This work was performed at the Steklov International Mathematical Center and supported by the Ministry of Science and Higher Education of the Russian Federation (agreement no. 075-15-2025-303).
Received: 23.05.2025 Presented:S. O. Gorchinskiy Accepted: 23.07.2025
Let ${\mathcal T}$ be a triangulated category, and let ${\mathcal I}_1$ and ${\mathcal I}_2$ be two full subcategories of ${\mathcal T}$. Denote by ${\mathcal I}_1*{\mathcal I}_2$ the full subcategory of ${\mathcal T}$ consisting of all objects such that there is a distinguished triangle $M_1\to M\to M_2$ with $M_i\in {\mathcal I}_i$. For any subcategory ${\mathcal I}\subset{\mathcal T}$, denote by $\langle {\mathcal I}\rangle$ (respectively, $\overline{\langle {\mathcal I}\rangle}$) the smallest full subcategory of ${\mathcal T}$ containing ${\mathcal I}$ and closed under finite direct sums, direct summands, and shifts (respectively, all direct sums and shifts). We put ${\mathcal I}_1 \diamond{\mathcal I}_2= \langle {\mathcal I}_1*{\mathcal I}_2\rangle$, and we define inductively $\langle {\mathcal I}\rangle_k=\langle{\mathcal I}\rangle_{k-1}\diamond \langle {\mathcal I}\rangle$ and $\overline{\langle {\mathcal I}\rangle}_k= \overline{\langle{\mathcal I}\rangle}_{k-1}\diamond \overline{\langle {\mathcal I}\rangle}$. If ${\mathcal I}$ consists of a single object $E$, we denote $\langle {\mathcal I}\rangle$ by $\langle E\rangle$ and put inductively $\langle E\rangle _k=\langle E\rangle_{k-1}\diamond\langle E\rangle$.
Let ${\mathscr A}$ be a differential graded algebra ( = DGA) over a base field $\Bbbk$. Denote by ${\mathcal D}({\mathscr A})$ the derived category of all right DG ${\mathscr A}$-modules, and by $\mathrm{perf}\text{-}{\mathscr A}\subset {\mathcal D}({\mathscr A})$ the triangulated subcategory of perfect modules. The latter is generated by ${\mathscr A}$, that is, $\mathrm{perf}\text{-}{\mathscr A}= \bigcup_{k}\langle{\mathscr A}\rangle_k$, and coincides with the subcategory of compact objects ${\mathcal D}({\mathscr A})^c\subset {\mathcal D}({\mathscr A})$ (see [3]).
A DGA ${\mathscr A}$ is called regular if $\mathrm{perf}\text{-}{\mathscr A}=\langle {\mathscr A}\rangle _k$ for some $k$. It is equivalent to the property ${\mathcal D}({\mathscr A})= \overline{\langle {\mathscr A}\rangle}_k$ (see [1], Proposition 2.2.4). A DGA ${\mathscr A}$ is called $\Bbbk$- smooth if ${\mathscr A}\in \mathrm{perf}\text{-}({\mathscr A}^{\circ}\otimes_{\Bbbk}{\mathscr A})$. A DGA ${\mathscr A}$ is called proper if ${\mathscr A}\in\mathrm{perf}\text{-}\Bbbk$. Denote by ${\mathcal D}_{\mathsf{cf}}({\mathscr A})\subset {\mathcal D}({\mathscr A})$ the full triangulated subcategory of all DG modules which are perfect as DG $\Bbbk$-modules. By [1], Theorem 1.3, for any regular and proper DGA ${\mathscr A}$ we have an equivalence ${\mathcal D}_{\mathsf{cf}}({\mathscr A})\cong \mathrm{perf}\text{-}{\mathscr A}$.
Let $h\colon{\mathscr A}\to{\mathscr B}$ be a homomorphism ( = map) of DGAs. Denote by $h^*\colon{\mathcal D}({\mathscr A})\to{\mathcal D}({\mathscr B})$ and $h_*\colon{\mathcal D}({\mathscr B})\to {\mathcal D}({\mathscr A})$ the induction and restriction functors, respectively. Every DG ${\mathscr A}\text{-}{\mathscr B}$-bimodule ${\mathsf T}$ defines a derived tensor functor $F_{\mathsf T}:=(-)\overset{\mathbf L} {\otimes}_{\mathscr A} {\mathsf T}\colon {\mathcal D}({\mathscr A}) \to {\mathcal D}({\mathscr B})$, which commutes with arbitrary direct sums. Denote by $\operatorname{Im} F_{\mathsf T}$ the full subcategory of ${\mathcal D}({\mathscr B})$ that consists of all objects isomorphic to $F_{\mathsf T}(X)$ for some $X\in{\mathcal D}({\mathscr A})$.
Definition 1. The functor $F_{\mathsf T}$ is called paving if ${\mathcal D}({\mathscr B})= \langle \operatorname{Im} F_{\mathsf T}\rangle_{n}$ for some $n\in{\mathbb Z}$.
Note that $\langle \operatorname{Im} F_{\mathsf T}\rangle_{n}= \overline{\langle \operatorname{Im} F_{\mathsf T}\rangle}_{n}$ since $\operatorname{Im} F_{\mathsf T}\subset{\mathcal D}({\mathscr B})$ is closed under direct sums and shifts.
Proposition 2. The composition of paving functors is paving. A DGA ${\mathscr A}$ is regular if and only if a functor ${\mathcal D}(\Bbbk)\to {\mathcal D}({\mathscr A})$ that takes $\Bbbk$ to ${\mathscr A}$ is paving.
Proposition 3. Let ${\mathscr A}$ and ${\mathscr B}$ be DGAs and ${\mathsf T}$ be a DG ${\mathscr A}\text{-}{\mathscr B}$-bimodule. Suppose ${\mathsf T}\!\in\! \overline{\langle {\mathscr B}\rangle}_{n}$ for some $n$ as a DG ${\mathscr B}$-module and $F_{\mathsf T}$ is paving. If ${\mathscr A}$ is regular, then ${\mathscr B}$ is also regular.
Proof. Since ${\mathscr A}$ is regular and $F_{\mathsf T}$ is paving, we get that ${\mathcal D}({\mathscr B})=\overline{\langle {\mathsf T}\rangle}_{m}$ for some $m$. Now, the property ${\mathsf T}\in \overline{\langle {\mathscr B}\rangle}_{n}$ implies that ${\mathcal D}({\mathscr B})=\overline{\langle {\mathscr B}\rangle}_{k}$ for some $k$, that is, that DGA ${\mathscr B}$ is regular.
Let ${\mathsf I}\subset{\mathscr A}$ be a two-sided DG ideal. Denote by ${\mathscr R}$ the quotient DGA ${\mathscr A}/{\mathsf I}$ and by $q\colon{\mathscr A}\to{\mathscr R}$ the natural map of DGAs.
Theorem 4. Let ${\mathscr A}$ be a DGA, ${\mathsf I}\subset{\mathscr A}$ be a two-sided DG ideal, and let ${\mathscr R}={\mathscr A}/{\mathsf I}$. Assume that ${\mathsf I}$ is nilpotent, that is, ${\mathsf I}^{p}=0$ for some $p$, and ${\mathscr R}\in\mathrm{perf}\text{-}{\mathscr A}$ as a right DG ${\mathscr A}$-module. If ${\mathscr R}$ is regular, then ${\mathscr A}$ is also regular. If ${\mathscr A}$ is proper and ${\mathscr R}$ is smooth, then ${\mathscr A}$ is also smooth.
Proof. For any DG ${\mathscr A}$-module ${\mathsf M}$ there is a filtration $0={\mathsf I}^p{\mathsf M}\subseteq{\mathsf I}^{p-1}{\mathsf M}\subseteq\dots\subseteq {\mathsf I}{\mathsf M}\subseteq \mathsf M$. Any quotient ${\mathsf I}^k{\mathsf M}/{\mathsf I}^{k-1}{\mathsf M}$ belongs to $\operatorname{Im} q_*$, where $q_*\colon{\mathcal D}({\mathscr R})\to{\mathcal D}({\mathscr A})$ is the restriction functor. Thus, $q_*$ is paving. Since ${\mathscr R}\in\mathrm{perf}\text{-}{\mathscr A}$, we have ${\mathscr R}\in\langle {\mathscr A}\rangle_{k}$ for some $k$. By Proposition 3 the regularity of ${\mathscr R}$ implies the regularity of ${\mathscr A}$. If ${\mathscr R}$ is smooth, then ${\mathscr R}^{\circ}\otimes_{\Bbbk}{\mathscr R}$ is also smooth; thus, it is regular. Applying the same reasoning to the map ${\mathscr A}^{\circ}\otimes_{\Bbbk}{\mathscr A}\to {\mathscr R}^{\circ}\otimes_{\Bbbk}{\mathscr R}$, we conclude that ${\mathscr A}^{\circ}\otimes_{\Bbbk}{\mathscr A}$ is regular. If ${\mathscr A}$ is proper, then ${\mathcal D}_{\mathsf{cf}}({\mathscr A}^{\circ}\otimes_{\Bbbk}{\mathscr A}) \cong\mathrm{perf}\text{-}({\mathscr A}^{\circ}\otimes_{\Bbbk}{\mathscr A})$. In particular, ${\mathscr A}\in\mathrm{perf}\text{-}({\mathscr A}^{\circ} \otimes_{\Bbbk}{\mathscr A})$, that is, ${\mathscr A}$ is smooth.
Let $h\colon{\mathscr A}\to{\mathscr B}$ be a map of DGAs, and let ${\mathsf T}_{h}$ be the DG ${\mathscr A}\text{-}{\mathscr A}$-bimodule which is the cone of the map $h$ in the DG category of all DG ${\mathscr A}\text{-}{\mathscr A}$-bimodules. Denote by ${\mathsf T}_h^{\otimes k}\in{\mathcal D}({\mathscr A}^{\circ} \otimes_{\Bbbk}{\mathscr A})$ the derived tensor product ${\mathsf T}_h\overset{\mathbf L}{\otimes}_{\mathscr A}\cdots \overset{\mathbf L}{\otimes}_{\mathscr A}{\mathsf T}_h$ of $k$ copies of ${\mathsf T}_h$.
Theorem 5. Let $h\colon{\mathscr A}\to{\mathscr B}$ be a map of DGAs. Suppose ${\mathsf T}_{h}^{\otimes p}\cong 0$ for some $p$. If ${\mathscr A}$ is regular, then ${\mathscr B}$ is also regular. If ${\mathscr B}\in\mathrm{perf}\text{-}{\mathscr A}$ and ${\mathscr B}$ is regular, then ${\mathscr A}$ is also regular.
Proof. For any DG ${\mathscr B}$-module ${\mathsf M}={\mathsf M}_0$ define ${\mathsf M}_k$ inductively as the cone of the map $h^*h_*{\mathsf M}_{k-1}\to {\mathsf M}_{k-1}$. The DG ${\mathscr A}$-module $h_*{\mathsf M}_k$ is isomorphic to ${\mathsf M}\overset{\mathbf L}{\otimes}_{\mathscr A} {\mathsf T}_h^{\otimes k}[k]$. Since ${\mathsf T}_h^{\otimes p}\cong 0$, we obtain ${\mathsf M}_p\cong 0$ for any ${\mathsf M}$. Hence ${\mathcal D}({\mathscr B})= \overline{\langle \operatorname{Im} h^*\rangle}_p$, that is, the functor $h^*$ is paving. Thus, by Proposition 3, if ${\mathscr A}$ is regular, then ${\mathscr B}$ is also regular. For any ${\mathsf N}={\mathsf N}_0\in{\mathcal D}({\mathscr A})$ define ${\mathsf N}_k$ inductively as the cone of the map ${\mathsf N}_{k-1}\to h_*h^*{\mathsf N}_{k-1}$. We have ${\mathsf N}_k\cong{\mathsf N}\overset{\mathbf L}{\otimes}_{\mathscr A} {\mathsf T}_h^{\otimes k}$. The property ${\mathsf T}_h^{\otimes p}\cong 0$ implies that $h_*$ is paving. Since ${\mathscr B}\in\mathrm{perf}\text{-}{\mathscr A}$, we have ${\mathscr B}\in\langle{\mathscr A}\rangle_k$ for some $k$. Thus, if ${\mathscr B}$ is regular, then ${\mathscr A}$ is also regular.
Theorem 6. Let $h\colon{\mathscr A}\to{\mathscr B}$ be a map of DGAs. Suppose ${\mathsf T}_{h}^{\otimes p}\cong 0$ for some $p$ and ${\mathscr B}\in\mathrm{perf}\text{-}{\mathscr A}$. Then the functor $h^*$ induces an isomorphism of K-theories $K_*({\mathscr A})\cong K_*({\mathscr B})$.
Proof. The functor $h_*h^*$ induces a map $[h_*h^*]$ from $K_*({\mathscr A})$ to itself. The map $[h_*h^*]$ is equal to $\operatorname{id}_{K_*({\mathscr A})}+t$, where $t=[-\overset{\mathbf L}{\otimes}_{\mathscr A}{\mathsf T}_h]$. Since ${\mathsf T}_{h}^{\otimes p}\cong 0$, we obtain $t^p=0$. Hence the map $[h_*h^*]$ is invertible. The arguments from the proof of Theorem 5 show that the map $[h^*h_*]$ from $K_*({\mathscr B})$ to itself is also invertible. We thus obtain an isomorphism $K_*({\mathscr A})\cong K_*({\mathscr B})$. Note that in this case we obtain an isomorphism for all additive invariants of ${\mathscr A}$ and ${\mathscr B}$.
Lemma 7 ([4], Lemma 2.8). A DGA ${\mathscr A}$ is smooth if and only if for each DGA ${\mathscr B}$, every DG ${\mathscr A}\text{-}{\mathscr B}$-bimodule ${\mathsf M}$, perfect as a DG ${\mathscr B}$-module, is perfect as a DG ${\mathscr A}\text{-}{\mathscr B}$-bimodule.
Theorem 8. Let $h\colon{\mathscr A}\to{\mathscr B}$ be a map of DGAs. Suppose ${\mathscr B}\in\mathrm{perf}\text{-}{\mathscr A}$ and ${\mathsf T}_{h}^{\otimes p}\cong 0$ for some $p$. If ${\mathscr A}$ is smooth, then ${\mathscr B}$ is smooth too.
Proof. By [2] we can replace $h$ by $h'\colon{\mathscr A}\to{\mathscr B}'$, with ${\mathscr B}'$ quasi-isomorphic to ${\mathscr B}$, such that $h'$ is mono and ${\mathsf T}_{h'}$ is a semi-free DG ${\mathscr A}$-module. Take the reduced Bar-complex:
Since ${\mathscr A}$ is smooth and ${\mathsf T}_{h}\in \mathrm{perf}\text{-}{\mathscr A}$, by Lemma 7, ${\mathscr A}$ and ${\mathsf T}_{h'}^{\otimes k}$ are perfect for each $k$ as DG ${\mathscr A}\text{-}{\mathscr A}$-bimodules. Hence the ${\mathscr B}'\otimes_{\mathscr A}{\mathsf T}_{h'}^{\otimes k} \otimes_{\mathscr A}{\mathscr B}'$ are perfect DG ${\mathscr B}'\text{-}{\mathscr B}'$-bimodules for all $k\geqslant 0$. Since all of them are acyclic for $k\geqslant p$, we conclude that ${\mathscr B}'$ is also perfect as a DG ${\mathscr B}'\text{-}{\mathscr B}'$-bimodule, that is, the DGAs ${\mathscr B}'$ and ${\mathscr B}$ are smooth.
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Citation:
A. I. Efimov, D. O. Orlov, “Remarks on regular and smooth DG algebras”, Russian Math. Surveys, 80:4 (2025), 723–725