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](https://stacks.math.columbia.edu/tag/0A7A) 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.



Definition of dualizing complex
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. 


