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sdey
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Let $R$ be a Commutative Noetherian ring. Let $M,X,Y$ be finitely generated $R$-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): \mod R \to \mod R$$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.

Please help.

Update: I have accepted Mark Grant's answer which gives a reference for additivity of the functor. If someone has an explicit reference for the fact that multiplication by ring elements is preserved too, please do post an answer; I'd really appreciate it and upvote your answer too!

Let $R$ be a Commutative Noetherian ring. Let $M,X,Y$ be finitely generated $R$-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 $Ext^1(M,-); Ext^1(-,M): \mod R \to \mod R$, 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.

Please help.

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.

Please help.

Update: I have accepted Mark Grant's answer which gives a reference for additivity of the functor. If someone has an explicit reference for the fact that multiplication by ring elements is preserved too, please do post an answer; I'd really appreciate it and upvote your answer too!

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

Let $R$ be a Commutative Noetherian ring. Let $M,X,Y$ be finitely generated $R$-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 $Ext^1(M,-); Ext^1(-,M): \mod R \to \mod R$, 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.

Please help.