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 15.99.2 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 15.69.7. 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 15.99.2 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 15.69.6. 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, 15.43, and 15.45 and Algebra, Sections 10.96 and 10.97 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 10.160.8). Let $A$ be a Dedekind domain. Then every ideal $I$ is a finite projective $A$-module (follows from Algebra, Lemma 10.78.2 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 15.69.2. It follows easily that $A[0]$ is a dualizing complex. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DWB. Beware of the difference between the letter 'O' and the digit '0'.