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This is a problem that I thought at first was obvious but that became less clear the more I thought about it. Assume we have a finitely generated algebra $A$ over a field $k$, and a short exact sequence of projective $A$-modules $l_1:\quad0\rightarrow P_1 \rightarrow P_2 \rightarrow P_3 \rightarrow 0$. Let's assume we can lift this short exact sequence to their generators, i.e., let $F_i$ be a free module that surjects onto $P_i$ and assume there is a short exact sequence $l_2:\quad0\rightarrow F_1 \rightarrow F_2 \rightarrow F_3 \rightarrow 0$ that surjects onto $l_1$. Furthermore assume that all morphisms in $l_2$ are given by matrices in $M_{n\times m}(k)$ for suitable values of $m$ and $n$. This implies that we can choose splitting morphisms for $l_2$ that are also in $M_{n\times m}(k)$. My question: is there a such a choice of a splitting for $l_2$ that induces a splitting on $l_1$? This is equivalent to asking whether such a splitting will map the kernels of $F_i\rightarrow P_i$ to each other. (If the answer depends on $A$, for what $A$'s is it “yes”?)

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  • $\begingroup$ Are the following fixed: $l_1$ and the choice of bases for the $F_i$ with respect to which the matrices have entries in $k$? Or can we choose them to make the choice of splitting easier? $\endgroup$ – Jeremy Rickard Jan 4 at 9:13
  • $\begingroup$ Yes the objects and morphisms in the exact sequences are fixed. So the basis that the morphisms between $F_i$'s are in the specific form is also fixed. The surjection from $l_2$ to $l_1$ is also fixed. $\endgroup$ – user127776 Jan 4 at 9:18
  • $\begingroup$ If the following can be shown, that is also good: Given the conditions of the problem maybe find another $l_2$ with similar properties that admits the desired splitting. $\endgroup$ – user127776 Jan 4 at 9:31
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The answer is "no" unless $A=k$.

Let $a\in A\setminus k$, and let $l_2$ and $l_1$ be the first and second rows of the commutative diagram $$\require{AMScd} \begin{CD} 0@>>>A@>\begin{pmatrix}1\\0\end{pmatrix}>>A^2 @>\begin{pmatrix}0&1\end{pmatrix}>>A@>>>0\\ @.@VV1V@VV\begin{pmatrix}1&a\end{pmatrix}V@VVV\\ 0@>>>A@>1>>A@>>>0@>>>0\\ @.@VVV@VVV@VVV\\ @.0@.0@.0@. \end{CD} $$

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Choose a splitting of $l_1$ then lift it to a splitting of $l_2$. Then of course this splitting of $l_2$ will induce a splitting of $l_1$.

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    $\begingroup$ Why is this lifting given by matrices over $k$, as required in the question? $\endgroup$ – LSpice Jan 4 at 4:36
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    $\begingroup$ Ups, I missed this condition (thought it is about matrices over $A$). $\endgroup$ – Sasha Jan 4 at 14:11

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