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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 thing 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!

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!

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 thing 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!

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 thing 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!