most recent 30 from http://mathoverflow.net 2013-05-22T02:08:51Z http://mathoverflow.net/feeds/question/74654 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74654/torsion-free-modules-over-general-ring student 2011-09-06T14:27:08Z 2011-09-10T14:09:02Z

<p>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! </p>

http://mathoverflow.net/questions/74654/torsion-free-modules-over-general-ring/74657#74657 David White 2011-09-06T15:12:00Z 2011-09-06T15:17:36Z

<p>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.</p> <p>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 $y\neq 0$.</p>

http://mathoverflow.net/questions/74654/torsion-free-modules-over-general-ring/75102#75102 student 2011-09-10T14:09:02Z 2011-09-10T14:09:02Z

<p>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"</p> <p>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.</p>