Suppose $R$ is a commutative ring, $I\subseteq R$ a principal ideal, and we're given split short exact sequences $ R \to R^n \to R^{n-1}$ and $ R/I \to (R/I)^n \to (R/I)^{n-1}$ the first inducing the second on quotients. Suppose further that there is another splitting of $R^n \cong S \times T$ with $S= R^{n-1}$ and $T= R$, such that on quotients $(R/I)^n \cong S/I \times T/I$, with $S/I =(R/I)^{n-1}$ and $T/I=R/I$.
A priori the direct summand $R$ specified in the first s.e.s. can embed into $S$, or into $T$ or diagonally into both.
Assume now that on quotients the rank one copy $R/I$ embeds as direct summand into $S/I$. This eliminates the possibility that $R$ embeds only into $T$ and restricts a diagonal embedding to the situation where the projection of $R$ to $T$ has to be contained in the ideal $I$.
Question: Can one show that, in case of a diagonal embedding, the projection of $R$ to $S$ is a direct summand of $S$?
I've put the question on MO, where Steven Landsburg provided a counter example: https://mathoverflow.net/questions/184387/lifting-a-direct-summand-of-a-free-module/184399#184399