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Let $R$ be a commutative ring, $I$ its ideal, and $\bar{R}=R/I$. For which flat $\bar{R}$-modules $\bar{F}$ is there a flat $R$-module $F$ such that $F \otimes_R R/I \simeq \bar{F}$?

By Lazard's Theorem, each flat module is a direct limit of free modules of finite rank. If the underlying direct system is $\omega$, we can just lift the maps of the direct system (as they are just matrices), and that does the trick. This gives an affirmative answer for all countably presented flat $\bar{R}$-modules.

For general directed systems however, I see no reason for this to work. Am I wrong? Do you know about a counterexample, reference...? Thanks!

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    $\begingroup$ There are projective modules (=finitely generated flat modules) over $\overline{R}$ which do not lift to projective modules over $R$, so I am skeptical of your argument using Lazard's theorem. $\endgroup$
    – Mohan
    Commented Jan 3, 2018 at 14:42
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    $\begingroup$ @Mohan An example of a projective that doesn’t lift to a projective is $R=\mathbb{Z}$, $I=6\mathbb{Z}$, $P=\mathbb{Z}/2\mathbb{Z}$. But applying Fred.Fred’s construction, $P$ does lift to a flat $\mathbb{Z}$-module, namely $\mathbb{Z}[\frac{1}{3}]$. Is this example enough to defuse your skepticism? $\endgroup$
    – Jeremy Rickard
    Commented Jan 3, 2018 at 14:52
  • $\begingroup$ Where does the argument using Lazard's Theorem break down for a general directed system? $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Jan 5, 2018 at 23:05
  • $\begingroup$ @MahdiMajidi-Zolbanin If the directed system is not well-ordered, lifting the maps between the finitely generated free $\bar{R}$-modules of the system arbitrarily may not yield a directed system of $R$-modules, or at least, I do not see a way how to ensure that. $\endgroup$
    – Fred.Fred
    Commented Jan 6, 2018 at 18:04

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