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 


$$Ext^i\_{S}(l,Ext^{t}\_{R}(S,\omega)) \cong Hom\_{k}(l,k) \otimes\_{k}Ext^{i+t}\_{k}(k,\omega)$$



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> 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?
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The same question is at [here](https://math.stackexchange.com/questions/1215843/decide-two-indices-of-ext-functor) 



Decide two indices of Ext functor
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 $i$).


