A-valued points of projective space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:45:05Z http://mathoverflow.net/feeds/question/46116 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46116/a-valued-points-of-projective-space A-valued points of projective space C S 2010-11-15T14:52:39Z 2010-11-16T18:09:12Z <p>I have been reading "The Geometry of Schemes" by Eisenbud and Harris and have a question about Exercise III-43. There, one should show that there is a bijection between the sets</p> <p>$\{(n+1)\mbox{-tuples of elements of }A\mbox{ that generate the unit ideal }\}$ and $\{ \mbox{maps} \mbox{ Spec} A \to \mathbb{P}^n_A$ such that the composite $\mbox{Spec} A \to \mathbb{P}^n_A\to \mbox{Spec}A=id\}$, i.e. $A$-valued points of $\mathbb{P}^n_A$.</p> <p>Now, of course, $(n+1)$-tuples of elements of $A$ give $A$-valued points, but if $A$ is not a ring such that every invertible $A$-module is free of rank one, I don't see why the converse should work:</p> <p>Let us take, e.g. a number field $K$ such that <code>$A=\mathcal{O}_K$</code> is not a PID. Then, up to multiplication by a unit, an $A$-valued point corresponds to an invertible $A$-module $P$ and an epimorphism $A^{n+1}\to P$ by the characterization of morphisms from $\mbox{Spec}A$ to <code>$\mathbb{P}^n_{\mathbb{Z}}$</code> (Corollary III/42 in Eisenbud+Harris).</p> <p>Starting with an $(n+1)$-tuple generating $A$, I clearly get an epimorphism $A^{n+1}\to A$ and $A$ is a projective $A$-module, so I get an $A$-valued point.</p> <p>However, if I start with an $A$-valued point corresponding to an epimorphism $A^{n+1}\to P$ and the invertible module $P$ is not free, how can I choose an $(n+1)$-tuple of points of $A$ which generate the unit ideal? Moreover, don't these $A$-valued points give "additional" points, which do not come from $(n+1)$-tuples of elements of $A$?</p> <p>Most books just consider the case when $A$ is a field, there everything works just fine.</p> <p>Thanks!</p> http://mathoverflow.net/questions/46116/a-valued-points-of-projective-space/46129#46129 Answer by profilesdroxford54 for A-valued points of projective space profilesdroxford54 2010-11-15T18:03:37Z 2010-11-15T18:22:54Z <p>Indeed, the functor from commutative rings to sets: $A\mapsto {(n+1)\mbox{-tuples of elements of }A\mbox{ that generate the unit ideal }}$ is the scheme <code>$\mathbb{A}_\mathbb{Z}^{n+1}\backslash \{0\}$</code>.</p> <p>The exercise is not even true over field -- after all you have to impose an equivalence relation on the <code>$(n+1)$</code>-tuples; for general rings we have:</p> <p>The <em>quotient</em> of <code>$\mathbb{A}_\mathbb{Z}^{n+1}\backslash \{0\}$</code> by the obvious action of $\mathbb{G}_m$ is <code>$\mathbb{P}^n_\mathbb{Z}$</code></p> <p>This is what the exercise should have asked.</p> http://mathoverflow.net/questions/46116/a-valued-points-of-projective-space/46175#46175 Answer by VA for A-valued points of projective space VA 2010-11-16T02:20:10Z 2010-11-16T02:20:10Z <p>Yes, you (and BCnrd) are absolutely correct and the quoted statement is wrong.</p> <p>Over any scheme $S$, the $S$-points of $\mathbb P^n$ are the surjections $\mathcal O_S^{\oplus n+1} \to F$ with invertible $\mathcal O_S$-module $F$. More generally, the $S$-points of the grassmannian $Gr(m,k)$ are the surjections $\mathcal O_S^{\oplus m}\to F$ with $F$ locally free of rank $k$. </p> <p>Note: no Noetherian assumptions on $S$ are necessary. This is the first step for Grothendieck's construction of Hilbert schemes, without Noetherian assumption.</p> <p>So, for a ring $A$, the $A$-points of $\mathbb P^n$ are the surjections $A^{n+1}\to P$ with $P$ locally free (equivalently, projective) $A$-modules of rank 1.</p> http://mathoverflow.net/questions/46116/a-valued-points-of-projective-space/46235#46235 Answer by C S for A-valued points of projective space C S 2010-11-16T13:56:52Z 2010-11-16T18:09:12Z <p>Thanks for your answers!!</p> <p>If you take $A=\mathcal{O}_K$, the invertible $A$-modules are exactly the non-zero fractional ideals of $A$, so I guess we can reformulate the exercise to be</p> <p>There is a bijection of sets $\{(n+1)$-tuples of elements of A such that $\exists i: a_i\neq 0 \}$ modulo equivalence, where equivalence is multiplication by a non-zero element of $K$</p> <p>and </p> <p>$\{A$-valued points of $\mathbb{P}^n_A\}$</p> <p>Thus, for $\mathcal{O}_K$, the "classical" definition of points of projective $n$-space coincides with the definition of $\mathcal{O}_K$-valued points of <code>$\mathbb{P}^n_{\mathcal{O}_K}$</code>.</p>