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Let $R$ be a commutative ring. Suppose $R$-modules $X,A,B,C$ and $Y$ are given such that the outer two rows and the outer two columns in the following diagram are exact.

$\hskip1in$ Diagram

Does it follow that there exists an $R$-module $D$ such that the augmented diagram commutes and the dotted row and column are exact? Or are there counterexamples?

Such a $D$ would provide a simultaneous extension of $C$ by $X$ and of $Y$ by $A$.

This is easily seen to be true if either the left column splits (in which case we may take $D=C\oplus X$) or if the bottom row splits (in which case we may take $D=A\oplus Y$). So the answer is positive if $R$ is a field, for instance. I'm not sure how to prove the general case, neither have I been able to produce a counterexample.

Any idea or reference would be welcome.

Edit: In light of Jason Starr's comment providing a counterexample over $R=\mathbb Z[t]$, I would also be interested in sufficient conditions on $R$ such that the claim is true. For instance:

Does the claim hold if $R$ is a PID? Hereditary?

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    $\begingroup$ That is not true. Let $C$ be a projective $R$-module, and let $B\subset C$ be an $R$-submodule that is not projective, e.g., $C=R=\mathbb{Z}[t]$ and $B=\langle 2,t \rangle$. Let $A\to B$ a surjection from a projective $R$-module. If such a diagram exists, then the middle column would split ($C$ is projective), implying the first column splits, implying that $B$ is projective. $\endgroup$
    – Jason Starr
    Commented Jun 29, 2016 at 17:36
  • $\begingroup$ @JasonStarr: Thanks, this is certainly useful information. Do you perhaps know of any sufficient conditions on $R$ under which the claim holds? (Aside from $R$ being a field?) $\endgroup$
    – Dejan Govc
    Commented Jun 29, 2016 at 18:03

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As pointed out by Jason Starr in his comment, this is not true in general.

However, it is true if $R$ is hereditary. To see this, notice that the exact sequence $$0\to X\to A\to B\to0$$ gives rise to the following long exact sequence of $\operatorname{Ext}$-modules: $$0\to\operatorname{Hom}(Y,X)\to\operatorname{Hom}(Y,A)\to\operatorname{Hom}(Y,B)\to\phantom{0}\\\phantom{0}\to\rlap{\;\operatorname{Ext}(Y,X)}\phantom{\operatorname{Hom}(Y,X)}\to\rlap{\;\operatorname{Ext}(Y,A)}\phantom{\operatorname{Hom}(Y,A)}\to\rlap{\;\operatorname{Ext}(Y,B)}\phantom{\operatorname{Hom}(Y,B)}\to0$$ Note that all higher $\operatorname{Ext}$-modules vanish since $R$ is hereditary.

In particular, $\operatorname{Ext}(Y,A)\to\operatorname{Ext}(Y,B)$ is an epimorphism. By the classical interpretation of elements of $\operatorname{Ext}$-modules as extensions, this means that there is a map of extensions as follows:

$\hskip1in$ enter image description here

By construction, the two outermost vertical arrows agree with the ones given in the question. Applying the Snake Lemma now concludes the proof.

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