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 (2)

Comment #1684 by on

It's perhaps better to say that (5) follows from 0A7K in general. In the current proof there's just "any ring essentially of finite type over a regular local ring [is Gorenstein]". This is a bit confusing.

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