$\operatorname{Hom}_R(R/fR,X)$ is an exact functor

Question

In the article "Injective dimension in Noetherian rings", Bass said (without proof) that: Let $R$ be a Noetherian ring (I do not think this assumption is necessary here), $f$ is $R$-regular, $X$ is an exact sequence such that $fX=X$. Then $\operatorname{Hom}_R(R/fR,X)$ is an exact sequence.

I set $ B \xrightarrow{g} C \to 0$ as an exact sequence such that $fB=B$ and $fC=C$, then I try to prove that $\operatorname{Hom}_R(R/fR,-)$ preserves the exactness. Let $h \in \operatorname{Hom}(R/fR,C)$ then $h(1)=g(b)$ for some $b\in B$. Then I set $z \colon R/fR\to B$ such that $z(1)=b$.

I think it suffices to show that $fb=0$, but this is where I'm stuck.

Am I on the right track? How do i continue?

Answer 1

If $f$ is regular then for any module $\operatorname{Ext}^1(R/Rf,M) = M/fM$. This means that the sequence is exact by the LES of Ext as soon as the first term in the SEC is such that $fX'=X'$.

To expand: consider the exact sequence (using that $f$ is regular).

$$\tag{1} 0\to R\stackrel{f}\to R\to R/Rf\to 0$$

which gives a resolution of $R/Rf$, and an arbitrary exact sequence

$$\tag{2} 0\to X'\to X\to X''\to 0$$

The LES for this and $\hom_R(R/Rf,?)$ has the following form

$$0\to \hom_R(R/Rf,X') \to \hom_R(R/Rf,X)\to \hom_R(R/Rf,X'')\to \operatorname{Ext}^1_R(R/Rf,X')\to \cdots$$

Now from the first resolution you get that for any module $M$, $\operatorname{Ext}_R(R/Rf,M)$ is the homology of

$$0\to M\stackrel{f} \to M\to 0$$

i.e $\hom_R(R/Rf,M) = \{m\in M : fm=0\}$ and $\operatorname{Ext}^1_R(R/Rf,M) = M/fM$ and all other Ext are zero. Thus you obtain the following claim:

Let $f$ be regular and consider an exact sequence $S$ as in $(2)$. Then $\hom_R(R/Rf,S)$ is exact if and only if $fX'=X'$.

Answer 2

$\require{AMScd}$ Another option is the following, which uses the snake lemma but does not mention extensions. As observed in the comments, $\hom_R(R/Rf,M)$ is just the kernel of multiplication by $f$ on $M$, $M_f$. Thus consider an exact sequence $(2)$ as in the other answer, and form a diagram

\begin{CD} {}&& 0 &&0 &&0\\ &&@VVV @VVV @VVV \\ 0 @>>>X'_f @>>> X_f @>>> X''_f\\ {}&&@VVV @VVV @VVV \\ {}&&X' @>>> X @>>> X'' \\ {}&&@VfVV @VfVV @VfVV \\ {}&&X' @>>> X @>>> X'' \\ {}&&@VVV &&&&\\ {}&&0 \end{CD}

Since $\hom_R$ is left exact, the first row is exact, and you just want to show that the map $a$ is onto. Now if $fX'=X'$ then the $0$ at the bottom is justified. The snake lemma gives en exact sequence from the first exact row to the $0$ (the cokernel) and shows what you what. Again, it is necessary and sufficient that $fX' =X'$.