Minimal injective resolution Let $R$ be a local commutative Noetherian ring with residue field $k$. Let $M$ be a finitely generated $R$-module and $I$ a minimal injective resolution of $M$. Is it true that $\mathrm{H}(\mathrm{Hom}_R(k,I))=\mathrm{Hom}_R(k,I)$? (Where $\mathrm{H}$ denotes (co)homology)
 A: To answer your question let me state a definition and lemma. 

Definition: For $p$ a prime of $R$, 
  $$\mu_i(p,M) = \dim_{k(p)} \operatorname{Ext}^i_{R_p} (k(p),M_p)$$
  is called the i-th Bass number of $M$ with respect to $p$.



Lemma: 
  (i) Let $I^\bullet$ be a $\textbf{minimal}$ injective resolution of $M$. Then 
  $$
I^j \cong \underset{p \text{ prime}}\oplus E_R(R/p)^{\mu_j(p,M)},
$$
  where $E_R(-)$ denotes the injective envelope. 
(ii) $\operatorname{Hom}_R(k, E_R(R/p)) \cong \begin{cases} 0 & if~p\neq m \\  k & if~p=m. \end{cases}$

By the lemma, one sees that 
$$\operatorname{Hom}_R(k, I^j) \cong \oplus k^{\mu_j(m,M)}.$$
Recall that $\mu_j(m,M) = \dim_k \operatorname{Ext}^j_R (k,M)$ and $\operatorname{Ext}^j_R (k,M) = H^i(\operatorname{Hom_R} (k, I^\bullet) )$. In other words, 
$$
\operatorname{Hom}_R(k, I^j) \cong H^i(\operatorname{Hom}_R(k, I^\bullet))
$$
as $k$-vector spaces. Thus, unless the differential maps, which are maps between $k$-vector spaces, are zero, the vector space dimensions will drop which is impossible by the isomorphism above. Hence, you obtain the equality in your question.
