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Let $R$ be a Noetherian integral domain. Let $x\in R$ be a prime element. Let $\overline{R}=R/Rx$.

Let $P$ be a finitely-generated projective $R$-module.

Assume that $\frac{P}{xP}$ is a free $\overline{R}$-module with a finite free basis $v_1,\dots,v_m$, where $m\ge0$. What are some conditions under which there are pre-images $u_1,\dots,u_m$ of $v_1,\dots,v_m$ under the natural $R$-epimorphism $\nu:P\to\frac{P}{xP}$ such that $P$ is generated by $u_1,\dots,u_m$? I am looking for a condition other than "let $x$ be in the Jacobson radical."

Edit: (1) I will let you assume that $C$, the submodule generated by $u_1,\dots,u_m$, is free.
(2) When can we embed $P/C$ into a free module?

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  • $\begingroup$ Even if you want this to be true for $m=1$, then you need to assume injectivity of the restriction map on class groups, $\text{Cl}(R)\to \text{Cl}(\overline{R})$. This is going to rule out most Dedekind domains, right off the bat. $\endgroup$
    – Jason Starr
    Commented Jan 25, 2016 at 21:40
  • $\begingroup$ Jason Starr, Thank you for your comment, but I am confused. What I am seeking seems to be true if $x$ is in the Jacobson radical (Lemma 6.4 of the following manuscript). I am not asking for it to be true for all $x$. www2.warwick.ac.uk/fac/sci/maths/people/staff/bouyer/… $\endgroup$
    – Tri
    Commented Jan 26, 2016 at 1:47
  • $\begingroup$ You wrote that you do not want to assume that $x$ is in the Jacobson radical. Have you changed your mind? For a Dedekind domain, unless it is semilocal, rarely will a nonzero element be contained in the Jacobson radical. Just thinking about rings of integers of number fields or coordinate rings of affine curves. For semilocal Noetherian rings, every finitely presented, projective module is free. $\endgroup$
    – Jason Starr
    Commented Jan 26, 2016 at 11:29
  • $\begingroup$ No, I just meant that there are general conditions under which the result will be true. I would like to find some conditions other than $x$ being in the Jacobson radical. $\endgroup$
    – Tri
    Commented Jan 26, 2016 at 19:35
  • $\begingroup$ I am confused with your edit. If you want the $u_i$s generate $P$, doesn't that mean $C=P$? Also, doesn't what you require force $P$ to be free? $\endgroup$
    – Mohan
    Commented Mar 16, 2018 at 18:10

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