$\DeclareMathOperator\Ext{Ext}$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) $.


1 Answer 1


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 $\mathfrak m \mathrm{Ext}(X,G)$.

  • $\begingroup$ 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 $\mathfrak m \text{Ext}^1_R(X,G)?$ $\endgroup$
    – uno
    Commented Sep 22, 2022 at 18:32
  • 1
    $\begingroup$ @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 $X = R / (x,y)$. $\endgroup$ Commented Sep 22, 2022 at 18:51

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