Let $(R,m)$ be a local ring and $M$ be a finitely generated $R$-module. Let $N$ be an $R$-module with $pd_R(N)=t$ (projective dimension) and the minimal free resolution
$$ 0\to F_t\xrightarrow{\partial_t} F_{t-1}\to...\to F_0 \to N \to 0$$
Then we have a presentation of $Ext_R^t(N,M)$
$$(\nabla) Hom_R(F_{t-1},M)\xrightarrow{\partial^*_t} Hom_R(F_t,M)\to Ext_R^t(N,M)\to 0$$
Let $\mathbb{A}$ denote the presentation matrix of the homomorphism $F_t\xrightarrow{\partial_t} F_{t-1}$ w.r.t. some free bases of $F_t$ and $F_{t-1}$. Then because every entry of $\mathbb{A}$ is contained in $m$ (because the free resolution is minimal), the presentation $\nabla $ is minimal, that is $R/m \otimes_R \partial^*_t=0$.
so that $Ext_R^t(N,M)\neq 0$ because $Hom_R(F_t,M)\cong M^{dim_R F_t}\neq 0$?????
My question is the final conclusion, how exactly $Ext_R^t(N,M)\neq 0$? I believe the answer would appear when we assume $Ext_R^t(N,M)= 0$, but still not see it yet.
Thank you for your help.
Let me denote $F_k=R^{\oplus n_k}$. Then, you have $M^{\oplus n_{t-1}}\stackrel {\partial_t^*}{\to} M^{\oplus n_t}\to Ext^t(N,M)\to 0$. $\partial_t^*$ has all its entries in the maximal ideal (it is the transpose of $\partial_t$). So, the image of $\partial_t^*$ is contained in $m M^{\oplus n_t}$ and thus $\partial_t^*$ can not be surjective. So, $Ext^t(N,M)\neq 0$.