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I want to know how to prove that a torsion-free module over a general ring is flat. In Lectures on Rings and Modules, T.Y. Lam proves this in the case where your ring is an integral domain. Can you please help me prove it or recommend some books or articles concerning this? Thanks!

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    $\begingroup$ Even over a domain a torsion free module is certainly not flat in general. $\endgroup$ Sep 6, 2011 at 14:33
  • $\begingroup$ @unknown(google) I'm not trying to pressure you, but if you think my post below answers your question and that you won't get a better answer, could you click the green check mark next to it? That way the MathOverflow software registers that this question has been answered and won't put it back up on the front-page in the future. I normally wouldn't say this, but as it's your first ever question I wanted to be sure you knew about how to accept an answer. If there's something more you want explained, please leave a comment and I'll say more. $\endgroup$ Sep 6, 2011 at 20:21
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    $\begingroup$ The integral domains that satisfy this property (every torsionfree module is flat) are exactly the so called Prüfer domains. $\endgroup$ Sep 6, 2011 at 21:33
  • $\begingroup$ I'm not sure there's a consensus for a definition of torsion-free module over a ring which is not a domain. Please include a definition, or restrict to domains. $\endgroup$
    – YCor
    Sep 23, 2016 at 8:59

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The best book for such questions in my opinion is the one you're already reading: "Lectures on Modules and Rings" by Lam. Indeed, on page 127 he provides a counter-example to your claim that torsion-free implies flat. Probably you meant the converse, which does hold: Any flat module is torsion-free. This is also on page 127.

Here's Lam's counter-example...Let $R=k[x,y]$ where $k$ is any commutative domain. Then $M=(x,y)$ is torsion-free because there are no relations on $x$ or $y$. However, $M$ is not flat. To see this set $S=R/(x)\cong k[y]$ so that $M\otimes_R S = M\otimes_R R/(x) \cong M/xM \cong (x,y)/(x^2,yx)$. If $M$ is flat over $R$ then $M\otimes_R S$ is flat over $S$ and hence torsion-free. This is a contradiction because $yx=0$ but $y\neq 0$.

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    $\begingroup$ Here is a geometric proof of non-flatness of the ideal $M\subset R$.If $M$ were flat, it would be projective (flat+finitely presented $\Rightarrow$ projective), necessarily invertible since it is an ideal. But an invertible ideal has height one: contradiction, since $M$ trivially has height $\geq 2$ [actually $2$, of course]. A variant: a rank one projective module on $\mathbb A_k^2$ is trivial, so $M$ would be a free ideal, which means a principal ideal: this is obviously false. $\endgroup$ Sep 6, 2011 at 19:02
  • $\begingroup$ Thanks, Georges, that is very nice. I forgot about the connection between invertible ideals and projective ideals, but with it I think your proof is much more elegant $\endgroup$ Sep 6, 2011 at 19:36
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    $\begingroup$ I, on the other hand, like the directness of your astute base change to $S$, David! And, also, Einstein remarked "If you are out to describe the truth, leave elegance to the tailor. " thinkexist.com/quotation/… $\endgroup$ Sep 6, 2011 at 20:19
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thank for your all answers! Here is my ideas: How can i prove the following proposition: " If every finitely generated ideal of R is principal, then a torsion - free R-module is flat"

Because most of books i have prove this property when R is integral domain, while i want to know how can we prove when R is general ring.

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    $\begingroup$ You can prove the proposition in the usual way. Let $M$ be your module. You need to show that the canonical map $f: I\otimes M\to IM$ is an isomorphism for any ideal $I$ of $R$ of finite type. Write $I=aR$ and consider the map $M\to I\otimes M$, $x\mapsto a\otimes x$ and $h=fg : x\to ax$. Then $f,g$ are surjective by construction, and $h$ is an isomorphism because $M$ is torsion free. So $f$ is an isomorphism. $\endgroup$
    – Qing Liu
    Sep 10, 2011 at 17:38
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    $\begingroup$ Please do not use answers to do anything except answering questions. $\endgroup$ Oct 19, 2013 at 19:00

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