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Let $A$ be a ring and let $\mathbb{P}^n = \operatorname{Proj} \mathbb{Z} [x_0, \ldots, x_n]$.

Question. What does $\mathbb{P}^n$ classify? In other words, is there some kind of algebraic structure (related to $A$), definable without reference to $\mathbb{P}^n$, such that there is a natural bijection between such structures and morphisms $\operatorname{Spec} A \to \mathbb{P}^n$?

The geometric answer seems to be well-known: $\mathbb{P}^n$ classifies isomorphism classes of line bundles on $\operatorname{Spec} A$ equipped with $n + 1$ nowhere-vanishing global sections up to simultaneous rescaling. (I think this is in turn the same as isomorphism classes of line bundles on $\operatorname{Spec} A$ equipped with a fibrewise-linear embedding into $\mathbb{A}^{n+1} \times \operatorname{Spec} A$ that admits a fibrewise-linear retraction.) But how does one express this in purely algebraic language?

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It seems like the answer to your last question is: very ample invertible sheaves globally generated by $n+1$ sections (up to a common constant multiple). –  Andrew Feb 6 '13 at 14:08
@Andrew If I recall correctly, such a thing defines an embedding into $\mathbb{P}^n$. I'm only asking for morphisms $X \to \mathbb{P}^n$. I've since convinced myself that these should be the same thing as line bundles over $X$ equipped with a chosen embedding into the trivial bundle $X \times k^{n+1}$. –  Zhen Lin Feb 6 '13 at 14:17
Whoops, of course you are correct -- I shouldn't have added the adjective "very ample". –  Andrew Feb 6 '13 at 14:41
Just out of curiosity, what would "purely algebraic" entail? Even though $\mathbb{P}^n$ being the "moduli space for line bundles with a specified set of ($n+1$) nowhere all-vanishing sections" is geometric, it's also algebraic, no? If it were, say, the quotient of $\mathbb{A}^{n+1}$ by the action of $\mathbb{G}_m$, would that make you happy? :) –  Alex Youcis Mar 13 at 9:40
I suppose one could interpret "purely algebraic" as meaning "in the language of commutative algebra". But an elementary description in terms of homogeneous coordinates would also qualify. –  Zhen Lin Mar 13 at 9:57

1 Answer 1

You already answered this question yourself in the question, but I'll spell it out here. A morphism from a scheme $X$ to $\mathbf{P}^n_\mathbf{Z}$ is given by a system $(\mathcal{L}, s_0, \ldots, s_n)$ where $\mathcal{L}$ is an invertible $\mathcal{O}_X$-module and $s_0, \ldots, s_n$ are global sections which generate $\mathcal{L}$. Two systems $(\mathcal{L}, s_0, \ldots, s_n)$ and $(\mathcal{N}, t_0, \ldots, t_n)$ determine the same morphism if and only if there exists an isomorphism $\alpha : \mathcal{L} \to \mathcal{N}$ of invertible sheaves such that $\alpha(s_i) = t_i$. See Lemma Tag 01NE.

Translating this into algebra when $X = \text{Spec}(A)$ we obtain: A morphism $\text{Spec}(A) \to \mathbf{P}^n_\mathbf{Z}$ is given by a system $(M, x_0, \ldots, x_n)$ where $M$ is a locally free $A$-module (see Definition Tag 00NW) of rank $1$ and $x_0, \ldots, x_n$ are elements of $N$ which generate it as an $A$-module. Two systems $(M, x_0, \ldots, x_n)$ and $(N, y_0, \ldots, y_n)$ define the same morphism if and only if there exists an $A$-module isomorphism $M \to N$ mapping $x_i$ to $y_i$.

A special kind of system is one of the form $(A, f_0, \ldots, f_n)$ where for some $i$ we have $f_i = 1$. Then the associated morphism maps into the standard open $D_+(X_i) = \mathbf{A}^n_\mathbf{Z}$ and is given by $f_0, \ldots, \hat f_i, \ldots f_n$ in coordinates. Every system is locally isomorphic to one of this form (exactly because the given elements generate!), so this tells you what the morphism is locally.

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