Limits of reduced schemes question from Eisenbud and Harris - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T04:28:22Z http://mathoverflow.net/feeds/question/70130 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70130/limits-of-reduced-schemes-question-from-eisenbud-and-harris Limits of reduced schemes question from Eisenbud and Harris Tait 2011-07-12T15:12:36Z 2011-07-13T13:03:42Z <p>My question pertains to exercise II-16 in Eisenbud and Harris' "The geometry of Schemes". For an algebraically closed field $K$ the question is as follows:</p> <blockquote> <p>Consider zero-dimensional subschemes $\Gamma \subset \mathbb{A}_K^4$ of degree 21 such that $$V(m^3)\subset\Gamma \subset V(m^4)$$ where $m$ is the maximal ideal of the origin in $\mathbb{A}_K^4$. Show that there is an 84-dimensional family of such subschemes, and conclude that in general one is not a limit of a reduced scheme.</p> </blockquote> <p>What does it mean for a family of subschemes to have dimension 84? I can only think it means up to isomorphism there are 84 such subschemes, but this doesn't seem to work.</p> http://mathoverflow.net/questions/70130/limits-of-reduced-schemes-question-from-eisenbud-and-harris/70221#70221 Answer by Sasha for Limits of reduced schemes question from Eisenbud and Harris Sasha 2011-07-13T13:03:42Z 2011-07-13T13:03:42Z <p>Such a subscheme is given by an ideal $I$ such that $m^4 \subset I \subset m^3$. In fact, any vector subspace in $m^3$ containing $m^4$ is an ideal (this is a simple exercise). Since $\dim O/m^3 = 15$ and $\dim O/m^4 = 35$, and we are interested in subspaces $I \subset O$ such that $\dim O/I = 21$, that is in subspaces $I/m^4$ of dimension $35 - 21 = 14$ of the space $m^3/m^4$ of dimension $35 - 15 = 20$. So, the family in question is just the Grassmannian $Gr(14,20)$. Its dimension is $14(20-14) = 14\cdot 6 = 84$.</p>