Hartshorne's associated scheme for a variety - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T16:14:57Z http://mathoverflow.net/feeds/question/69278 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69278/hartshornes-associated-scheme-for-a-variety Hartshorne's associated scheme for a variety abourdon 2011-07-01T17:20:57Z 2011-07-02T01:34:21Z <p>This question comes from Proposition 2.6 in Chapter 2 of Hartshorne's <em>Algebraic Geometry</em>. In my edition, that's on page 78.</p> <p>For a variety $V$, Hartshorne defines the topological space $t(V)$ to consist of the nonempty closed irreducible subsets of $V$, where the closed sets of $t(V)$ are of the form $t(Y)$ for $Y$ closed in $V$. He then defines a map $\alpha: V \rightarrow t(V)$ where P gets sent to {P} in $t(V)$. The claim is that <code>$(t(V), \alpha_*(\mathcal{O}_V))$</code> is a scheme. I understand why this is true if $V$ is affine, but I have been unable to show <code>$(t(V), \alpha_*(\mathcal{O}_V))$</code> is a scheme for an arbitrary variety $V$.</p> <p>I had hoped to show that if $U$ is an affine open subset of $V$, then $t(U)$ is isomorphic to an open subset of $t(V)$. I used the map from $t(U)$ into $t(V)$ where we send an irreducible subset $W$ in $U$ to the smallest irreducible subset of $V$ containing $W$. However, although the image of of $t(U)$ is contained in $[t(U^c)]^c$, I don't believe these are equal.</p> http://mathoverflow.net/questions/69278/hartshornes-associated-scheme-for-a-variety/69285#69285 Answer by Ramsey for Hartshorne's associated scheme for a variety Ramsey 2011-07-01T18:22:41Z 2011-07-01T18:22:41Z <p>To show that $(t(V),\alpha_*\mathcal{O}(V))$ is a scheme, you must show that $t(V)$ has an open cover on which this ringed space is isomorphic to an affine scheme.</p> <p>Take an affine open cover $\{U_i\}$ of $V$. Since you believe the affine case, it suffices to show that $\{t(U_i)\}$ is an open cover of $t(V)$, and</p> <p><code>$(t(V),\alpha_*\mathcal{O}(V))|_{t(U_i)} \cong (t(U_i),\alpha_*\mathcal{O}(U_i))$</code></p> <p>for each $i$. Given your last paragraph, it sounds like the first of these points is your difficulty. Let $Y$ be a nonempty irreducible closed subset $Y\subseteq U_i$. For each $j$, $Y\cap U_j$ is (when nonempty) a nonempty irreducible closed subset of $U_i\cap U_j$ (since an open subset of an irreducible is irreducible). The intersection $U_i\cap U_j$ is an affine open subset of $U_j$, and it's not hard to see (look at the pre-image of the corresponding prime ideals!) that $Y\cap U_j$ extends in a natural way to an irreducible closed subset of $U_j$. These extensions glue for varying $j$ to give an irreducible closed subset of $V$, since a locally irreducible subset of a (connected) space is irreducible. This furnishes the map $t(U_i)\to t(V)$ (which, in particular, I think addresses the issue you raise in the last paragraph). </p> <p>It remains to see that this is an open subset and gives an open cover of $t(V)$, and to prove the above isomorphism. So now try from here...</p> http://mathoverflow.net/questions/69278/hartshornes-associated-scheme-for-a-variety/69305#69305 Answer by Poldavian for Hartshorne's associated scheme for a variety Poldavian 2011-07-02T01:34:21Z 2011-07-02T01:34:21Z <p>I can't comment yet, so this is a brief comment about some intuition, not the exact answer. At least on affine schemes, the points (i.e. prime ideals) are in one-to-one correspondence with irreducible closed subsets of the affine scheme (I have just read very little EGA, so please excuse my ignorance if this holds generally for any scheme - which is actually great). This really explains the "somewhat unintuitive" construction of the map. </p>