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Rigid Dualizing Complexes over Commutative Rings and their Functorial Properties

Mattia Ornaghi, Saurabh Singh, Amnon Yekutieli

Abstract

In this paper we treat Grothendieck Duality for noetherian rings via rigid dualizing complexes. In particular, we prove that every ring, essentially finite type over a regular base ring, has a unique rigid dualizing complex. The rigid dualizing complexes have strong functorial properties, allowing us to construct the twisted induction pseudofunctor, which is our ring-theoretic version of the twisted inverse pseudofunctor $f^{!}$. This is the first article of a bigger project, whose final goal is establishing Grothendieck Duality, including global duality for proper maps, for Deligne-Mumford stacks.

Rigid Dualizing Complexes over Commutative Rings and their Functorial Properties

Abstract

In this paper we treat Grothendieck Duality for noetherian rings via rigid dualizing complexes. In particular, we prove that every ring, essentially finite type over a regular base ring, has a unique rigid dualizing complex. The rigid dualizing complexes have strong functorial properties, allowing us to construct the twisted induction pseudofunctor, which is our ring-theoretic version of the twisted inverse pseudofunctor . This is the first article of a bigger project, whose final goal is establishing Grothendieck Duality, including global duality for proper maps, for Deligne-Mumford stacks.
Paper Structure (16 sections, 135 theorems, 350 equations, 7 figures)

This paper contains 16 sections, 135 theorems, 350 equations, 7 figures.

Key Result

Proposition 1.21

Let $u : A \to B$ be a DG ring homomorphism, and for $i = 1, 2$ let $M_i \in \operatorname{\mathsf{D}}(A)$, $N_i \in \operatorname{\mathsf{D}}(B)$, and let $\theta_i : N_i \to M_i$ be nondegenerate backward morphisms in $\operatorname{\mathsf{D}}(A)$ over $u$.

Figures (7)

  • Figure 1: An illustration for Remark \ref{['rem:1130']}.
  • Figure 2: Illustration for the proof of Theorem \ref{['thm:1015']} and for Remark \ref{['rem:1560']}.
  • Figure 3: $y \in Y' \subseteq Y$, and $Y' \subseteq \tilde{Y}'$.
  • Figure 4: $Y' \times_X Y'$ is a topological subspace of $\tilde{Y}' \times_X \tilde{Y}'$, and $\operatorname{diag}(Y') = \operatorname{diag}(\tilde{Y}') \cap (Y' \times_X Y')$.
  • Figure 5: For the proof of Theorem \ref{['thm:1141']}: $V = \tilde{V} \cap (Y' \times_X Y')$ and $y \in V$.
  • ...and 2 more figures

Theorems & Definitions (345)

  • Definition 1.1
  • Definition 1.3
  • Definition 1.4
  • Definition 1.6
  • Remark 1.7
  • Definition 1.19
  • Proposition 1.21
  • proof
  • Definition 1.26
  • Proposition 1.28
  • ...and 335 more