Proposition 3.3.3 in Bruns and Herzog The proposition says that
If $(R,m,k)$ is a Cohen-Macaulay (CM) local ring of dimension $d$ and $C$ is a maximal Cohen-Macaulay (MCM) $R$-Module, then
a) Suppose $M$ is a MCM $R$-module with $Ext^j_R(M,C)=0$ for all $j>0$. Then $Hom_R(M,C)$ is MCM $R$-module, and for any $R$-sequence $X$ we have $Hom_R(M,C)\otimes R/XR \cong Hom_{R/XR}(M/XM,C/XC)$.
b) Assume in addition that $C$ has finite injective dimension and $M$ is a CM $R$-module of dimension $t$. Then
i) $Ext^j_R(M,C)=0$ for all $j\neq d-t$.
ii) $Ext^{d-t}_R(M,C)$ is a CM module of dimension $t$.
I could prove the isomorphism in the statement a) but was unable to understand how to verify that the $Hom$ set is MCM.
For part b) i) the book refers to Theorem 1.2.10 (e) which says that if $I$ is an ideal of $R$ and $M, N$ are finite $R$-modules with $Supp(N)=V(I)$ then $grade(I,M)=inf\{i: Ext^i(N,M)\neq 0\}$. The author claims that statement b) i) follows from the above statement for $j<d-t$, which I am not able to figure out.
I am not equipped with much familiarity with these things so a detailed answer will be helpful to me. I am also looking for a lucid reference from where I can learn the proof of the statements.
I really appreciate any help you can provide.
 A: a) Use induction on dimension of $R$.
If dim $R$ = 0, then dim $Hom(M,C)$ as $R$-module 0 as $ {\textrm{dim} Hom(M,C)} = {\textrm{dim} R/ ann(Hom(M,C)} \leq {\textrm{dim}} R = 0$. Hence $Hom(M,C)$ is a MCM $R$-module.
If dim $R$ = 1. Let $x_1 \in {\mathfrak{m}}$ be an $R$-regular element. Since $C$ is MCM, $x_1$ is also a $C$-regular element. Hence, we have the following exact sequence
$$\require{AMScd}
\begin{CD}
0 @>{}>>  C @>{x_1}>> C @>{}>> C/x_1C @>{}>> 0.
\end{CD}$$
This will give the following long exact sequence
$$\require{AMScd}
\begin{CD}
0 @>{}>>  Hom_R(M,C) @>{x_1}>> Hom_R(M,C) @>{}>> Hom_R(M,C/x_1C) @>{}>> 0. 
\end{CD}$$
as $Ext^1(M,C) = 0$.
Hence $x_1$ is a Hom(M,C) regular element. Hence $ 1 \leq depth(Hom(M,C)) \leq dim (Hom(M,C)) \leq dim R = 1$. Hence Hom(M,C) is MCM.
Use the fact that if $x_1, x_2, \ldots, x_n$ is an $R$-sequence, then it is a $C$ sequence and $M$ sequence as $C, M$ are maximal Cohen-Macaulay Modules.
Observe that we have an exact sequence $$\require{AMScd}
\begin{CD}
0 @>{}>>  M @>{x_1}>> M @>{}>> M/x_1M @>{}>> 0,
\end{CD}$$
which gives a long exact sequence
$$\require{AMScd}
\begin{CD}
\cdots  @>{}>>  Ext_R^i(M,C) @>{}>> Ext_R^i(M/x_1M, C) @>{}>> Ext_R^{i+1}(M,C) @>{}>> \cdots
\end{CD}$$
Since $Ext_R^i(M,C) = 0$, we get $Ext_R^i(M/x_1M , C) = 0$ for all $i>0$. Hence $$Ext_{R/x_1R}^i(M/x_1M, C/x_1C) \cong Ext_R^{i+1}(M/x_1M,C) = 0$$ for all $i>0$.
Moreover, for ${\bf{x}} = x_1, x_2, \ldots, x_n$ an $R$-sequence, we get $Ext_{R/{\bf x}}^i(M/{\bf x}M , C/ {\bf x}C) = 0$ for all $i>0$.
Now inductively deduce that $x_1, x_2, \ldots, x_n$ is $Hom(M,N)$ sequence, which proves $Hom(M,C)$ is MCM.
b) Use $1.2.10(e)$ inductively on dimension of $M$ to obtain the result.
If $dim(M) = 0$, then from $1.2.10(e)$, result is clear as $Supp(M) = \{\mathfrak{m}\}$.
If $dim(M) =1$, then there exists $x \in \mathfrak{m}$ which is $M$ regular. Then $M/xM$ is a dimension zero $R$-module. Using dimension zero case, we get $Ext^i(M/xM, C) = 0$ for all $i <d$. Now consider the exact sequence
$$\require{AMScd}
\begin{CD}
0 @>{}>>  M @>{x_1}>> M @>{}>> M/x_1M @>{}>> 0,
\end{CD}$$  and use the long exact sequence in $Ext$ gives
$$Ext^j(M,C) \cong x Ext^j(M,C)$$ for all $j < d-1$. Use Nakayama lemma to get the result when $dim(M) =1$.
Now similarly apply induction to obtain the result.
