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Let $M$, $N$ be two modules over a ring $R$, suppose $M$ (resp. $N$) is a non-split extension of $M_2$ by $M_1$ (resp. of $N_2$ by $N_1$) as $R$-modules. We make the following assumption:

(1) $\mathrm{Ext}^1_R(N_1,M_1)=0$ and $\mathrm{Ext}^1_R(N_1, M) \cong \mathrm{Ext}^1_R(N_1,M_2)$;

(2) $\mathrm{Ext}^2_R(N_2,M_2)=0$ and $\mathrm{Ext}^2_R(N_2,M)\cong \mathrm{Ext}^2_R(N_2,M_1)$.

Consider the adjunction map $\mathrm{Ext}^1_R(N_1,M) \rightarrow \mathrm{Ext}^2_R(N_2,M)$ (induced by $N_1\rightarrow N\rightarrow N_2$), and I'm wondering if this map is always zero, since, by our assumption, any non-zero element in $\mathrm{Ext}^1_R(N_1,M)$ is from $\mathrm{Ext}^1_R(N_1,M_2)$ while any non-zero element in $\mathrm{Ext}^2_R(N_2,M)$ is from $\mathrm{Ext}^2_R(N_2,M_1)$?

Many thanks!

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    $\begingroup$ Do you assume $R$ is commutative (since you are using the ac.commutative-algebra tag)? $\endgroup$
    – Dag Oskar Madsen
    Commented Mar 12, 2017 at 23:22
  • $\begingroup$ Yes, $R$ can be assumed to be commutative. But I suspect now the conclusion does not hold... $\endgroup$
    – Echeru
    Commented Mar 15, 2017 at 11:58
  • $\begingroup$ It is certainly false without the commutativity assumption. Take $R$ to be the quiver algebra $1 \stackrel{\alpha}\to 2 \stackrel \beta \to 3 \stackrel \gamma \to 4$ modulo $\beta\alpha=0$, $N_1=S_2, N_2=S_1, M_1=S_4, M_2=S_3$. $\endgroup$
    – M T
    Commented Jun 14, 2017 at 12:30

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