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Let $R$ be an $I$-adically separated and complete valuation ring, with $I$ finitely generated.

It is used a few times in Bosch, Lectures on Formal and Rigid Geometry e.g. first lines of pg. 164, Cor. 5 and Cor. 6 (their condition (V) is what I stated above) that

If an $A$ module has no $I$ torsion then it is flat over $R$.

I don't see why this is true. Any suggestions / references would be appreciated.

What I thought: We know $A$ is flat over domain $R$ iff it is $R$-torsion free.

If the statement were true: $I$-torsion free $\Rightarrow$$R$-torsion free.

This would hold if $I$ is a maximal ideal but otherwise I don't see why.

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  • $\begingroup$ What book by Bosch? Anyway, maybe the answer is here: mathoverflow.net/questions/51095/… $\endgroup$
    – Drew Heard
    Commented Jul 24, 2021 at 10:28
  • $\begingroup$ Sorry, I added it, didn't know why the hyper link didn't show up. and unfortunately I can't find which part of the llink addresses this problem $\endgroup$
    – Bryan Shih
    Commented Jul 24, 2021 at 14:29

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In Section 7.3 it is assumed that $I$ is the ideal of definition or $R$. It follows that the $I$-adic topology is separated, so (because $R$ is a valuation ring) every nonzero ideal of $R$ contains some power of $I$, so everything that is $R$-torsion is $I$-torsion.

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  • $\begingroup$ Now looking back: is having no $I$-torsion the same as being flat when $R$ is of type (N) in the text? (N)=Noetherian adic with ideal of defitiion $I$. (page 162). In otherwords is iii of definition 3 equivalent to being flat? $\endgroup$
    – Bryan Shih
    Commented Jul 25, 2021 at 10:01
  • $\begingroup$ No, the maximal ideal of $k[[x,y]]$ is torsion-free but not flat. $\endgroup$
    – Piotr Achinger
    Commented Sep 25 at 16:58

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