Prove that $\operatorname{Ext}^{d+1}(A, B)\cong \operatorname{Ext}^1(M_d,B)$

Question

So, given a resolution, with $P_{i}$ projective modules: $$0\longrightarrow M_d\longrightarrow P_{d-1} \longrightarrow \cdots \longrightarrow P_0 \longrightarrow A\longrightarrow 0,$$ I'm trying to prove that $\operatorname{EXT}^{d+1}(A, B)\cong \operatorname{EXT}^1(M_d,B)$(Weibel, pg 93). The thing is that, if I've get the functor correctly, we first of all apply $\operatorname{Hom}(-,B)$ and so we have: $$0\longrightarrow \operatorname{Hom}(P_0,B)\longrightarrow \operatorname{Hom}(P_1,B)\longrightarrow\cdots\longrightarrow \operatorname{Hom}(M_d,B)\longrightarrow 0$$

Now we have to make the homology asociated with $P_i$ to obtain $\operatorname{EXT}^i(A,B)$, but I'm stuck here. What do I do to see that isomorphism? Thanks.

Answer 1

Hint: Proceed by induction on the length of the resolution to prove that, for any $i\ge1$, one has $$\DeclareMathOperator{\Ext}{Ext}\Ext^{d+i}(A,B)\simeq\Ext^i(M_d,B).$$

For the inductive step, split a resolution of length $d+1$: $$0\to M_{d+1}\to P_d\to P_{d-1}\to\dots\to P_0\to A\to 0 \tag{1}$$ into the short exact sequence $\;0\to M_{d-1}\to P_d\to M_{d-1}\to 0$, and the shorter resolution: $$0\to M_d\to P_{d-1}\to\dots\to P_0\to A\to 0,$$ where $M_{d-1}=\operatorname{Im}(P_d\to P_{d-1})$. Then use the inductive hypothesis and the long homology sequence deduced from $(1)$ to show that, for all $i\ge 1$, $$\Ext^{d+1+i}(A,B)\simeq\Ext^i(M_d,B).$$

Answer 2

Hint: Choose a projective resolution$\DeclareMathOperator{\Ext}{Ext}$ $$\cdots\to Q_1\to Q_0 \to M_d\to 0$$ of $M_d$.

Then the sequence $$\cdots\to Q_1\to Q_0\to P_{d-1}\to P_{d-2}\to \cdots \to P_0\to A\to 0$$ is exact (prove it), hence a projective resolution of $A$. Compute the $(d+1)$-th $\Ext$-group of $A$ with that resolution and then compute the first $\Ext$-group of $M_d$ with the above resolution.