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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!

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9  
Yes, non-trivial line bundles on Spec($A$) which admit $n+1$ generators provide "additional" Spec($A$)-valued points of projective $n$-space $\mathbf{P}^n_{\mathbf{Z}}$. The same issue comes up when describing $C^{\infty}$-maps from smooth manifolds to real projective $n$-space (using $C^{\infty}$-line bundles), holomorphic maps from complex manifolds to complex projective $n$-space (using holomorphic line bundles), and continuous maps from topological spaces to real (resp. complex) projective $n$-space (using real (resp. complex) topological line bundles), so it's not specific to schemes. –  BCnrd Nov 15 '10 at 15:01
2  
A warning tangentially relevant to the question: around the same part of that book, they attempt to define Proj(F) for a coherent sheaf F in terms of "prime ideal sheaves." There's some fishy stuff going on there, and I think the lesson is that one should not try to define things that way: read Hartshorne (or EGA) for this instead. –  Dave Anderson Nov 15 '10 at 19:05

3 Answers 3

up vote 9 down vote accepted

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.

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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.

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Yes, of course you are right. In the case of a PID I automatically meant the set of $(n+1)$-tuples of $A$ that generate the unit ideal modulo equivalence -- multiplication by a unit. I was particularly concerned about the "generate the unit ideal" part. –  C S Nov 16 '10 at 13:56

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

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... $\mathcal O_K$, the ring of integers in a number field $K$. –  VA. Nov 16 '10 at 14:07
    
... and "multiplication by a nonzero element"? For $A=\mathbb Z$, (1,1)=(2,2) in $\mathbb P^1$. –  VA. Nov 16 '10 at 15:04
    
Yes, thank you. I also forgot "fractional" in the above. I editted it, I hope it is correct now. –  C S Nov 16 '10 at 17:42

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