User mohan kumar - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T12:38:41Z http://mathoverflow.net/feeds/user/6703 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27424/a-ring-such-that-all-projectives-are-stably-free-but-not-all-projectives-are-free/27611#27611 Answer by Mohan Kumar for A ring such that all projectives are stably free but not all projectives are free? Mohan Kumar 2010-06-09T19:34:06Z 2010-06-09T19:34:06Z <p>The more canonical example probably is the standard universal example for such a question. So, let $R_n=k[x_i,y_i]/\sum x_iy_i=1$ where $k$ is any field and there are $2n$ variables. By localization one easily checks that $K_0(R_n)=\mathbb{Z}$ for any $n$. But the projective module given by the presentation, $$0\to R_n\stackrel{(x_i,\ldots,x_n)}{\to} R_n^n\to P\to 0$$ is clearly stably free but not free if $n\geq 3$. An algebraic proof (given by myself and Madhav Nori) can be found in the article of Swan (Annals of Math studues, vol 113, pp 432-522). </p> http://mathoverflow.net/questions/38994/normal-compactification-of-an-algebraic-variety Comment by Mohan Kumar Mohan Kumar 2010-09-16T17:15:42Z 2010-09-16T17:15:42Z The answer is no. For example, $X$ could be a normal(or smooth) complete (and thus compact) algebraic variety which is not projective. Not too many options for compactifications there. http://mathoverflow.net/questions/38599/ampleness-of-the-universal-subbundle Comment by Mohan Kumar Mohan Kumar 2010-09-13T16:59:03Z 2010-09-13T16:59:03Z I would assume never. If $U$ is ample on $G$, then its restriction to a fiber over $x\in X$ must be too, but this just takes you back to the universal subbundle of the Grasmmannian of $r+1$ dimensional subspaces in a vector space and thus can not be ample. http://mathoverflow.net/questions/33417/kahler-forms-on-cohen-macaulay-spaces/33429#33429 Comment by Mohan Kumar Mohan Kumar 2010-07-26T19:26:22Z 2010-07-26T19:26:22Z If one takes a reduced irreducible curve (hence Cohen-Macaulay), most of the time (and easy to check), the module of 1-forms will have torsion. In fact, a well known conjecture of Berger posits that if this module is torsion free then the curve is smooth. The conjecture is open to the best of my knowledge.