Finitely presented modules lead to injective map between higher Ext groups

Question

We know that an $A$-module $M$ is finitely presented if and only if for every filtered system $(N_{\lambda}, \phi_{\lambda \mu})$, the natural morphism $\lim \mathrm{Hom}(M,N_{\lambda})\rightarrow \mathrm{Hom}(M,\lim N_{\lambda})$ is bijective. My question is, how to know the natural morphism $\lim \mathrm{Ext}^1(M,N_{\lambda})\rightarrow \mathrm{Ext}^1(M,\lim N_{\lambda})$ is injective when $M$ is finitely presented? The hint is the filtered system $(N_{\lambda}, \phi_{\lambda \mu})$ can be injected into a filtered system consisting of injective $A$-modules. But I do not know how to get it in this way.

Answer 1

I think what you say $lim$ is filtered direct limit.

In fact, if $X$ is finite generated $A$-module, then $lim Hom(X,N_\lambda)\rightarrow Hom(X,lim N_\lambda)$ is injective if $(N_\lambda,\phi_{\lambda \mu})$ is filtered system.

Then for your question, take s.e.s $0\rightarrow K\rightarrow A^n\rightarrow M\rightarrow 0$. Hence we know $K$ is finite generated since $M$ is finite presented. Then we have long exact sequence: $0\rightarrow limHom(M,N_\lambda)\rightarrow lim (A^n,N_\lambda)\rightarrow lim(K,N_\lambda)\rightarrow limExt^1(M,N_\lambda)\rightarrow 0$;

on the other hand, we also have long exact sequence: $0\rightarrow Hom(M,limN_\lambda)\rightarrow (A^n,limN_\lambda)\rightarrow (K,limN_\lambda)\rightarrow Ext^1(M,limN_\lambda)\rightarrow 0$.

Then you have a commutative diagram about these two exact sequences, with first two columns are isomorphic and the third column is injective. By five lemma, there is an injection: $limExt^1(M,N_\lambda)\rightarrow Ext^1(M,limN_\lambda)$.