most recent 30 from http://mathoverflow.net 2013-05-24T00:37:01Z http://mathoverflow.net/feeds/user/7455 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83943/a-module-with-extim-r-0-for-all-i-0 Amit 2011-12-20T14:04:28Z 2013-02-25T15:13:00Z

<p>Let $M$ be a finitely generated module over a noetherian local ring $R$. We can take our ring to be Cohen-Macaulay. Suppose $M$ satisfies the condition $Ext^i(M,R) = 0$ for all $i > 0$. We want to know if $M$ is projective?</p> <p>One can easily show from the given condition that for any module $N$ of finite projective dimension, we do have $Ext^1(M,N) = 0$. Thus if (for example) our ring $R$ were regular local ring (which means any f.g module will have finite projective dimension), then we get the desired result (that $M$ is projective). </p> <p>Now, Regular local => Cohen-Macaulay. So my first question is can we say the same with only Cohen-Macaulay condition on $R$ (that $M$ is projective)?</p> <p>If it helps, we may assume that $M$ itself has finite projective dimension. </p> <p>My second question is that can we write any f.g. module on (say) a noetherian local ring, as direct limit of modules having finite projective dimension?</p>

http://mathoverflow.net/questions/93069/when-are-conformal-maps-holomorphic/93070#93070 Amit 2012-04-04T04:21:11Z 2012-04-04T04:21:11Z

<p>If we define <strong>$f' : \mathbb{R}^2 \rightarrow \mathbb{R}^2$</strong> by <strong>$f'(x,y) = (x, -y)$</strong>, then it is conformal, but the corresponding map $f_1 + i f_2$ is not holomorphic.</p>

http://mathoverflow.net/questions/82453/ampleness-of-a-big-line-bundle Amit 2011-12-02T13:35:03Z 2011-12-02T14:40:22Z

<p>Let $X$ be a non-singular complex variety with a big line and base point free bundle $M$ on it. My question is can we say that for any locally free sheaf $F$ on $X$, $F \otimes M^n$ is globally generated for $n \gg 0$. </p> <p>Motivation: If $M$ were an ample line bundle then all we need is that $F$ is coherent sheaf. But since we are given a much stronger condition on $F$ (which is local freeness) can we say the same thing with $M$ just being big and base point free.</p> <p>I tried to use the fact that any big line bundle is tensor product of an ample line bundle and an effective line bundle. </p> <p>I am not even sure that this has to be true but am unable to find a counterexample. </p>

http://mathoverflow.net/questions/123459/when-adding-a-constant-makes-a-multivariate-polynomial-reducible Amit 2013-03-03T07:32:06Z 2013-03-03T07:32:06Z

About your first question - there need not be any simple criteria to say when such $m$ does not exist. Take e.g polynomial $x^2 + ny + m$. As long as $n \neq 0$, no value of $m$ will make this factorisable. Examples of this kind suggests that a simple criteria in terms of coefficients of the polynomial may not exist.

http://mathoverflow.net/questions/83943/a-module-with-extim-r-0-for-all-i-0 Amit 2011-12-21T16:33:22Z 2011-12-21T16:33:22Z

@Mahdi, thanks. @Benjamin, thanks. I was not aware of this conjecture.

http://mathoverflow.net/questions/83943/a-module-with-extim-r-0-for-all-i-0/83958#83958 Amit 2011-12-21T10:54:50Z 2011-12-21T10:54:50Z

Thanks for this example.

http://mathoverflow.net/questions/82453/ampleness-of-a-big-line-bundle/82464#82464 Amit 2011-12-03T10:51:27Z 2011-12-03T10:51:27Z

Thanks for this reference

http://mathoverflow.net/questions/82453/ampleness-of-a-big-line-bundle/82458#82458 Amit 2011-12-02T15:17:29Z 2011-12-02T15:17:29Z

Thanks for this concrete counterexample.