The Stacks project

47.22 The ubiquity of dualizing complexes

Many Noetherian rings have dualizing complexes.

Lemma 47.22.1. Let $A \to B$ be a local homomorphism of Noetherian local rings. Let $\omega _ A^\bullet $ be a normalized dualizing complex. If $A \to B$ is flat and $\mathfrak m_ A B = \mathfrak m_ B$, then $\omega _ A^\bullet \otimes _ A B$ is a normalized dualizing complex for $B$.

Proof. It is clear that $\omega _ A^\bullet \otimes _ A B$ is in $D^ b_{\textit{Coh}}(B)$. Let $\kappa _ A$ and $\kappa _ B$ be the residue fields of $A$ and $B$. By More on Algebra, Lemma 15.92.2 we see that

\[ R\mathop{\mathrm{Hom}}\nolimits _ B(\kappa _ B, \omega _ A^\bullet \otimes _ A B) = R\mathop{\mathrm{Hom}}\nolimits _ A(\kappa _ A, \omega _ A^\bullet ) \otimes _ A B = \kappa _ A[0] \otimes _ A B = \kappa _ B[0] \]

Thus $\omega _ A^\bullet \otimes _ A B$ has finite injective dimension by More on Algebra, Lemma 15.66.7. Finally, we can use the same arguments to see that

\[ R\mathop{\mathrm{Hom}}\nolimits _ B(\omega _ A^\bullet \otimes _ A B, \omega _ A^\bullet \otimes _ A B) = R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , \omega _ A^\bullet ) \otimes _ A B = A \otimes _ A B = B \]

as desired. $\square$

Lemma 47.22.2. Let $A \to B$ be a flat map of Noetherian rings. Let $I \subset A$ be an ideal such that $A/I = B/IB$ and such that $IB$ is contained in the Jacobson radical of $B$. Let $\omega _ A^\bullet $ be a dualizing complex. Then $\omega _ A^\bullet \otimes _ A B$ is a dualizing complex for $B$.

Proof. It is clear that $\omega _ A^\bullet \otimes _ A B$ is in $D^ b_{\textit{Coh}}(B)$. By More on Algebra, Lemma 15.92.2 we see that

\[ R\mathop{\mathrm{Hom}}\nolimits _ B(K \otimes _ A B, \omega _ A^\bullet \otimes _ A B) = R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet ) \otimes _ A B \]

for any $K \in D^ b_{\textit{Coh}}(A)$. For any ideal $IB \subset J \subset B$ there is a unique ideal $I \subset J' \subset A$ such that $A/J' \otimes _ A B = B/J$. Thus $\omega _ A^\bullet \otimes _ A B$ has finite injective dimension by More on Algebra, Lemma 15.66.6. Finally, we also have

\[ R\mathop{\mathrm{Hom}}\nolimits _ B(\omega _ A^\bullet \otimes _ A B, \omega _ A^\bullet \otimes _ A B) = R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , \omega _ A^\bullet ) \otimes _ A B = A \otimes _ A B = B \]

as desired. $\square$

Lemma 47.22.3. Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Let $\omega _ A^\bullet $ be a dualizing complex.

  1. $\omega _ A^\bullet \otimes _ A A^ h$ is a dualizing complex on the henselization $(A^ h, I^ h)$ of the pair $(A, I)$,

  2. $\omega _ A^\bullet \otimes _ A A^\wedge $ is a dualizing complex on the $I$-adic completion $A^\wedge $, and

  3. if $A$ is local, then $\omega _ A^\bullet \otimes _ A A^ h$, resp. $\omega _ A^\bullet \otimes _ A A^{sh}$ is a dualzing complex on the henselization, resp. strict henselization of $A$.

Proof. Immediate from Lemmas 47.22.1 and 47.22.2. See More on Algebra, Sections 15.11, 15.42, and 15.44 and Algebra, Sections 10.95 and 10.96 for information on completions and henselizations. $\square$

Lemma 47.22.4. The following types of rings have a dualizing complex:

  1. fields,

  2. Noetherian complete local rings,

  3. $\mathbf{Z}$,

  4. Dedekind domains,

  5. any ring which is obtained from one of the rings above by taking an algebra essentially of finite type, or by taking an ideal-adic completion, or by taking a henselization, or by taking a strict henselization.

Proof. Part (5) follows from Proposition 47.15.11 and Lemma 47.22.3. By Lemma 47.21.3 a regular local ring has a dualizing complex. A complete Noetherian local ring is the quotient of a regular local ring by the Cohen structure theorem (Algebra, Theorem 10.158.8). Let $A$ be a Dedekind domain. Then every ideal $I$ is a finite projective $A$-module (follows from Algebra, Lemma 10.77.2 and the fact that the local rings of $A$ are discrete valuation ring and hence PIDs). Thus every $A$-module has finite injective dimension at most $1$ by More on Algebra, Lemma 15.66.2. It follows easily that $A[0]$ is a dualizing complex. $\square$


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