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This question is from the proof of Theorem 11.34 in the book: Twenty-four Hours of Local Cohomology.

Let $R$ and $S$ be CM local ring and $R\to S$ a local homomorphism such that $S$ is a finite generated $R$-module, $k$ and $l$ are residue fields of $R$ and $S$ respectively. Let $t = dim R - dim S$, and $\omega$ be a canonical module of $R$. Let $I$ be a minimal injective resolution of $\omega$.

Now we have natural isomorphism of l-vector space:

$$Hom_{l}(l,Hom_{R}(S,I) \cong Hom_{R}(l,I) \cong Hom_{k}(l,k) \otimes_{k}Hom_{k}(k,I)$$

Passing to cohomology yields isomorphisms

$$Ext^i_{S}(l,Ext^{t}_{R}(S,\omega)) \cong Hom_{k}(l,k) \otimes_{k}Ext^{i+t}_{k}(k,\omega)$$

Question: Can we change the LHS to $Ext^a_{S}(l,Ext^{b}_{R}(S,\omega))$, with $a+b=i+t$? If not, how to decide the indices $i$ and $t$ of $Ext$ functor in the LHS?

The same question is at here

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  • $\begingroup$ It seems that $t$ is constant but $i$ varies $\endgroup$
    – user 1
    Commented Apr 4, 2015 at 14:10
  • $\begingroup$ As you write, $t$ is given by dim(R)-dim(S). $\endgroup$
    – user43326
    Commented Apr 4, 2015 at 15:04

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I don't know if one can prove the isomorphisms for other $a$ and $b$; but about what we have in this proof, you can see that $t$ is an specified number, namely $\dim R - \dim S$ and the proof is about this specific number because as you can see in the book they have wrote "Thus, after shifting $t$ steps to the left, $Hom_R(S, I^•)$ is a finite injective resolution of the S-module $Ext^t_R(S, ω)$." so we can not change $t$ (in this proof).
On the other side $i$ varies: this is clear both from the expression of the book that says "Setting i = dim S implies...", and from definition of homology module (that we write an injective resolution and apply $Hom$ functor on it and find $i$-th homology module for all $i$).

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  • $\begingroup$ Yes, I know here t is special, but it seems that the step from the Hom to Ext maybe is a little jump. $\endgroup$
    – Strongart
    Commented Apr 5, 2015 at 5:43
  • $\begingroup$ set $T=Ext^t_R(S, ω)$. we want to have $Ext^i_{S}(l,T)$. for this we should write "an injective resolution of the S-module, $T$", apply $Hom(l,-)$ and calculate homology module at the i-th point. Now as you see in the answer (which is from book) "$Hom_R(S, I^•)$ is a finite injective resolution of the S-module T" $\endgroup$
    – user 1
    Commented Apr 5, 2015 at 6:15
  • $\begingroup$ If we set T=$Ext^{b}_{R}(S,\omega))$, 0≤b≤t+i, then do the same discussing, can we get the isomorphism for a and b, a+b=i+t? $\endgroup$
    – Strongart
    Commented Apr 6, 2015 at 6:48
  • $\begingroup$ no we (I, at least) cant get the isomorphism for a and b, a+b=i+t. see the 1st paragraph of proof. $\endgroup$
    – user 1
    Commented Apr 6, 2015 at 7:39
  • $\begingroup$ I see, if b≠t, the Ext functor with the index b is 0 by prop11.33, thanks. $\endgroup$
    – Strongart
    Commented Apr 7, 2015 at 13:37

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