Why does $\text{Ext}^i_A(N_j/N_{j+1},M)=0$ imply $\text{Ext}^i_A(N,M)=0?$

Question

This comes theorem 17.1 of commutative ring theory by Matsumura:

It is easy to see that if $\text{Ext}^i_A(N_j/N_{j+1},M)=0$ for each $j$ then $\text{Ext}^i_A(N,M)=0$...

I am not very familiar with identities of the Ext functor, why is this easy to see?

Answer 1

From the short exact sequence $$0\to N_1/N_2\to N_0/N_2\to N_0/N_1\to0$$ we get a long exact sequence $$ \cdots\to\text{Ext}_A^i(N_0/N_1,M) \to\text{Ext}_A^i(N_0/N_2,M) \to\text{Ext}_A^i(N_1/N_2,M)\to\cdots$$ and as the outer groups here are zero, so is the inner term. One now proves $\text{Ext}_A^i(N_0/N_3,M)=0$ etc.

Answer 2

Let we have a sequence $0=N_n\subset N_{n-1}\subset \cdots \subset N_1\subset N_0 =N $. Thus we have the following exact sequences: $$0\to N_1\to N_0\to N_0/N_1\to 0$$ and $$0\to N_2\to N_1\to N_1/N_2\to 0$$ and $$\vdots$$ $$0\to N_{n-1}\to N_{n-2}\to N_{n-2}/N_{n-1}\to 0$$ and $$0\to N_n=0\to N_{n-1}\to N_{n-1}/N_{n}\to 0$$

Hence, we have long exact sequences: $$\cdots\to 0=\text{Ext}^i_A(N_0/N_1, M)\to \text{Ext}^i_A(N_0, M)\to \text{Ext}^i_A(N_1, M)\to \text{Ext}^{n+1}_A(N_0/N_1, M)=0\to\cdots$$ and $$\cdots\to 0=\text{Ext}^i_A(N_1/N_2, M)\to \text{Ext}^i_A(N_1, M)\to \text{Ext}^i_A(N_2, M)\to \text{Ext}^{n+1}_A(N_1/N_2, M)=0\to\cdots$$ and $$\vdots$$ $$\cdots\to 0=\text{Ext}^i_A(N_{n-1}/N_{n-2}, M)\to \text{Ext}^i_A(N_{n-1}, M)\to \text{Ext}^i_A(N_{n-2}, M)\to \text{Ext}^{n+1}_A(N_{n-1}/N_{n-2}, M)=0\to\cdots$$ and $$\cdots\to 0=\text{Ext}^i_A(N_{n}/N_{n-1}, M)\to \text{Ext}^i_A(N_{n }=0, M)\to \text{Ext}^i_A(N_{n-1}, M)\to \text{Ext}^{n+1}_A(N_{n }/N_{n-1}, M)=0\to\cdots.$$ Thus, we have:

$$0=\text{Ext}^i_A(N_{n }=0, M)\cong \text{Ext}^i_A(N_{n-1}, M)\cong\text{Ext}^i_A(N_{n-2}, M)\cong\cdots\cong\text{Ext}^i_A(N_{1}, M)\cong\text{Ext}^i_A(N_{0}, M).$$ So $\text{Ext}^i_A(N, M)=0$.