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[EDIT]: After getting a nice counter example provided by Steven Landsburg I realize that I forgot to impose an important condition...namely $R$ is supposed to be complete w.r.t. the $I$-adic topology. (In particular, this implies that elements of the form $1+i$ for $i\in I$ are units.)

Suppose $R$ is a commutative ring complete w.r.t. the $I$-adic topology, where $I\subseteq R$ a principal ideal generated by a non-zero divisor. Assume we're given split short exact sequences \begin{align*} R \to S \times T &\to R^{n-1} \\ R/I \to S_I \times T_I &\to (R/I)^{n-1} \end{align*} the first inducing the second on quotients, consisting of free $R$-modules ($R/I$-modules) $S= R^{n-1}$ and $T= R$ (resp. $S_I =(R/I)^{n-1}$ 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. My question concerns its image under the assumption that on quotients the rank one copy $R/I$ embeds as direct summand into $S_I$.

I can see that 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: Is it possible that to show that, in case of a diagonal embedding, the projection of $R$ to $S$ is a direct summand of $S$?

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  • $\begingroup$ After your edit, the answwer is trivially yes: Split the map $R_I\rightarrow S_I$ with a map $f_I:S_I\rightarrow R_I$. Lift $f$ arbitrarily to a map $f:S\rightarrow R$. Then the composition $R\rightarrow S\rightarrow R$ is the identity mod $I$, hence an isomorphism. $\endgroup$
    – Steven Landsburg
    Commented Oct 20, 2014 at 13:30
  • $\begingroup$ Thanks again. I will accept your answer again since I can't accept this comment and I'm sorry that I can't up vote neither your answer nor your comment...looks that I don't have enough reputation, yet. $\endgroup$
    – Harry
    Commented Oct 20, 2014 at 15:50

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Let $v$ be any map at all from $R$ to $S\approx R^{n-1}$.

Then $v=qu$ where $u:R\rightarrow R^n$ is given by $r\mapsto (vr,r)$ and $q:R^n\rightarrow S$ is the obvious projection.

Therefore your map from $R\rightarrow S$ can be any map at all. You are therefore asking whether an arbitrary map that becomes a split injection mod $I$ must be a split injection.

For a counterexample, let $n=1$, and let $v$ be multiplication by some non-unit $1+i$ with $i\in I$.

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  • $\begingroup$ Thanks for your answer, but sorry I don't understand. I have the assumption that $R/I$ split injects into $S_I$ and the quotient maps respect all splittings. Hence the lift $R$ of can only embed as described above, and in particular not solely into $T$. So $\nu$ is not arbitrary...in fact I think I can show that $\nu$ is injective. So the question would reduce to whether or not it is split. $\endgroup$
    – Harry
    Commented Oct 14, 2014 at 16:53
  • $\begingroup$ Multiplication by $1+i$ becomes a split injection mod $I$. $\endgroup$
    – Steven Landsburg
    Commented Oct 14, 2014 at 17:42

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