# Linearity of covariant and contravariant $Ext^1$ functors defined via short exact sequences

Let $$R$$ be a Commutative ring. Let $$M,X,Y$$ be $$R$$-modules. Let $$f: X \to Y$$ be an $$R$$-linear map.

Then, given an exact sequence $$\eta: 0\to X \to Z_{\eta} \to M \to 0$$ in $$Ext^1(M,X)$$, the pushout of $$\eta$$ by $$f$$ gives an exact sequence $$f\eta: 0\to X \to Z'_{\eta} \to M \to 0$$ in $$Ext^1(M,Y)$$.

Similarly, given an exact sequence $$\eta: 0\to M \to W_{\zeta} \to Y \to 0$$ in $$Ext^1(Y,M)$$, the pullback of $$\zeta$$ by $$f$$ gives an exact sequence $$\zeta f: 0\to M \to W'_{\zeta} \to X \to 0$$ in $$Ext^1(X,M)$$.

It is well known that these assignments actually define functors $$Ext^1(M,-); Ext^1(-,M): \text{Mod } R \to \text{Mod } R$$, where for $$f \in Hom(X,Y)$$;

$$Ext^1(M,f): Ext^1(M,X) \xrightarrow{\eta \mapsto f\eta} Ext^1(M,Y)$$

$$Ext^1(f,M): Ext^1(Y,M) \xrightarrow{\zeta \mapsto \zeta f} Ext^1(X,M).$$

My question is: Are the functors $$Ext^1(M,-)$$ and $$Ext^1(-,M)$$ $$R$$-linear?

I think this should be well known, but I am looking for an explicit reference.

• @Maxime Ramzi: Indeed, I think you are very much right that any Noetherian hypothesis is irrelevant. The Commutativity is needed otherwise $Hom$ is not a module in general. Even commutativity could be disregarded I think if I only asked about "additivity" instead of "linearity" ... – sde Apr 26 at 8:45
in particular Chapter III, Theorem 2.1. The book seems to be treating rings $$R$$ which need not be commutative, hence there is no mention of the $$R$$-module structure since it need not exist!