if $\dim _k J/J^2 < \infty $ .

To see that (c) $\implies $ (a), it suffices to fix finite-dimensional subspaces $X \subseteq A$ and $V \subseteq J$ such that $A = X + J$ and $J = V + J^2$ , and to verify that $A = X + V + V^2 + V^3 + \cdots $ so that A is finite generated. One may also show that for each graded component $A_m$ , there exists $r \gg 0$ such that $A_m \subseteq X + V + \cdots + V^r$ , proving that A is locally finite.

To verify that (a) $\implies $ (c), begin with a finite-dimensional graded subspace W of A that generates A as a k-algebra. We may assume without loss of generality that W is an $(A_0, A_0)$ -bimodule. Then writing $W = A_0 + V$ for some $(A_0)^e$ -submodule $V \subseteq A_{\geq 1}$ and verifying that $A = A_0 + V + J^2$ , one concludes that $\dim _k(J/J^2) < \infty $ .

Proof. (1) If M is graded perfect, then it is clearly perfect. Conversely, assume that M has a projective resolution of finite type, say

$$ \begin{align*} 0 \to P^{-n} \to \cdots \to P^{-2} \overset{d^{-2}}{\longrightarrow} P^{-1} \overset{d^{-1}}{\longrightarrow} P^0 \to M \to 0, \end{align*} $$

so that M is quasi-isomorphic to the perfect complex $P^{\bullet }$ . Then one may construct a graded resolution $Q^{\bullet }$ of M consisting of finitely generated graded projective modules. Because M is finitely generated, it has a finite set of homogeneous generators, and thus there is a finitely generated graded projective module $Q^0$ with a graded surjection $Q^0 \twoheadrightarrow M$ . The rest of the graded resolution can now be constructed by an inductive argument, with the assistance of the generalized Schanuel’s lemma [Reference Lam24, Corollary 5.5].

(2) Since Q is a bounded above complex of graded projectives, $ {\mathrm {RHom}}_A(Q, N) = {\mathrm {{Hom}}}_A(Q,N)$ and $\text{R}\underline { {\mathrm {{Hom}}}}_A(Q,N) = \underline { {\mathrm {{Hom}}}}_A(Q,N)$ . Since each $Q^i$ is finitely generated, the inclusion $\underline { {\mathrm {{Hom}}}}_A(Q^i, N^j) \to {\mathrm {{Hom}}}_A(Q^i, N^j)$ is an equality for each $i, j$ ; it follows that the natural inclusion of complexes $\underline { {\mathrm {{Hom}}}}_A(Q, N) \to {\mathrm {{Hom}}}_A(Q, N)$ is an equality as required.

In particular, if M is a perfect module, it is quasi-isomorphic to a bounded complex Q of finitely generated graded projectives and we have $ {\mathrm {RHom}}_A(M, N) = {\mathrm {{Hom}}}_A(Q,N)$ and $\text{R}\underline { {\mathrm {{Hom}}}}_A(M,N) = \underline { {\mathrm {{Hom}}}}_A(Q,N)$ ; thus the second statement arises from the first by taking cohomology.

Proof. For notational convenience, we set $j = -i$ throughout this proof. Minimality of $P^\bullet $ implies that the boundary operators of the tensored complex $S \otimes _A P^\bullet $ are all zero. Setting $J = J(A)$ , the homology modules of this complex are equal to

$$ \begin{align*} {\mathrm{Tor}}^A_{-j}(S, M) = A/J \otimes_A P^j \cong P^j/JP^j. \end{align*} $$

Because M is bounded below, the same is true of each term $P^j$ . Similarly, because $P^\bullet $ is minimal, the boundary operators of $ {\mathrm {{Hom}}}_A(P^\bullet ,S)$ are all zero. Combining this observation with tensor-Hom adjointness yields

$$ \begin{align*} {\mathrm{{Ext}}}^{-j}_A(M,S) &\cong {\mathrm{{Hom}}}_A(P^j, S) \cong {\mathrm{{Hom}}}_A(P^j, {\mathrm{{Hom}}}_S(S,S)) \\ &\cong {\mathrm{{Hom}}}_S(S \otimes_A P^j, S) \cong {\mathrm{{Hom}}}_S(P^j/JP^j, S). \end{align*} $$

A direct comparison of these expressions gives $ {\mathrm {{Hom}}}_S( {\mathrm {Tor}}_i(S,M),S) \cong {\mathrm {{Ext}}}^i_A(M,S)$ .

By Lemma 2.2(2), any generating set for $P^j/JP^j$ lifts to a generating set for $P^j$ . Note that the action of A on $P^j/JP^j$ factors through the finite-dimensional algebra $S = A/J$ . Thus $P^j$ is finitely generated (respectively, zero) if and only if $P^j / JP^j \cong {\mathrm {Tor}}^A_{-j}(S,M)$ is finite-dimensional (respectively, zero), establishing (a) $\iff $ (b). Because the left and right S-dual functors $ {\mathrm {{Hom}}}_S(-,S)$ and $ {\mathrm {{Hom}}}_{S^{\mathrm {op}}}(-,S)$ restrict to inverse equivalences between finite-dimensional graded left and right S-modules, we obtain (b) $\iff $ (c).

Proof. Let $e = e_0 + e_1 + \dots + e_n$ be a idempotent, where $e_i \in A_i$ , and note that $e_0$ is a homogeneous idempotent. Suppose that $e \neq 0,1$ is nontrivial; we will prove that $e_0$ is nontrivial. First suppose that $e_0 = 0$ . Fix $d> 0$ minimal such that $e_d \neq 0$ . Then $0 \neq e_d = (e^2)_d = 0$ , a contradiction. Now suppose that $e_0 = 1$ . As $e \neq 1$ , let $d> 0$ be the minimal positive index such that $e_d \neq 0$ . Then $e_d = (e^2)_d = 2e_d$ , yielding the contradiction $e_d = 0$ . Thus $e_0$ is a homogeneous nontrivial idempotent. It is also easy to see that if e is central, then so is $e_0$ . Thus if A has a nontrivial central idempotent, then it has a nontrivial homogeneous central idempotent. The converse is trivially true.

Proof. (1) We prove the ungraded case; the proof in the graded case is similar. For any $(R, S)$ -bimodule M and $(R, T)$ -bimodule N, the proof that $ {\mathrm {{Hom}}}_R(Me, Nf) = e {\mathrm {{Hom}}}_R(M,N) f$ as $(eSe, fTf)$ -bimodules is routine. Then for any complexes $P, Q$ as in the statement one immediately obtains $ {\mathrm {{Hom}}}_R(Pe, Qf) = e {\mathrm {{Hom}}}_R(P,Q) f$ from the definition of the total Hom complex. Considering P as a complex of $R \otimes S^{\mathrm {op}}$ -modules, it may be replaced by a quasi-isomorphic bounded above complex of projective $R \otimes S^{\mathrm {op}}$ -modules, which are also projective as left R-modules. The modules in $Pe$ are then also projective on the left, so we get

$$ \begin{align*} {\mathrm{RHom}}_R(Pe, Qf) = {\mathrm{{Hom}}}_R(Pe, Qf) = e {\mathrm{{Hom}}}_R(P,Q) f = e {\mathrm{RHom}}_R(P, Q)f \end{align*} $$

as required.

(2) Note that P is a complex of $(R, S \otimes Z(R))$ -bimodules and Q a complex of $(R, T \otimes Z(R))$ -bimodules. Since $Pz = zP$ and $Qz = zQ$ , most of the equalities follow immediately from part (1) and the others are similarly easy.

Proof. Let $z_i \in \bigoplus _{j=1}^n A_{(j)}$ denote the element whose ith entry is the identity of $A_{(i)}$ and whose other entries are zero, so that $1 = \sum z_i$ is a sum of orthogonal central idempotents.

For (1), consider the central idempotent $z = z_i \otimes z_i^{\mathrm {op}} \in A^e$ . Note that as a left $A^e$ -module we have $A_{(i)} = zA_{(i)}$ , and also that $zA^e \cong A_{(i)}^e$ as algebras and as left $A^e$ -modules. Then using Lemma 2.8, we have that

$$ \begin{align*} {\mathrm{RHom}}_{A^e}(A_{(i)}, A^e) &= {\mathrm{RHom}}_{A^e}(zA_{(i)}, A^e) \\ &\cong {\mathrm{RHom}}_{zA^e}(zA_{(i)}, zA^e) \\ &\cong {\mathrm{RHom}}_{A_{(i)}^e}(A_{(i)}, A_{(i)}^e). \end{align*} $$

Since $A^e$ is a $(A^e, A^e)$ -bimodule, the isomorphisms above hold as complexes of right $A^e$ -modules.

We deduce (2) as follows, where (1) is invoked in the final isomorphism:

$$ \begin{align*} {\mathrm{RHom}}_{A^e}(A,A^e) &= {\mathrm{RHom}}_{A^e}\left( \bigoplus A_{(i)}, A^e \right) \\ &\cong \bigoplus {\mathrm{RHom}}_{A^e}(A_{(i)}, A^e) \cong \bigoplus {\mathrm{RHom}}_{A_{(i)}^e}(A_{(i)}, A_{(i)}^e). \end{align*} $$

In case the $A_i$ are graded algebras, analogous proofs to those above, using the $\text{R}\underline { {\mathrm {{Hom}}}}$ version of Lemma 2.8, yield the corresponding graded isomorphisms.

Proof. Because the bimodule U induces a Morita self-equivalence of A, it is finitely generated projective (and a generator) on each side. Thus the natural map $\underline { {\mathrm {{Hom}}}}_A(U,A) \to {\mathrm {{Hom}}}_A(U,A)$ is an isomorphism, and similarly for $\underline { {\mathrm {{Hom}}}}_{A^{\mathrm {op}}}(U,A)$ .

Now invertibility of U implies that the natural evaluation map

$$ \begin{align*} {\mathrm{{Hom}}}_{A^{\mathrm{op}}}(U,A) \otimes_A U \to A. \end{align*} $$

is an isomorphism of $(A, A)$ -bimodules. But this map preserves grading, so we in fact have a graded isomorphism

$$ \begin{align*} \underline{{\mathrm{{Hom}}}}_{A^{\mathrm{op}}}(U,A) {\mathrm{\underline{\otimes}}}_A U \overset{\sim}{\longrightarrow} A. \end{align*} $$

Allowing homomorphisms of left modules to act from the right (opposite of the scalars), the evaluation map $U {\mathrm {\underline {\otimes }}}_A \underline { {\mathrm {{Hom}}}}_A(U,A) \to A$ is also a graded isomorphism by symmetry. From this it is straightforward to deduce that U is graded-invertible with inverse bimodule $\underline { {\mathrm {{Hom}}}}_{A^{\mathrm {op}}}(U,A) \cong \underline { {\mathrm {{Hom}}}}_A(U,A)$ .

Proof. (1) Note that $- {\mathrm {\underline {\otimes }}}_A U$ induces an autoequivalence of the category $ {\mathrm {Gr-\!}} A$ , with quasi-inverse given by $- {\mathrm {\underline {\otimes }}}_A U^{-1}$ . Because $S_A$ is a semisimple graded right A-module, the same is true of $S \otimes _A U$ . Thus $(S \otimes _A U) J = 0$ . Because $S \otimes _A U \cong U/JU$ , this means that $(U/JU) J = 0$ and thus $UJ \subseteq JU$ . By a symmetric argument, we have $JU \subseteq UJ$ , giving $UJ = JU$ . This implies that

$$ \begin{align*} S \otimes_A U \cong U/JU = U/UJ \cong U \otimes_A S. \end{align*} $$

(2) Because $V = S \otimes _A U \cong U \otimes _A S$ satisfies $JV = VJ = 0$ , we may consider it as an $(S,S)$ -bimodule. It is easy to deduce from (1) that $U^{-1} \otimes _A S \cong S \otimes _A U^{-1}$ as well, so that this bimodule can also be considered as an $(S,S)$ -bimodule. We claim that $V' = S \otimes _A U^{-1}$ is a graded inverse to V. Indeed,

$$ \begin{align*} V \otimes_S V' &\cong (U \otimes_A S) \otimes_S (S \otimes_A U^{-1}) \\ &\cong U \otimes_A S \otimes_A U^{-1} \\ &\cong S \otimes_A U \otimes_A U^{-1} \\ &\cong S, \end{align*} $$

and similarly one may compute $V' \otimes _S V \cong S$ . Thus V is graded-invertible as an $(S,S)$ -bimodule.

Proof. Part (1) is proved, for instance, in [45, Tag 07VI], and the same proof carries over to the graded case. Then (2) follows directly from (1) because

$$ \begin{align*}& {\mathrm{RHom}}_A(L,M) \otimes^L_B N \cong {\mathrm{RHom}}_A(L,A) \otimes^L_A M \otimes^L_B N \cong {\mathrm{RHom}}_A(L,M \otimes^L_B N).\\[-38pt]\end{align*} $$

Proof. (1) This is well known. For a proof in the case where N consists of a single nonzero module, see [Reference Ginzburg17, p. 68].

(2) Since A is homologically smooth, we may fix a perfect $A^e$ -projective resolution $P^{\bullet }$ of A. By part (1), $P \otimes _A N$ is quasi-isomorphic to N. Because each $N^j$ is finite-dimensional and each $P^i$ is a finitely generated projective $A^e$ -module, $P^i$ is a summand of some $(A^e)^n$ , and then $P^i \otimes _A N^j$ is a summand of $(A^e)^n \otimes _A N^j \cong A \otimes _k N^j$ , which is finitely generated free on the left. So $P^i \otimes _A N^j$ is finitely generated projective. Thus $P \otimes _A N$ is a complex of finitely generated projective left A-modules and so N is perfect.

(3) Let $J = J(A)$ . In this case, $S = A/J$ is perfect as a left A-module by part (2), so that A is finitely generated as an algebra and locally finite according to Lemma 2.3.

Proof. See [Reference Bourbaki7, Section 13], [Reference Curtis and Reiner12, Section 71], [Reference DeMeyer and Ingraham13, Chapter II], or [Reference Weibel49, Section 9.2.1].

Proof. Clearly (a) $\implies $ (b); to prove the converse, assume $ {\mathrm {pdim}}({{\hspace {-0.0001pt}}}_{A^e} A) = d < \infty $ . Since $A^e$ is finite-dimensional and consequently noetherian, all terms in the minimal projective resolution of A in $A^e {\mathrm {\!-Mod}}$ are finitely generated. Thus (b) $\implies $ (a). The equivalence of (b)–(d) is demonstrated in [Reference Buchweitz, Green, Madsen and Solberg10, p. 807].

Proof. Let $k^s \subseteq K$ denote the separable closure of k. By [Reference Bourbaki7, Section 12.7, Corollaire 1 to Proposition 12.8], the extension $S^s = S \otimes k^s$ is semisimple. Thus $S^s \cong \bigoplus _{i=1}^n \mathbb {M}_{r_i}(D_i)$ for some finite-dimensional division $k^s$ -algebras $D_i$ . The centers $L_i = Z(D_i)$ form intermediary fields $k^s \subseteq L_i \subseteq K$ . Since $K/k^s$ is purely inseparable, the same is true for each $K/L_i$ . Thus each $L_i$ is separably closed and consequently has trivial Brauer group [Reference Farb and Dennis16, Corollary 4.6]. It follows that in fact $S^s \cong \bigoplus \mathbb {M}_{r_i}(L_i)$ is a sum of matrix algebras over fields. Now $S^K = S^s \otimes _{k^s} K \cong \bigoplus \mathbb {M}_{r_i}(L_i \otimes _{k^s} K)$ . Each ring $R_i = L_i \otimes _{k^s} K$ is a finite-dimensional commutative K-algebra and hence is Artinian. Then $R_i$ is a direct sum of finitely many commutative local K-algebras, say $R_i = \bigoplus _{j=1}^{n_i} R_{ij}$ . Thus $S^K = \bigoplus _i \bigoplus _{j=1}^{n_i} \mathbb {M}_{r_i}( R_{ij})$ is a finite direct sum of matrix rings over commutative local K-algebras.

Proof. Set $K = \overline {k}$ . It suffices to show that $S^K = S \otimes K$ is semisimple; see [Reference Curtis and Reiner12, Theorem 71.2] or [Reference DeMeyer and Ingraham13, Proposition II.1.8]. Thanks to Lemma 3.5 we have $ {\mathrm {{Ext}}}^1_{S^K}(U,V) = 0$ for any nonisomorphic simple left $S^K$ -modules U and V. Note that $A^K$ has finite global dimension by Lemma 3.4. If there were a nonsplit extension of a simple left $S^K$ -module U by itself, then we would have $ {\mathrm {{Ext}}}^1_{A^K}(U,U) \neq 0$ , contradicting the no-loops conjecture [Reference Igusa20]. It follows that all extensions of simple left $S^K$ -modules split, making $S^K$ semisimple.

Proof. Because $J(A \otimes B) \supseteq (A \otimes B)_{\geq 1} = A \otimes B_{\geq 1} + A_{\geq 1} \otimes B$ , we have $(A \otimes B)/J(A \otimes B) \cong (A \otimes B)_0/J((A \otimes B)_0) = (A_0 \otimes B_0)/J(A_0 \otimes B_0)$ . So without loss of generality, we may replace A and B by their degree-zero parts to assume that $A = A_0$ and $B = B_0$ are finite-dimensional algebras.

Let $J' = A \otimes J(B) + J(A) \otimes B$ . Because A and B are finite-dimensional, their respective Jacobson radicals $J(A)$ and $J(B)$ are nilpotent ideals. It follows that $J'$ is a nilpotent ideal of $A \otimes B$ , so that $J' \subseteq J(A \otimes B)$ .

On the other hand, if V and W are finite dimensional vector spaces with subspaces $V'$ and $W',$ respectively, then it is a standard fact that $V' \otimes W + V \otimes W'$ is the kernel of the natural map $V \otimes W \to V/V' \otimes W/W'$ . Thus the kernel of the natural map $A \otimes B \twoheadrightarrow S \otimes T$ , given by the tensor product of the natural surjections $A \twoheadrightarrow A/J(A)$ and $B \twoheadrightarrow B/J(B)$ , is equal to $J'$ . Because S is separable and T is semisimple, the algebra $S \otimes T \cong (A \otimes B)/J'$ is also semisimple by Lemma 3.3(2). We deduce (as in [Reference Lam25, Exercise 4.11]) that $J(A \otimes B) \subseteq J'$ . So in fact $J' = J(A \otimes B)$ , and the isomorphism $S \otimes T \cong (A \otimes B)/J' = (A \otimes B)/J(A \otimes B)$ follows.

Proof. Let $P^{\bullet } \to M \to 0$ be a minimal projective left A-module resolution of M. Recall that every left bounded graded left projective A-module is of the form $A \otimes _{A_0} Q$ for some graded left projective $A_0$ -module Q. Writing $P^i = A \otimes _{A_0} Q^i$ for some graded projective $A_0$ -modules $Q^i$ , because M is non-negatively graded and the resolution $P^{\bullet }$ is minimal, it easily follows that each $Q^i$ is non-negatively graded. Take the degree-zero part of the resolution of M to obtain $P^{\bullet }_0 \to M_0 \to 0$ , which is an exact sequence of $A_0$ -modules. Then $P^i_0 = Q^i_0$ , which is projective over $A_0$ for each i as it is a summand of $Q^i$ . So we have a projective resolution $P_0^\bullet $ of M over $A_0$ . If $P^\bullet $ is perfect or has length at most d, then the same is true of the $A_0$ -resolution $P_0^\bullet $ .

Proof. (1) We adapt the method of proof from the case of a connected algebra given in [Reference Yekutieli and Zhang51, Lemma 4.3(a)]. Using Lemma 2.13(1) in the second quasi-isomorphism below, we have

$$ \begin{align*} S \otimes^L_A S \cong S \otimes^L_A (A \otimes^L_A S) \cong S^e \otimes^L_{A^e} A, \end{align*} $$

as complexes of left $S^e$ -modules. Taking cohomology of the above yields an isomorphism $ {\mathrm {Tor}}^A_i(S,S) \cong {\mathrm {Tor}}^{A^e}_i(S^e, A)$ as desired.

(2) Because A is perfect as a left $A^e$ -module, it follows from Lemma 3.8 that $A_0$ is perfect as a module over $(A^e)_0 = (A_0)^e$ . Thus $A_0$ is homologically smooth.

Proof. Assume (a) holds. Then S is perfect as both a left and a right A-module by Lemma 3.2(2). Next, $A_0$ is homologically smooth by Lemma 3.9(2). Now Rickard’s Theorem 3.6 implies that $S = A_0/J(A_0)$ is separable over k. Thus (a) $\implies $ (b).

Next, assume that (b) holds; we will deduce (a). Because S is separable we have $A^e/J(A^e) \cong S^e$ by Lemma 3.7. Then Lemma 3.9 (1) yields

$$ \begin{align*} {\mathrm{Tor}}^A_i(A/J(A),A/J(A)) \cong {\mathrm{Tor}}^{A^e}_i(A^e/J(A^e),A). \end{align*} $$

Applying Lemma 2.6 (and its right sided variant) for both A and $A^e$ , we have that S is perfect as a left (respectively, right) A-module if and only if the $ {\mathrm {Tor}}$ vector spaces above are all finite-dimensional and eventually zero for $i \gg 0$ , if and only if A is perfect as a left $A^e$ -module. Thus (b) $\implies $ (a).

If condition (b) (or equivalently condition (a)) holds, then Lemma 2.3 implies that A is a finitely generated and locally finite algebra. To establish the final claim about the splitting of $A \twoheadrightarrow S$ , simply note that this factors through the split surjection $A \twoheadrightarrow A_0$ via the surjection $A_0 \twoheadrightarrow S$ , and that the latter is split when S is separable by a classical result of Wedderburn and Malcev [Reference Curtis and Reiner12, Theorem 72.19].

Proof. First suppose that A and B are homologically smooth. For convenience, denote $A_{(1)} = A$ and $A_{(2)} = B$ . Then $R^e = (A_{(1)} \oplus A_{(2)}) \otimes (A_{(1)} \oplus A_{(2)})^{\mathrm {op}} = \bigoplus _{i,j\in \{1,2\}} A_{(i)} \otimes A_{(j)}^{\mathrm {op}}$ as k-algebras. Let $P_{(i)}$ denote a perfect left $A_{(i)}^e$ -resolution of each $A_{(i)}$ ; then it is also a perfect left $R^e$ -resolution via the projection $R^e \twoheadrightarrow A_{(i)} \otimes A_{(i)}^{\mathrm {op}} = A_{(i)}^e$ . Thus $P_{(1)} \oplus P_{(2)}$ is a perfect left $A^e$ -resolution of $A = A_{(1)} \oplus A_{(2)}$ , and so A is homologically smooth as desired.

Conversely, suppose that R is homologically smooth. Consider the central idempotent $z = (1,0) \in A \oplus B = R$ . If $P^{\bullet }$ is a finite resolution of R by finitely generated $R^e$ -projectives, then $(z \otimes z) P^{\bullet }$ is an $A^e$ -projective resolution of A with the same properties. Thus A is homologically smooth, and the same is true of B by symmetry.

Proof. (1) For any $M, N \in A {\mathrm {\!-Mod}}$ there is a natural map $\psi : {\mathrm {{Hom}}}_A(M,N)^K \to {\mathrm {{Hom}}}_{A^K}(M^K, N^K)$ which respects any bimodule structures involved. The map $\psi $ is clearly an isomorphism when $M = A$ , and one can readily deduce that the same is true when M is finitely generated projective.

Since P is bounded above, we can replace it with a quasi-isomorphic bounded above complex of projective modules. Then $P^K$ is a bounded above complex of projective $A^K$ -modules, so $ {\mathrm {RHom}}_A(P, Q)^K = {\mathrm {{Hom}}}_A(P, Q)^K$ and $ {\mathrm {RHom}}_{A^K}(P^K, Q^K) = {\mathrm {{Hom}}}_{A^K}(P^K, Q^K)$ . Thus one defines $\Psi $ to be the map on complexes whose nth component is the natural map

$$ \begin{align*}\Psi^n: \bigg[\prod_i {\mathrm{{Hom}}}_A(P^i, Q^{i+n})\bigg] \otimes K \to \prod_i {\mathrm{{Hom}}}_{A^K}(P^i \otimes K, Q^{i+n} \otimes K) \end{align*} $$

induced by $\psi $ . When P is perfect, we can assume that P is a bounded complex of finitely generated projectives. In this case in $\Psi ^n$ above, the product $\prod _i$ is actually finite and each $P^i$ is finitely generated. Since tensor products commute with finite products (i.e., direct sums), we see that $\Psi ^n$ is an isomorphism from the first paragraph.

(2) This is a similar argument as part (1), but no additional hypotheses are needed since tensor products commute with arbitrary direct sums. We leave the details to the reader.

The proofs in the graded case are the same.

Proof. (1) Let $P^{\bullet } \to M$ be a projective resolution of M as a left A-module. If M is perfect, then we can take $P^i$ to be finitely generated for all i and zero for $i \ll 0$ . The functor $- \otimes _k K$ is exact, and is easily seen to preserve the properties of being projective and being finitely generated as a module. Then $(P^{\bullet })^K$ is an $A^K$ -projective resolution of $A^K$ , for which $(P^i)^K$ is finitely generated for all i and $0$ for $i \ll 0$ . Thus $M^K$ is perfect.

(2) By part (1), it suffices to assume that $M^K$ is perfect and prove that M is. The conditions on A and M ensure that we can take a minimal graded projective resolution $P^{\bullet } \to M$ of M as a left A-module. Similarly as above, $(P^{\bullet })^K$ is a graded $A^K$ -projective resolution of $M^K$ . We claim that this is also a minimal projective resolution. Suppose that $N, P \in A {\mathrm {\!-Gr}}$ are left bounded graded projective modules. Let $f: N \to P$ be a graded A-module homomorphism with $\ker f \in J(A) N$ . Then $f \otimes 1: N^K \to P^K$ is a graded $(A^K)$ -module homomorphism and clearly $\ker (f \otimes 1) = (\ker f) \otimes K$ since $- \otimes _k K$ is exact. Recall that $J(A) = A_{\geq 1} \oplus J(A_0)$ . Clearly $A_{\geq 1}^K = (A^K)_{\geq 1} \subseteq J(A^K)$ . Since $J(A_0)$ is nilpotent, so is $J(A_0)^K$ and hence $J(A_0)^K \subseteq J(A^K)$ as well. Thus $J(A)^K \subseteq J(A^K)$ . This easily extends to projective modules to prove that for any left bounded graded projective A-module Q, $(J(A) Q)^K \subseteq J(A^K)(Q^K)$ . Thus

$$ \begin{align*} \ker (f \otimes 1) = (\ker f) \otimes K \subseteq (J(A) N)^K \subseteq J(A^K)N^K. \end{align*} $$

Since the condition that $P^{\bullet }$ be minimal is that for each $d^i: P^i \to P^{i-1}$ we have $\ker d^i \subseteq J(A) P^i$ , we see that $(P^K)^{\bullet }$ is also minimal, as claimed. Now suppose that $M^K$ is perfect. Then since $M^K$ has a finite $(A^K)$ -resolution by finitely generated projectives, the minimal graded projective resolution of $M^K$ must have this property. So each $(P^i)^K$ is a finitely generated $A^K$ -module, with $(P^i)^K = 0$ for $i \ll 0$ . It follows that $P^i = 0$ for $i \ll 0$ . It is also easy to see that $(P^i)^K$ finitely generated implies that $P^i$ is a finitely generated A-module, as follows. Let $Q = P^i$ and suppose that $Q = \bigcup Q_j$ is a directed union of submodules. Then $Q^K = \bigcup Q_j^K$ is a directed union of the extended submodules. By finite generation of $Q^K$ , we have some $Q_j^K = Q^K$ . Thus the quotient module $Q^K/Q_j^K \cong (Q/Q_j)^K$ is zero. As K is faithfully flat over k we have $Q/Q_j = 0$ , whence $Q = Q_j$ . Thus $Q = P^i$ is finitely generated.

We see now that M is perfect. The final statement on the lengths of the minimal projective resolutions is also clear from the proof above.

Proof. Note that $(A \otimes K)^e \cong A^e \otimes K$ as (graded) K-algebras. Thus the result follows immediately from Proposition 3.13, applied to the algebra $A^e$ and the module A.

Proof. Denote $R = A \otimes B$ , so that $R^e = A^e \otimes B^e$ . If A and B are homologically smooth, we may fix finite length resolutions $P^{\bullet } \to A \to 0$ and $Q^\bullet \to B \to 0$ , whose terms are finitely generated projective left modules over $A^e$ and $B^e$ , respectively. By Remark 3.15, the complex $P \otimes Q$ forms a projective $R^e$ -resolution of R, from which we deduce that R is homologically smooth.

Proof. Fix a k-central invertible $(A,B)$ -bimodule U, so that $U \otimes _B - \colon B {\mathrm {\!-Mod}} \to A {\mathrm {\!-Mod}}$ and $- \otimes _A U \colon {\mathrm {Mod-\!}} A \to {\mathrm {Mod-\!}} B$ are k-linear equivalences. Then the functor $U \otimes _B - \otimes _B U^{-1}$ gives a k-linear equivalence of categories from k-central $(B,B)$ -bimodules to k-central $(A,A)$ -bimodules; that is, we have a linear equivalence from $B^e {\mathrm {\!-Mod}}$ to $A^e {\mathrm {\!-Mod}}$ . This is readily verified to preserve the tensor product of bimodules up to natural isomorphism, yielding a monoidal equivalence. In particular, the tensor units B and A correspond under the equivalence.

It follows that A has a perfect resolution in $A^e {\mathrm {\!-Mod}}$ if and only if B has a perfect resolution in $B^e {\mathrm {\!-Mod}}$ , establishing the claim.

Proof. For (1), note that $\sup \{n \mid {\mathrm {Tor}}^A_n(S,M) \neq 0\} \leq {\mathrm {pdim}}(M) \leq {\mathrm {gr.pdim}}(M)$ . Now let $P^\bullet \to M \to 0$ be a minimal graded projective resolution of M. Then by Lemma 2.6, we see that $ {\mathrm {Tor}}^A_n(S,M) = 0$ if and only if the term $P^{-n}$ in the complex is zero, giving $\sup \{n \mid {\mathrm {Tor}}^A_n(S,M) \neq 0\} = {\mathrm {gr.pdim}}(M)$ . The inequality $ {\mathrm {gr.gldim}}_l(A) \leq {\mathrm {gldim}}_l(A)$ readily follows.

Part (2) follows from the results of [Reference Eilenberg14, Sections 5 and 6]. Note that similar results appear in [Reference Huishi28] and [Reference Minamoto and Mori33, Proposition 2.7].

To prove (3), we adapt the argument from the connected graded case given in [Reference Berger4]. Note that

$$ \begin{align*} {\mathrm{pdim}}({{\hspace{-0.0001pt}}}_A S) = {\mathrm{gr.gldim}}_l(A) \leq {\mathrm{gldim}}_l(A) \leq {\mathrm{pdim}}({{\hspace{-0.0001pt}}}_{A^e} A), \end{align*} $$

where the first equality follows from part (2), the middle inequality from part (1), and the final inequality from Lemma 3.2. Applying (1) to the left $A^e$ -module A, along with Lemmas 3.7 and 3.9(1), we see that

$$ \begin{align*} {\mathrm{pdim}}({{\hspace{-0.0001pt}}}_{A^e} A) = \sup\{n \mid {\mathrm{Tor}}^{A^e}_n(A^e/J(A^e), A) \neq 0\} = \sup\{n \mid {\mathrm{Tor}}^A_n(S,S) \neq 0\}. \end{align*} $$

As we saw in the proof of part (2), the quantity on the right-hand-side above is $ {\mathrm {pdim}}({{\hspace {-0.0001pt}}}_A S) = {\mathrm {pdim}}(S_A)$ . Thus we obtain

$$ \begin{align*} {\mathrm{pdim}}({{\hspace{-0.0001pt}}}_A S) = {\mathrm{gr.gldim}}_l(A) = {\mathrm{gldim}}_l(A) = {\mathrm{pdim}}({{\hspace{-0.0001pt}}}_{A^e} A) = {\mathrm{pdim}}(S_A), \end{align*} $$

and the rest of the equalities follow by symmetry.

Proof. Considering A as a graded algebra concentrated in degree zero, this follows immediately from Proposition 3.18(3).

Proof. By Theorem 3.10, we know that A is homologically smooth if and only if S is a separable algebra and S is perfect as a left A-module. Thus it is enough to show, under the assumption that S is separable, that A has finite global dimension if and only if S is perfect. Because A is left noetherian, the finite-dimensional module ${{\hspace {-0.0001pt}}}_A S$ is perfect if and only if it has finite projective dimension. The latter holds (since S is separable) if and only if $ {\mathrm {gldim}}(A) = {\mathrm {gr.gldim}}(A)$ is finite, thanks to Proposition 3.18, establishing the claim.

Proof. As in Proposition 3.18(2), the left and right graded global dimensions of A coincide, and similarly for $A_0$ . Thus it suffices to prove that $ {\mathrm {gldim}}_l(A_0) \leq d$ . Given a left $A_0$ -module M, consider it as a graded left A-module concentrated in degree zero. Because the projective dimension of M as a left A-module is at most d, it follows from Lemma 3.8 that it also has projective dimension at most d as a left $A_0$ -module, establishing $ {\mathrm {gldim}}_l(A_0) \leq d$ .

Proof. First note that by Lemma 2.4, A is homologically smooth if and only if it is graded homologically smooth, and in case either condition is satisfied the inclusions $\underline { {\mathrm {{Ext}}}}^i_{A^e}(A, A^e) \subseteq {\mathrm {{Ext}}}^i_{A^e}(A,A^e)$ of right $A^e$ -modules are equalities for all i by Lemma 2.4. For homologically smooth A, the graded bimodule $U = {\mathrm {{Ext}}}^d_{A^e}(A, A^e)$ is invertible if and only if it is graded-invertible, by Lemma 2.10. It follows immediately that A is graded twisted Calabi–Yau algebra of dimension d if and only if it is (ungraded) twisted Calabi–Yau of dimension d.

Proof. (1) This is a standard result. Assume that $M \neq 0$ . By definition, we have a finite projective resolution $P^{\bullet } \to M$ , with all $P^i$ finitely generated. Let $c = {\mathrm {pdim}}_{A}(M)$ . We have

$$ \begin{align*} c = \max \{ i | {\mathrm{{Ext}}}^i_{A}(M, L) \neq 0\ \text{for some}\ A\text{-module}\ L \}. \end{align*} $$

If F is a free A-module with a surjection onto L, from the long exact sequence we obtain $ {\mathrm {{Ext}}}^c_{A}(M, F) \neq 0$ . Since the terms $P^i$ in the resolution of M are finitely generated, it follows from computing $ {\mathrm {{Ext}}}$ with the resolution P of M that $ {\mathrm {{Ext}}}^c_A$ commutes with direct sums in the second coordinate, so that $ {\mathrm {{Ext}}}^c_{A}(M, A) \neq 0$ . Since $ {\mathrm {{Ext}}}^i_A(M, A) = 0$ for $i \geq {\mathrm {pdim}}_A(M)$ , the result follows.

(2) The result is trivial if $M = 0$ , so assume that $M \neq 0$ . By part (1), we have $ {\mathrm {pdim}}(M) = d$ . Thus we can choose a projective resolution of M of the form $P^{\bullet } \to M$ , where

$$ \begin{align*} P^{\bullet} = P^{-d} \to \dots \to P^0. \end{align*} $$

Then $ {\mathrm {RHom}}_A(M, A)= {\mathrm {{Hom}}}_A(P, A)$ is a complex which lives in complex degrees $0$ to d. By assumption, $ {\mathrm {RHom}}_A(M, A)$ has cohomology only in degree d, where the cohomology is N. It is now easy to see that there is a quasi-isomorphism $ {\mathrm {RHom}}_A(M, A) \cong N[-d]$ .

(3) We may again assume that $M \neq 0$ . Then $ {\mathrm {pdim}}(M) = d$ , and it follows from [Reference Minamoto and Mori33, Proposition 3.4] that M is (graded) perfect. Now $ {\mathrm {RHom}}_A(M,A) = {\mathrm{R}}\underline { {\mathrm {{Hom}}}}_A(M,A)$ by Lemma 2.4 so that the quasi-isomorphism in part (2) automatically preserves the grading.

Proof. We prove the ungraded version; the proof in the graded case is the same. It is enough to assume that A is homologically smooth and prove that under this assumption, for an invertible bimodule U, the following conditions are equivalent:

1. (ii) $ {\mathrm {{Ext}}}^i_{A^e}(A,A^e) \cong \begin {cases} 0, & i \neq d \\ U, & i = d \end {cases}$

2. (ii′) $ {\mathrm {RHom}}_{A^e}(A,A^e) \cong U[-d]$ in $D^b(A^e {\mathrm {\!-Mod}})$ .

If (ii ${}^{\prime}$ ) holds, then by taking cohomology we obtain (ii). If (ii) holds, then since A is a perfect $A^e$ -module we obtain (ii ${}^{\prime}$ ) from Lemma 4.3(2).

The last statement follows from Lemma 4.3(1).

Proof. Fix $z \in Z(A)$ and denote $r = z \otimes 1 - 1 \otimes z^{\mathrm {op}} \in Z(A^e)$ . Let $\rho _r \colon A \to A$ denote the $A^e$ -module homomorphism given by (left) multiplication by r. Because A is a central $(A,A)$ -bimodule, we have $\rho _r = 0$ . It follows from [Reference Rotman42, Proposition 6.18] that the $Z(A^e)$ -module endomorphism of $ {\mathrm {{Ext}}}^i_{A^e}(A,A^e)$ given by multiplication by r is equal to

$$ \begin{align*} {\mathrm{{Ext}}}^i_{A^e}(\rho_r, A^e) \colon {\mathrm{{Ext}}}^i_{A^e}(A,A^e) \to {\mathrm{{Ext}}}^i_{A^e}(A,A^e). \end{align*} $$

Because $\rho _r = 0$ , this is the zero morphism. Therefore $U = {\mathrm {{Ext}}}^d_{A^e}(A,A^e)$ is a central $(A,A)$ -bimodule.

Proof. First, suppose that both A and B are twisted Calabi–Yau of dimension d with respective Nakayama bimodules $U_1$ and $U_2$ . Then R is homologically smooth by Proposition 3.11. Note that $U_1 \oplus U_2$ is an invertible $(R,R)$ -bimodule with inverse $U_1^{-1} \oplus U_2^{-1}$ . Applying Lemma 2.9 we have

$$ \begin{align*} {\mathrm{RHom}}_R(R, R^e) &\cong {\mathrm{RHom}}_{A^e}(A, A^e) \oplus {\mathrm{RHom}}_{B^e}(B, B^e) \\ &\cong U_1[-d] \oplus U_2[-d] \\ &\cong (U_1 \oplus U_2)[-d], \end{align*} $$

so that R is twisted Calabi–Yau of dimension d with Nakayama bimodule $U_1 \oplus U_2$ . In case $U_1 = {{\hspace {-0.0001pt}}}^1 A^{\mu _A}$ and $U_2 = {{\hspace {-0.0001pt}}}^1 B^{\mu _B}$ for Nakayama automorphisms $\mu _A$ and $\mu _B$ , then we obtain $U_1 \oplus U_2 = {{\hspace {-0.0001pt}}}^1 R^{\mu _R}$ for $\mu _R = \mu _A \oplus \mu _B$ .

Conversely, suppose that R is twisted Calabi–Yau of dimension d and denote $U = {\mathrm {{Ext}}}^d_{R^e}(R,R^e)$ ; it follows from Proposition 3.11 that both A and B are homologically smooth. Lemma 4.5 implies that the central idempotent $z_1 = (1,0) \in A \oplus B = R$ centralizes U, so that also $U \cong U_1 \oplus U_2$ for $U_1 = z_1U$ and $U_2 = (1-z_1)U$ , which are, respectively, invertible bimodules over A and B. Now, using Lemma 2.8, we have for the central idempotent $z = z_1 \otimes z_1^{\mathrm {op}} \in R^e$ that

$$ \begin{align*} {\mathrm{RHom}}_{A^e}(A,A^e) \cong {\mathrm{RHom}}_{R^e}(zR,R^e) \cong z{\mathrm{RHom}}_{R^e}(R,R^e) = zU[-d] = U_1[-d]. \end{align*} $$

Thus A is twisted Calabi–Yau of dimension d, and the same is true of B by symmetry.

Proof. (1) Suppose that A is twisted Calabi–Yau of dimension d. Then $A^K$ is homologically smooth by Corollary 3.14. Set $U = {\mathrm {{Ext}}}^d_{A^e}(A,A^e)$ . Then using Lemma 3.12(2), since A is perfect as an $A^e$ module we have

$$ \begin{align*} {\mathrm{RHom}}_{(A^K)^e}(A^K,(A^K)^e) &\cong {\mathrm{RHom}}_{(A^e) \otimes K}(A \otimes K, (A^e) \otimes K) \\ &\cong {\mathrm{RHom}}_{A^e}(A,A^e) \otimes K \quad \\ &\cong U \otimes K[-d]. \end{align*} $$

It follows from taking $0$ th cohomology in Lemma 3.12(2) that $(M \otimes _A N)^K \cong M^K \otimes _{A^K} N^K$ for any modules $M \in {\mathrm {Mod-\!}} A$ and $N \in A {\mathrm {\!-Mod}}$ . In particular, $U^K$ is an invertible $A^K$ -bimodule (with inverse $(U^{-1})^K$ ), so that $A^K$ is twisted Calabi–Yau of dimension d.

(2) One direction is a special case of part (1). To prove the converse, suppose that $A^K$ is twisted Calabi–Yau of dimension d. Again, Proposition 3.13 tells us that A is homologically smooth. Now the argument above shows that

$$ \begin{align*} {\mathrm{RHom}}_{(A^K)^e}(A^K,(A^K)^e) \cong {\mathrm{RHom}}_{A^e}(A,A^e) \otimes K, \end{align*} $$

which by assumption is isomorphic to $V[-d]$ for a graded invertible $A^K$ -bimodule V. In particular, taking cohomology we see that $ {\mathrm {RHom}}_{A^e}(A, A^e) \otimes K$ has cohomology only in degree d, so the same is true of $ {\mathrm {RHom}}_{A^e}(A, A^e)$ since the functor $- \otimes K$ is exact. Thus we have $ {\mathrm {{Ext}}}_{A^e}^i(A, A^e) = 0$ for $i \neq d$ and since A is perfect as an $A^e$ -module, Lemma 4.3(2) implies that $ {\mathrm {RHom}}_{A^e}(A, A^e) \cong U[-d]$ for some $(A, A)$ -bimodule U. Clearly $U \otimes K \cong V$ .

Now there are natural evaluation maps of bimodules $\phi _l: U \otimes _A {\mathrm {{Hom}}}_A(U, A) \to A$ and $\phi _r: {\mathrm {{Hom}}}_{A^{\mathrm {op}}}(U, A) \otimes _A U \to A$ , and U is an invertible bimodule if and only if both $\phi _l$ and $\phi _r$ are bijections. Since $U^K = V$ is an invertible $A^K$ -module, it is finitely generated and projective on both sides, and so is certainly perfect as a left and right $A^K$ -module. Then U is perfect as a left and right A-module, by Proposition 3.13(2). By taking $0$ th cohomology in Lemma 3.12(1), we get that the natural map $( {\mathrm {{Hom}}}_A(U, A))^K \to {\mathrm {{Hom}}}_{A^K}(U^K, A^K)$ is an isomorphism, and similarly on the right. Using Lemma 3.12(1)(2) again, extending the base field in $\phi _l$ and $\phi _r$ now yields the morphisms of $(A^K, A^K)$ -bimodules $U^K \otimes _{A^K} {\mathrm {{Hom}}}_{A^K}(U^K, A^K) \to A^K$ and $ {\mathrm {{Hom}}}_{(A^{\mathrm {op}})^K}(U^K, A^K) \otimes _{A^K} U^K \to A^K$ , which are bijections since $U^K = V$ is invertible. By the exactness of base field extension, $\phi _l$ and $\phi _r$ are also bijections, so that U is invertible as required. Thus A is also (graded) twisted Calabi–Yau.

Proof. For left R-modules M and N and S-modules H and J, it is easy to see that there is a natural map

$$ \begin{align*} \phi: {\mathrm{{Hom}}}_R(M, N) \otimes {\mathrm{{Hom}}}_S(H, J) \to {\mathrm{{Hom}}}_{R \otimes S}(M \otimes H, N \otimes J), \end{align*} $$

respecting any bimodule structures that arise. A similar argument as in the first paragraph of the proof of Lemma 3.12 shows that $\phi $ is an isomorphism in case M and H are finitely generated projective modules.

One may now replace P and Q by bounded above complexes of projectives and proceed in the same spirit as the proof Lemma 3.12, obtaining the desired isomorphism if P and Q are perfect. We omit the details.

Proof. Proposition 3.16 implies that $R = A \otimes B$ is homologically smooth. Let $U_1$ and $U_2,$ respectively, denote the Nakayama bimodules of A and B, which are invertible bimodules and therefore projective on each side. Now applying Lemma 4.8 in the second isomorphism below, which gives an isomorphism since A is perfect over $A^e$ and B is perfect over $B^e$ , we obtain

$$ \begin{align*} {\mathrm{RHom}}_{R^e}(R,R^e) &\cong {\mathrm{RHom}}_{A^e \otimes B^e}(A \otimes B, A^e \otimes B^e) \\ &\cong {\mathrm{RHom}}_{A^e}(A, A^e) \otimes {\mathrm{RHom}}_{B^e}(B, B^e) \\ &\cong U_1[-d_1] \otimes U_2[-d_2] = (U_1 \otimes U_2)[-(d_1 + d_2)], \end{align*} $$

where these isomorphisms hold as complexes of right $A^e$ -modules. Thus we see that A is twisted Calabi–Yau of the desired dimension, and it is easy to verify the behavior of Nakayama automorphisms from the above.

Proof. This is the same as in [Reference van den Bergh47, Theorem 1], but without passing to cohomology. Beginning with Lemma 2.12, we have:

$$ \begin{align*} {\mathrm{RHom}}_{A^e}(A,M) &\cong {\mathrm{RHom}}_{A^e}(A,A^e) \otimes^L_{A^e} M \\ &\cong U[-d] \otimes^L_{A^e} M \\ &\cong (U \otimes^L_{A^e} M)[-d].\\[-38pt] \end{align*} $$

Proof. If A is twisted Calabi–Yau of dimension d, then A is homologically smooth by definition, and satisfies the above isomorphisms by passing to cohomology in Lemma 4.10. Conversely, suppose that A is perfect as an $A^e$ -module and satisfies the isomorphisms above. Evaluating these functors at the $(A^e,A^e)$ -bimodule $A^e$ yields isomorphisms $ {\mathrm {{Ext}}}^{i}_{A^e}(A,A^e) \cong {\mathrm {Tor}}_{d-i}^{A^e}(U,A^e)$ as right $A^e$ -modules for $0 \leq i \leq d$ . For $i < d$ one has

$$ \begin{align*} {\mathrm{{Ext}}}^i_{A^e}(A,A^e) \cong {\mathrm{Tor}}_{d-i}^{A^e}(U,A^e) = 0 \end{align*} $$

by flatness of $A^e$ . For $i = d$ we have

$$ \begin{align*} {\mathrm{{Ext}}}^d_{A^e}(A,A^e) \cong {\mathrm{Tor}}_0^{A^e}(U,A^e) = U \otimes_{A^e} A^e \cong U, \end{align*} $$

as right $A^e$ -modules. Thus A is twisted Calabi–Yau of dimension d with Nakayama bimodule U.

Now suppose that A is twisted Calabi–Yau of dimension d with Nakayama bimodule U, and suppose that an algebra B is k-linearly Morita equivalent to A. Then B is homologically smooth by Proposition 3.17. As discussed in the proof of that proposition, a k-linear equivalence of categories $A {\mathrm {\!-Mod}} \to B {\mathrm {\!-Mod}}$ induces a monoidal k-linear equivalence $F \colon A^e {\mathrm {\!-Mod}} \to B^e {\mathrm {\!-Mod}}$ . Because A and B are the respective tensor units, we have $F(A) \cong B$ ; also, $V = F(U)$ must be a k-central invertible $(B,B)$ -bimodule. Given an algebra C, F also induces an equivalence of categories $(A^e \otimes C^{\mathrm {op}}) {\mathrm {\!-Mod}} \to (B^e \otimes C^{\mathrm {op}}) {\mathrm {\!-Mod}}$ by simply transporting the right C-action on a left $A^e$ -module X via F as an algebra morphism $C^{\mathrm {op}} \to {\mathrm {{End}}}_{A^e}(X) \to {\mathrm {{End}}}_{B^e}(F(X))$ ; with a slight abuse of notation we use F to denote this equivalence as well. Now the isomorphism in the statement induces a natural isomorphism

$$ \begin{align*} {\mathrm{Tor}}_i^{A^e}(U,F^{-1}(-)) \cong {\mathrm{{Ext}}}^{d-i}_{A^e}(A,F^{-1}(-)) \colon (B^e \otimes C^{\mathrm{op}}) {\mathrm{\!-Mod}} \to {\mathrm{Mod-\!}} C. \end{align*} $$

Because F is an equivalence, it also preserves the construction of $ {\mathrm {{Ext}}}$ and $ {\mathrm {Tor}}$ spaces, so that applying F to the previous isomorphism yields a natural isomorphism of functors

$$ \begin{align*} {\mathrm{Tor}}_i^{B^e}(V,-) \cong {\mathrm{{Ext}}}^{d-i}_{B^e}(B,-) \colon (B^e \otimes C^{\mathrm{op}}) {\mathrm{\!-Mod}} \to {\mathrm{Mod-\!}} C. \end{align*} $$

Thus B is twisted Calabi–Yau of dimension d, as desired.

Proof. Because M is finite-dimensional over k, there is a natural isomorphism of $(A^e, B^{\mathrm {op}} \otimes C)$ -bimodules $ {\mathrm {{Hom}}}_k(M,N) \cong N \otimes M^*$ . Combining this observation with Lemma 2.13 and Lemma 4.10 yields

$$ \begin{align*} {\mathrm{RHom}}_A(M,N) &\cong {\mathrm{RHom}}_{A^e}(A, {\mathrm{{Hom}}}_k(M,N)) \qquad ( \text{by Lemma 2.13(2)}) \\ &\cong {\mathrm{RHom}}_{A^e}(A, N \otimes M^*) \\ & \cong U \otimes^L_{A^e} (N \otimes M^*)[-d] \qquad (\text{by Lemma 4.10})\\ &\cong M^* \otimes^L_A (U \otimes^L_A N)[-d] \qquad (\text{by Lemma 2.13(1)}), \end{align*} $$

and these are quasi-isomorphisms of complexes of right $B^{\mathrm {op}} \otimes C$ -modules. By the definition of twisted Calabi–Yau, U is invertible and hence projective as both a left and a right A-module, so we obtain

$$ \begin{align*} M^* \otimes^L_A (U \otimes_A N) \cong M^* \otimes^L_A U \otimes^L_A N \cong (M^* \otimes_A U) \otimes^L_A N. \end{align*} $$

The isomorphisms between $ {\mathrm {{Ext}}}$ and $ {\mathrm {Tor}}$ are obtained by taking cohomology.

The proof in the graded case is exactly the same. Note that because M is finite-dimensional and A is homologically smooth, M is perfect by Lemma 3.2. Thus in the graded case there is no difference betweeen $ {\mathrm {RHom}}_A(M,N)$ and ${\mathrm{R}}\underline { {\mathrm {{Hom}}}}_A(M,N)$ , justifying the uniform statement of the result.

Proof. Invoking Proposition 4.12, we compute as follows:

$$ \begin{align*} {\mathrm{RHom}}_A(M,N)^* &\cong {\mathrm{RHom}}_k({\mathrm{RHom}}_A(M,N),k) \\ &\cong {\mathrm{RHom}}_k(M^* \otimes^L_A (U \otimes_A N), k)[d] \\ &\cong {\mathrm{RHom}}_A(U \otimes_A N, {\mathrm{RHom}}_k(M^*,k))[d] \\ &\cong {\mathrm{RHom}}_A(U \otimes_AN, M)[d]. \end{align*} $$

As before, the isomorphism relating the $ {\mathrm {{Ext}}}$ groups is obtained by passing to cohomology.

Again, the same proof applies in the graded case.

Proof. Recall from Lemma 4.4 that $ {\mathrm {pdim}}({{\hspace {-0.0001pt}}}_{A^e} A) = d$ , so that the left and right global dimensions of A are both at most d by Lemma 3.2(1). If A has a finite-dimensional left module $M \neq 0$ , then it also has a finite-dimensional right module $M^* \neq 0$ . So by symmetry, it suffices to show that a finite-dimensional left A-module $M \neq 0$ has projective dimension d. But this follows from Proposition 4.12, since $U = {\mathrm {{Ext}}}^d_{A^e}(A,A^e)$ is invertible and

$$ \begin{align*} {\mathrm{{Ext}}}^d_A(M,A) \cong {\mathrm{Tor}}_0^A(M^*, U) = M^* \otimes_A U \neq 0. \end{align*} $$

If A is graded with $\dim _k A_0 < \infty $ , then $A_0 = A/A_{\geq 1}$ is a nonzero finite-dimensional module and the argument above applies. Alternatively, one may conclude that $d = {\mathrm {pdim}}({{\hspace {-0.0001pt}}}_{A^e} A) = {\mathrm {gldim}}_l(A) = {\mathrm {gldim}}_r(A)$ from Proposition 3.18(3), since $S = A/J(A)$ must be separable by Theorem 3.6.

Proof. See [Reference Lam24, Section 16F].

Proof. The finite-dimensional semisimple algebra S satisfies $S^* \cong S$ as an $(S,S)$ -bimodule by Lemma 4.16, hence also as a graded left A-module. Because ${{\hspace {-0.0001pt}}}_A S$ is annihilated by $J(A)$ , any graded module homomorphism $S \to M$ has image annihilated by $J(A)$ , and therefore has image in $ {\mathrm {soc}}(M)$ . Thus we have $ {\mathrm {{Hom}}}_A(S, M) = {\mathrm {{Hom}}}_S(S, {\mathrm {soc}}(M)) = {\mathrm {soc}}(M)$ . Now applying Proposition 4.12, we have

$$ \begin{align*} {\mathrm{Tor}}^A_d(S_A , {{\hspace{-0.0001pt}}}_A M) &\cong {\mathrm{{Ext}}}^0_A(S^*, U^{-1} \otimes_A M) \\ &\cong {\mathrm{{Ext}}}^0_A(S, U^{-1} \otimes_A M) \\ &= {\mathrm{soc}}(U^{-1} \otimes_A M) \\ &= U^{-1} \otimes_A {\mathrm{soc}}(M) \end{align*} $$

as left S-modules, where we use in the last step that the graded autoequivalence $U^{-1} \otimes _A -$ must preserve the graded socle of a module.

Proof. (1) Invoking Proposition 4.17 in the case where $d> 0$ yields $U^{-1} \otimes _A {\mathrm {soc}}(A) \cong {\mathrm {Tor}}^A_d(S,A) = 0$ . Because U is invertible, we obtain $ {\mathrm {soc}}(A) = 0$ .

(2) If A is finite-dimensional, then we may consider A as a graded algebra with $A = A_0$ . Because A is artinian, its (graded) socle is nonzero. It follows from part (1) above that $d = 0$ .

Proof. Clearly (c) $\implies $ (a). For (a) $\implies $ (d), note that if A is twisted Calabi–Yau of dimension 0, then $ {\mathrm {pdim}}({{\hspace {-0.0001pt}}}_{A^e} A) = 0$ by Lemma 4.4, making A separable by Lemma 3.3(3).

To see that (d) $\implies $ (c), suppose that A is separable. Then ${{\hspace {-0.0001pt}}}_{A^e} A$ is projective, and so A is certainly homologically smooth. As noted earlier, Lemma 3.3(5) shows that a separable algebra must be finite-dimensional. Then we also know that $A^e$ is a finite-dimensional semisimple k-algebra, by Lemma 3.3(4). Since a separable algebra must be semisimple by definition, we have $A^* \cong A$ as $(A, A)$ -bimodules in Lemma 4.16. We now calculate that

$$ \begin{align*} {\mathrm{{Hom}}}_{A^e}(A,A^e) &\cong {\mathrm{{Hom}}}_{A^e}(A, {\mathrm{{Hom}}}_k(A^e,k)) \\ &\cong {\mathrm{{Hom}}}_k(A^e \otimes_{A^e} A, k) \\ &\cong {\mathrm{{Hom}}}_k(A,k) \cong A \end{align*} $$

as right $A^e$ -modules. Furthermore, $ {\mathrm {{Ext}}}_{A^e}^i(A,A^e) = 0$ for $i> 0$ as ${{\hspace {-0.0001pt}}}_{A^e} A$ is projective. Thus A is Calabi–Yau of dimension 0, as desired. Since we already recalled that a separable algebra is finite-dimensional, (d) $\implies $ (b) as well. Finally, (b) $\implies $ (a) follows from Corollary 4.18(2).

Proof. If $A = A_0$ is separable, then it is Calabi–Yau of dimension 0 by Theorem 4.19; because the grading of A is trivial, it is clear that A is graded twisted Calabi–Yau of dimension 0.

Conversely, suppose that A is graded twisted Calabi–Yau of dimension 0. Theorem 4.2(1) shows that A is twisted Calabi–Yau of dimension 0. It follows from Theorem 4.19 that A is separable, hence a finite-dimensional semisimple k-algebra. Thus $J(A) = 0$ ; because $A_{\geq 1} \subseteq J(A)$ for an $\mathbb {N}$ -graded algebra, we obtain $A = A_0$ .

Proof. (1) Since M is bounded below, we have $ {\mathrm {pdim}}(M) = {\mathrm {gr.pdim}}(M) \leq d$ by Proposition 3.18. This now follows directly from Lemma 4.3(3) and its right-sided analog.

(2) This is an easy consequence of the definitions of $\mathcal {X}$ and $\mathcal {Y}$ and the long exact sequence in Ext.

(3) Let $M \in \mathcal {X}$ . Since M is perfect of projective dimension at most d by part (1), M is quasi-isomorphic to a complex of graded A-modules $P^{\bullet }$ of the form $0 \to P^{-d} \to \dots \to P^{-1} \to P^0 \to 0$ , where each $P^i$ is a finitely generated graded projective A-module. Then $N = {\mathrm {RHom}}_A(M, A)[d]$ is identified with a complex $Q^\bullet $ of the form $0 \to Q^0 \to Q^1 \to \dots \to Q^d \to 0$ , where $Q^i = {\mathrm {{Hom}}}_A(P^i, A)$ . Since the $P^i$ are finitely generated, it readily follows that $ {\mathrm {RHom}}_{A^{\mathrm {op}}}( {\mathrm {RHom}}_A(P^{\bullet },A), A)) \cong P^{\bullet }$ , which means that $ {\mathrm {RHom}}_{A^{\mathrm {op}}}(N[-d], A) \cong M$ , or equivalently $ {\mathrm {RHom}}_{A^{\mathrm {op}}}(N, A)[d] \cong M$ . Thus we have $N \in \mathcal {Y}$ .

The whole argument can be repeated starting with $N \in \mathcal {Y}$ and $Q^\bullet $ to get that $M = {\mathrm {RHom}}_{A^{\mathrm {op}}}(N, A)[d]$ is in $\mathcal {X}$ , with $ {\mathrm {RHom}}_A(M, A)[d] \cong N$ . It is now easy to see that the functors $ {\mathrm {RHom}}_A(-, A)[d]$ and $ {\mathrm {RHom}}_{A^{\mathrm {op}}}(-, A)[d]$ yield a contravariant equivalence between $\mathcal {X}$ and $\mathcal {Y}$ as claimed.