On image of map $\text{Ext}^1_R(X,F)\to \text{Ext}^1_R(X,G)$ induced by $R$-linear map of free modules $F\to G$ with entries in the maximal ideal

$$\DeclareMathOperator\Ext{Ext}$$Let $$(R,\mathfrak m)$$ be a Noetherian local ring. Let $$F,G$$ be finitely generated free $$R$$-modules and $$f:F\to G$$ be an $$R$$-linear map such that $$f(F)\subseteq \mathfrak m G$$. Let $$X$$ be a finitely generated $$R$$-module, and let $$\alpha : 0\to F \to A_{\alpha} \to X \to 0$$ be a short exact sequence i.e. $$[\alpha]\in \Ext^1_R(X,F)$$. We have a following push-out diagram with $$[\beta] \in \Ext^1_R(X,G)$$.

$$\require{AMScd}\begin{CD} \sigma : 0 @>>> F @>>> A_\alpha @>>> X @>>> 0 \\ @. @VVfV @VVV @| \\ \beta : 0 @>>> G @>>> A_\beta @>>> X @>>> 0. \end{CD}$$ My question is: must it be necessarily true that $$[\beta] \in \mathfrak m \Ext^1_R(X,G)$$?

Some thoughts: The answer is affirmative if $$F\cong G\cong R$$. Indeed, in this case, $$f:R\to R$$ must be given by multiplication by some $$x\in R$$. Since $$f(F)\subseteq \mathfrak m G$$, so $$x\in \mathfrak m$$. Then in $$[\beta]=x[\alpha]\in \Ext^1_R(X,R)$$, hence $$[\beta]\in x\Ext^1_R(X,R)\subseteq \mathfrak m \Ext^1_R(X,G)$$.

This is true for any $$\mathrm{Ext}$$-degree (and, in fact, without many hypotheses except that $$\mathfrak m$$ is finitely generated and $$F$$ is free).
Let $$(x_1,\dots,x_n)$$ be generators of the maximal ideal $$\mathfrak m$$. Then there is a surjection $$G^n \to \mathfrak m G$$ given by $$(g_1,\dots,g_n) \mapsto \sum x_i g_i.$$ Because $$F$$ is free, the map $$F \to \mathfrak m G$$ lifts to a map $$F \to G^n$$.
Now $$\mathrm{Ext}(X,-)$$ respects sums and takes the map multiplication-by-$$x$$ map $$M \to M$$ to the multiplication-by-$$x$$ map on $$\mathrm{Ext}(X,M)$$ (I'm assuming $$R$$ is commutative here, because otherwise $$\mathrm{Ext}$$ doesn't necessarily take values in $$R$$-modules). Therefore, the composite $$\mathrm{Ext}(X,G)^n \cong \mathrm{Ext}(X,G^n) \to \mathrm{Ext}(X,\mathfrak m G) \to \mathrm{Ext}(X,G)$$ is given by $$(w_1,\dots,w_n) \mapsto \sum x_i w_i$$, taking values in $$\mathfrak m \mathrm{Ext}(X,G)$$.
As a result, because the map $$\mathrm{Ext}(X,F) \to \mathrm{Ext}(X,G)$$ factors through this map $$\mathrm{Ext}(X,G^n)$$, it also factors through $$\mathfrak m \mathrm{Ext}(X,G)$$.
• Thank you for your answer. I think I understand it, but I do have a few questions ... in general, if $G$ is a finitely generated free $R$-module, would the map $\text{Ext}^1_R(X,\mathfrak m G) \to \text{Ext}^1_R(X,G)$ induced by inclusion $\mathfrak m G \to G$ still have image in $\mathfrak m \text{Ext}^1_R(X,G)?$
• @uno No, I don't think that's true. I believe an example is if $R = k[x,y] / (x^2, xy, y^2)$, $G = R$, and $X = R / (x,y)$. Commented Sep 22, 2022 at 18:51