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i want to know how to prove a torsion free modules over general ring is flat. (in "lecture on ring and modules, T.Y.Lam prove in case R is interal domain). please help me prove it or give me some books or article concern this problem. Thanks!

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Even over a domain a torsion free module is certainly not flat in general. –  Georges Elencwajg Sep 6 '11 at 14:33
    
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The integral domains that satisfy this property (every torsionfree module is flat) are exactly the so called Prüfer domains. –  Johannes Hahn Sep 6 '11 at 21:33
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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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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. –  Georges Elencwajg Sep 6 '11 at 19:02
    
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 –  David White Sep 6 '11 at 19:36
    
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/… –  Georges Elencwajg Sep 6 '11 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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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. –  Qing Liu Sep 10 '11 at 17:38
    
thank you very much. let me try to prove –  student Sep 16 '11 at 9:25
    
Please do not use answers to do anything except answering questions. –  Mariano Suárez-Alvarez Oct 19 '13 at 19:00
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