Given a notherian ring $A$ and a $A$-module $M$, let $I$ be an ideal of $A$ with $IM\neq M$. Then in Matsumura's, there is a statement that:
If there exists an $M$-regular sequence $a_1,a_2,...,a_n$ of length $n$ in $I$, then $\mathrm{Ext}^i_A (N,M)=0(i<n)$ for every finite $A$-module $N$ with $\mathrm{Supp}(N)\subseteq V(I)$.
In his argument, he first proved that if $Ext^0_A(N,M)=0$, then $M$ must exists a $M$-regular element. Then, he putted $M_1=M/a_1M$, and said that $Ext^i_A(N,M_1)=0$ for $i<n-1$ by induction on $n$.
I am not quite sure how this step follows? And here is my attempt.
Let's denote $M_n=M_{n-1}/a_nM$ and $M_1=M/a_1$ sa definiton. Then we have short exact sequences $0\rightarrow M_i\to M_i\to M_{i+1}\to0$, where the first map is multiply by $a_n$. With these, we can form a long exact sequence
$0\to M_1\to M_1\to M_2\to M_3 \to...\to M_{n-1} \to M_n \to 0$
Applying $Hom(N,.)$, and notice the map we defined above, it seems we could conclude that $Ext^i_A(N,M_1)=0$ for $i<n-1$. And since $M_n$ has no $M$-regular elements, all elements are zero-divisors, hence $Hom(N,M_n)=0$. Thus $Ext^{n-1}_A(N,M)=Hom(N,M_{n-1})$ is nonzero.
Hope someone could help. Thanks!