Sorry for a not research level question asking for a definition but unfortunately I nowhere found a source which explains the construction presented below in a satisfactory way.

This question refers to definition of dualizing complex $\omega_A ^{\bullet}$ presented in the Stacks project Stacks as an object satisfying following properties:

(1) $ω^∙_A$ has finite injective dimension,

(2) $H^i(ω^∙_A)$ is a finite A-module for all i, and

(3) $A→RHomA(ω^∙_A,ω^∙_A)$ is a quasi-isomorphism.

My rudimentary question is what is the complex $\mathrm{RHom}_A(ω^∙_A,ω^∙_A)$ and how is is it given in each degree? what is the for example the $k$-degree value of $\mathrm{RHom}_A(ω^∙_A,ω^∙_A)$ concretely?

Naively if I try to imitate the calculations of higher derived functors then I would try following: I would take injective resolution $I^∙_k$ of $ω^∙_A[k]$, apply the functor $\mathrm{Hom}(ω^∙_A[k],-)$ to the resolution $I^∙_k \toω^∙_A[k]$ and take the first homology. does this give me $R^n\mathrm{Hom}_A(ω^∙_A,ω^∙_A)[k]$ or is my approach wrong? in case of $R^n\mathrm{Hom}_A(ω^∙_A,ω^∙_A)$ I would do same procedure but take the $n$-th homology.

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    $\begingroup$ If you don't know what is a derived category, you'll be in trouble to understand the dualizing complex. You might try Hartshorne's Residues and duality which starts from scratch. $\endgroup$
    – abx
    Jul 1, 2019 at 16:58
  • $\begingroup$ I have elementary knowledge of derived catories so if the explaination not usues too many deep considerations from dc than I shall try to understand $\endgroup$ Jul 1, 2019 at 17:06
  • $\begingroup$ in Hartshorne $RF$ is described as follows: if $F: A \to B$ is a functor then the induced functor $RF: D(A) \to D(B)$ has the characteristical property $H^i(RF(X))= R^iF(X)$. on the left hand side $X$ is interpreted as a class of a complex $X \to I^{\bullet}$ with it's injective resolution, on the right it's the conventional derived functor. so in our case $F=Hom(-,-)$ is the bifunctor. the relation $H^i(RF(X))= R^iF(X)$ helps to determine $RHomA(ω^∙_A,ω^∙_A)$ if $ω^∙_A$ would be homotopic to a complex of the shape $M \to I^{\bullet}$ but generally this cannot be expected. $\endgroup$ Jul 1, 2019 at 17:44
  • $\begingroup$ is there general expression what is $RHomA(C^∙,D^∙)$ in $k$-th degree for arbitrary two representant complexes $C^∙, D^∙$? $\endgroup$ Jul 1, 2019 at 17:44

1 Answer 1


In the derived category, one doesn't want to directly compute say the $k$-th slot of a complex as complexes can have many different quasi-isomorphic representations. That is in the derived category, there is no distinction between say a module M thought out as a complex supported in degree zero and a projective resolution.

Instead, you might want to know what the cohomology is of the dualizing complex. Indeed, in general $h^n( \mathbf{R}\textrm{Hom}_R(L,M)) = \textrm{Hom}_{D(R)}(L,M[n])$ where $R$ is a commutative ring and $L$ and $M$ are complexes of $R$-modules, see SP TAG 0A64, however this is not always easy to apply to an example. Sometimes it is easier to wrap your head around $\mathbf{R}\textrm{Hom}$ by noting that one has derived hom-tensor adjunction, that is $$\textrm{Hom}_{D(R)}(K, \mathbf{R}\textrm{Hom}_R(L,M)) \cong \textrm{Hom}_{D(R)}(K \otimes_R^{\mathbf{L}} L, M)$$ for complexes $K,L,$ and $M$. Maybe the derived tensor product feels easier to understand.

However, dualizing complexes for say modules over local rings have specific interpretation via local duality. I'll leave you to find the corresponding global scheme-theoretic versions. If $(R,\mathfrak{m})$ is a local ring and $\omega_R^{\bullet}$ is a dualizing complex, which is only unique up to quasi-isomorphism and shift, so it often assumed to be normalized so that $h^{-\textrm{dim} R}( \omega_R^{\bullet})$ is the first non-zero cohomology module, then $\omega_R = h^{-\textrm{dim} R}\omega_R^{\bullet}$ is a canonical module for $R$. Local duality gives in the complete case a quasi-isomorphism $$\mathbf{R}\textrm{Hom}_R( K, \omega_R^{\bullet}) \cong \text{Hom}_R( \mathbf{R}\Gamma_{\mathfrak{m}}(K),E)$$ where $E$ is the injective hull of the residue field. That is,$R$ is Cohen-Macualay if and only if $\omega_R^{\bullet}$ has only one non-zero module of support at $-\textrm{dim} R$ and when $R$ is Gorenstein, this module is isomorphic to $R$.

As an example of how this is used, let's show a quick proof that when $R$ is a local Noetherian domain, $\textrm{Ann} H_\mathfrak{m}^i(R)$ is not zero for $i < \textrm{dim} R$. Indeed, you just need to find an element $r$ so that $r h^{-i}(\omega_R^{\bullet}) = 0$ by local duality. Each $ h^{-i}(\omega_R^{\bullet})$ is finitely generated, so if we localize at the unique minimal prime, we end up with a dualizing complex over a field. The latter lives in exactly one degree, $-\textrm{dim} R$.

Finally, an important computational note is that when $S$ is a regular ring and $R = S/I$ then $\omega_R^{\bullet} := \mathbf{R}\textrm{Hom}_S(R,S)$ is a dualizing complex for $R$. Also, often dualizing complexes are used to control the study of singularities. Once one has such, they are most often studied by viewing them in various commutative diagrams and taking homology, so one almost never needs to know what the actual supporting modules are but much more interested in the cohomology of the complex.

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    $\begingroup$ I'm a bit confused, lets say I take a ring $R$ and two $R$-module $M$ and $N$. I can pick representing classes $C_{\bullet} := [M]$ and $D_{\bullet} := [N]$ supported in degree zero and compute $\sum_{i+j = n} C_i + D_j$ but this will definitely not compute $M \otimes^{\mathbf{L}} N$ in general, so I'm not sure what you mean by canonical representative. To compute this, we have to replace our choice of canonical representative with a very non-canonical choice, namely some projective resolution. $\endgroup$
    – lemiller
    Jul 2, 2019 at 2:57
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    $\begingroup$ For a similar reason, I don't see an easy way to take say the definition of $\text{Hom}(K,L)$ and get and from the actual terms of the complexes $K$ and $L$ have a way to product the terms of the complex $\mathbf{R}\text{Hom}(K,L)$ for the same reason as before, to compute a derived functor, one must replace an object with a resolution of the correct type, which is inherently not canonical. $\endgroup$
    – lemiller
    Jul 2, 2019 at 2:59
  • 1
    $\begingroup$ @TimGrosskreutz You seem to be confusing homotopy equivalences of complexes with quasi-isomorphisms of complexes. The definition of $\otimes^L$ in the Stacks project (Tag 064M) requires choosing flat resolutions (i.e. particular representatives) and the naively defined $-\otimes-$ is not invariant under quasi-isomorphisms. $\endgroup$ Jul 2, 2019 at 12:46
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    $\begingroup$ In particular, note that $\mathbb{Z}/2\otimes\mathbb{Z}/2\cong \mathbb{Z}/2$, but $\mathbb{Z}/2\otimes^L\mathbb{Z}/2\cong \mathbb{Z}/2\oplus \mathbb{Z}/2[1]$ in $\mathcal{D}(\mathbb{Z})$. $\endgroup$ Jul 2, 2019 at 12:47
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    $\begingroup$ @Denis, Thanks! Tim, yes when I write $[N]$ for a module $N$ I mean the complex supported in degree $0$ with support $N$. Recall that $h^{n}( [N] \otimes^{\mathbf{L}} [M]) \cong Tor_n(N,M)$ and roughly that $Hom_{D(R)}(N,M[n])$ calculates something like $Ext^n(N,M)$. This means you would not hope for the tensor product formula you had before. Another good and detailed place to read on this subject is Daniel Murfet's wonderfully detailed notes here where he explains what he calls the "new Ext" and RHom. $\endgroup$
    – lemiller
    Jul 2, 2019 at 14:25

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