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The following definition is the one, I found here http://stacks.math.columbia.edu/tag/0A7A. But let me recall it (out of consistency's sake):

Definition For $A$ a Noetherian ring, a dualizing complex for it is a complex of $A$-modules say $\omega^{\bullet}_A$ with the following properties

1.) $\omega^{\bullet}_A$ has finite injective dimension.

2.) $H^i(\omega^{\bullet}_A)$, is a finite $A$-modules for all $i$.

3.) $A \rightarrow RHom_A(\omega^{\bullet}_A,\omega^{\bullet}_A)$, is a quasi-isomoprhism.

Now many questions come along this definition. Because no other information is provided concerning it, I don't understand various things.

Firstly, what does it mean for a complex to have finite injective dimension? I think that probably means that since we're working over an abelian category, there is always a (co-chain) map $\omega^{\bullet}_A \rightarrow I^{\bullet}$, which is quasi-isomorphism and the right-hand side complex consisting of injective modules (straightforward from Cartan-Eilenberg resolution). Does finite mean that $I^{\bullet}$ is bounded in that case?

Secondly, concerning 3.), do we treat $A$ as a cochain complex on its own, where we have at $0$ and $1$ position only $A$, and zero elsewhere?

One last question, under which circumstances for the ring $A$ this dualizing complex exists? And why do we substitute the dualizing module with that instead in some cases?

P.S.1

I asked the same question on MSE but I think isn't very easy question for it (even if it is soft in general), so I posted here.

P.S.2

Of course as soon as I have a response here, will deleted by MSE.

Thank you!

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    $\begingroup$ A more illuminating condition to as #3 (explaining the name "dualizing complex", and originally used by Grothendieck) is expressed in terms of the "derived dual" functor $\mathbf{D}(M^{\bullet}) := \mathbf{R}Hom(M^{\bullet}, \omega^{\bullet}_A)$ that carries $D^b_c(A)$ to itself due #1 and #2: the natural map ${\rm{id}} \rightarrow \mathbf{D} \circ \mathbf{D}$ is an isomorphism. (One can deduce that these properties characterize $\omega^{\bullet}_A$, when it exists, uniquely up to tensoring against $L[n]$ for invertible $L$ and an integer $n$ on each connected component of Spec($A$).) $\endgroup$
    – nfdc23
    Commented Sep 7, 2017 at 5:52
  • $\begingroup$ There are some non-trivial consequences of existence of such $\omega^{\bullet}_A$, such as that $A$ must have finite Krull dimension. So noetherian rings with infinite Krull dimension (such as made by Nagata, but not encountered when walking down the street) don't have such. However, existence is very robust, inherited by quotient rings, and for regular rings the ring itself is a dualizing complex (as is $L[n]$ for any invertible module $L$ and any integer $n$). Since all rings you encounter walking down the street are quotients of regular rings, existence is rather ubiquitous. $\endgroup$
    – nfdc23
    Commented Sep 7, 2017 at 5:55
  • $\begingroup$ @nfdc23 thank you for your comments. What about the ubiquity of 1.and 3.? $\endgroup$
    – mayer_vietoris
    Commented Sep 7, 2017 at 7:04
  • $\begingroup$ Why do you want to drop #2? $\endgroup$
    – nfdc23
    Commented Sep 7, 2017 at 7:09
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    $\begingroup$ @mayer_vietoris A slicker definition of "finite injective dimension" (as alluded to by nfdc23) would be that taking the internal hom in the derived category $RHom(-,\omega_A)$ takes $D^b_c(A)$ (bounded derived category of complexes with coherent cohomologies) to itself. Regarding (3) I think that whenever you see a module in place of a complex with no shift signs near it you can safely assume that the complex it refers to is the module itself placed in degree 0 and 0 everywhere else. $\endgroup$
    – Saal Hardali
    Commented Sep 7, 2017 at 7:21

1 Answer 1

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A. Your guess as to what it means for a complex to have finite injective dimension is indeed the standard definition.

B. You treat $A$ as a complex by letting it be concentrated entirely in degree zero. That is, the zeroth entry is $A$, and all other entries are the zero module.

C. Existence is equivalent to $A$ being the homomorphic image of a Gorenstein ring of finite Krull dimension. See the mathoverflow question State of the art on a question on the existence of dualizing complex .

D. The 'dualizing module' is the right substitute for dualizing complex in the case where $A$ is Cohen-Macaulay. What I mean is this: When $A$ is Cohen-Macaulay, then a dualizing module exists if and only if $A$ is the homomorphic image of a Gorenstein ring. When $A$ is not Cohen-Macaulay, there is no dualizing module. When a dualizing module exists, its injective resolution satisfies the conditions of a dualizing complex. Furthermore, any dualizing complex for $A$ is quasi-isomorphic to the dualizing module concentrated in degree zero. I like the development of dualizing modules (also called 'canonical modules') in the book of Bruns and Herzog, chapter 3.

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