# When splitting of short exact sequence preserves the kernels

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”?)

• 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? – Jeremy Rickard Jan 4 at 9:13
• 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. – user127776 Jan 4 at 9:18
• 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. – user127776 Jan 4 at 9:31

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 $$\require{AMScd} \begin{CD} 0@>>>A@>\begin{pmatrix}1\\0\end{pmatrix}>>A^2 @>\begin{pmatrix}0&1\end{pmatrix}>>A@>>>0\\ @.@VV1V@VV\begin{pmatrix}1&a\end{pmatrix}V@VVV\\ 0@>>>A@>1>>A@>>>0@>>>0\\ @.@VVV@VVV@VVV\\ @.0@.0@.0@. \end{CD}$$
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 $$l_1$$.
• Why is this lifting given by matrices over $k$, as required in the question? – LSpice Jan 4 at 4:36
• Ups, I missed this condition (thought it is about matrices over $A$). – Sasha Jan 4 at 14:11