Rees algebra for non-radical ideals - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T14:42:26Z http://mathoverflow.net/feeds/question/46202 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46202/rees-algebra-for-non-radical-ideals Rees algebra for non-radical ideals Justin Campbell 2010-11-16T06:29:25Z 2010-11-17T02:17:54Z <p>Today in my introductory algebraic geometry class we defined the so-called Rees algebra associated with an ideal $I$ of a ring $R$ (with strong conditions on $R$, if you like: I don't mind restricting to finitely generated reduced algebras $R$ over an algebraically closed field $k$). If we want to think of (maximal) Proj applied to the Rees algebra as Spec $R$ blown up along the vanishing set of $I$, what happens when $I$ is not radical? </p> <p>In particular, if $R = k[X_1,\cdots,X_n]$ and $I = (X_1,\cdots,X_n)$ we get $\mathbb{A}^n$ blown up at the origin. Have people tried to interpret geometrically the Rees algebra of $I^m$ for $m > 1$, or even $I^2$? I asked my professor, and (in an unusual turn of events) he did not have an answer available off the top of his head.</p> http://mathoverflow.net/questions/46202/rees-algebra-for-non-radical-ideals/46233#46233 Answer by mdeland for Rees algebra for non-radical ideals mdeland 2010-11-16T13:41:27Z 2010-11-16T13:41:27Z <p>The blowup of $Spec R$ along $I$ and $I^m$ give isomorphic results. This is Hartshorne exercise II.7.11.a. In general, any birational projective morphism is realized as the blowing up of the target along some ideal sheaf, which in general will be quite complicated (e.g. not radical, but not a power of a radical ideal either).</p> http://mathoverflow.net/questions/46202/rees-algebra-for-non-radical-ideals/46318#46318 Answer by Karl Schwede for Rees algebra for non-radical ideals Karl Schwede 2010-11-17T02:17:54Z 2010-11-17T02:17:54Z <p>Here are some other examples that you might find interesting. Try blowing up the following ideals:</p> <p>$(x,y^2), (x^2, y^2), (x,y^2)(x,y) = (x^2, xy, y^3)$</p> <p>Playing around with the third one will likely lead you to understand what blowing up the product of ideals does.</p> <p>In some related contexts, you might ask what ideals $I, J$ have the same <em>normalized</em> blow-ups $\pi : \widetilde X \to X$ and the same pull-backs $I \cdot O_{\widetilde X} = J \cdot O_{\widetilde X}$. The answer is whenever $I$ and $J$ have the same <em>integral closure</em> (in some sense, the integral closure of an ideal is the largest ideal having the same normalized blow-up, and pull-back, as the given ideal).</p> <p>In particular, the ideal $(x^2, y^2)$ has the same integral closure as $(x^2,xy, y^2)$ because, while they have different blow-ups, the normalizations of those blow-ups is the same. </p> <p>The previously mentioned book by Swanson and Huneke certainly will have some discussion of this.</p>