Torsion-free modules over a general ring 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! 
 A: 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 $y\neq 0$.
A: 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.
