A-valued points of projective space - MathOverflow

A-valued points of projective space

2010-11-15T14:52:39Z

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

$\{(n+1)\mbox{-tuples of elements of }A\mbox{ that generate the unit ideal }\}$ 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 $\mathbb{P}^n_A$.

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:

Let us take, e.g. a number field $K$ such that $A=\mathcal{O}_K$ 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 $\mathbb{P}^n_{\mathbb{Z}}$ (Corollary III/42 in Eisenbud+Harris).

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.

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 $A$?

Most books just consider the case when $A$ is a field, there everything works just fine.

Thanks!

Answer by profilesdroxford54 for A-valued points of projective space

2010-11-15T18:03:37Z

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 $\mathbb{A}_\mathbb{Z}^{n+1}\backslash \{0\}$ .

The exercise is not even true over field -- after all you have to impose an equivalence relation on the $(n+1)$ -tuples; for general rings we have:

The quotient of $\mathbb{A}_\mathbb{Z}^{n+1}\backslash \{0\}$ by the obvious action of $\mathbb{G}_m$ is $\mathbb{P}^n_\mathbb{Z}$

This is what the exercise should have asked.

Answer by VA for A-valued points of projective space

2010-11-16T02:20:10Z

Yes, you (and BCnrd) are absolutely correct and the quoted statement is wrong.

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 $k$.

Note: no Noetherian assumptions on $S$ are necessary. This is the first step for Grothendieck's construction of Hilbert schemes, without Noetherian assumption.

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.

Answer by C S for A-valued points of projective space

2010-11-16T13:56:52Z

Thanks for your answers!!

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

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 $K$

and

$\{A$-valued points of $\mathbb{P}^n_A\}$

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 $\mathbb{P}^n_{\mathcal{O}_K}$ .