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I hope this question isn't too silly. It is certainly fundamental, so the answer is likely contained at least implicitly in most sources out there, but I haven't seen it done this way (that is, in this particular functorial manner) in a way which is overt enough for me to catch on. I am familiar with the classical $Proj$-construction for a graded ring, so that's not quite what I'm looking for.

Let $k$ be a ring. Let's call a covariant functor of sets on some category of $k$-algebras an algebraic functor (over $k$). The affine $I$-space over $k$ is the algebraic functor $\mathbb{A}^I:(k−alg)→(set)$ which takes a $k$-algebra $R$ to the set $\mathbb{R}^I$ of $I$-tuples of elements of $R$. This functor is (co)representable by the ring $k[T_i],i∈I$, so $\mathbb{A}^I$ is (represented by) an affine scheme.

I want the projective space over $k$ in terms of an algebraic functor over $k$. I'm thinking something like $R↦\{\mathbb{R}^{I+∞}/\mathbb{G}_m(R)\}$ (where $\mathbb{G}_m(R)$ is the multiplicative group of $R$), or as a functor sending $R$ to some set of modules of rank $1$. One should then be able to show that it has a cover by four copies of the affine $I$-space over $k$. Alternatively, it would likely make sense to consider it a functor on the category of graded $k$-algebras.

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up vote 6 down vote accepted

Classically, if $K$ is a field, then $\mathbb P^n(K)$ is the set of lines $L\subset K^{n+1}$ of the vector space $K^{n+1}$.

If $R$ is a ring, the correct generalization of a line in $R^{n+1}$ is a projective submodule $L\subset R^{n+1}$ of rank one, which is also a direct summand : $ R^{n+1}= L\oplus E$, where $E$ is projective of rank $n$.
We call such submodules supplemented line bundles and $\mathbb P^n(R)$ is the set of these.

Beware that it is not automatic that a projective submodule of rank one of $ R^{n+1} $ is a direct summand, even if it is free :
for example the submodule $2\mathbb Z\oplus 0\subset \mathbb Z^2$ is free of rank one but not a direct summandand and is thus not an element of $\mathbb P^1(\mathbb Z)$ .
However a free $R$-module $R(r_0,r_1,\cdots ,r_n)\subset R^{n+1}$ is a supplemented line bundle iff the $r_i$'s generate $R$ i.e. $\Sigma Rr_i=R$

Because of Grothendieck's vast generalization mentioned by Martin one also considers the dual definition of $\mathbb P^n(R)$ as equivalence classes of projective $R$-modules of rank one $Q$ equipped with a surjective morphism $R^{n+1}\to Q$.

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Thank you! Another good answer. Hard to know which one to pick, but I think I'll go for this one because it is closer to the generality I asked for. I also found this treated in a set of notes by Strickland on formal schemes. – Eivind Dahl Mar 17 '12 at 22:11
Ah, the condition that the $r_i$ generate $R$ got much clearer when considering that this means that they do not all vanish at any point. This plugs nicely into thinking about projective schemes as looking for strictly non-trivial solutions to homogeneous equations. Neat :-) – Eivind Dahl Apr 14 '12 at 21:19

If $S$ is a scheme and $\mathcal{E}$ is a locally free module on $S$, then the projective space bundle $\mathbb{P}(\mathcal{E}) \to S$ represents the following functor:

$\mathrm{Sch}/S \to \mathrm{Set}, (f:X \to S) \mapsto \{\text{invertible quotients of } f^* \mathcal{E}\}$

You can find this in every introduction to algebraic geometry, for example EGA I (1970), § 9. Actually this is the definition of $\mathbb{P}(\mathcal{E})$ and then it can be shown with a general principle (every Zariski sheaf, which is locally representable, is representable) that this functor is representable by a scheme.

In the special case $\mathcal{E} = \mathcal{O}_S^{d+1}$, one writes $\mathbb{P}^d_S$ for this $S$-scheme and it represents the functor

$\mathrm{Sch}/S \to \mathrm{Set}, (f:X \to S) \mapsto \{\text{invertible quotients of } \mathcal{O}_X^{d+1}\}.$

Of course you can even specify to $S=\mathrm{Spec}(k)$ for some ring $k$ and restrict to $k$-algebras. But in my opinion it is hard to really understand the projective space when only defined as a functor on $k$-algebras. By the way, I first understood Grassmannians when I learned the general "global" definition in loc. cit. - these chart definitions in topology are only confusing ...

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Thank you! I agree charts might not be all too great for understanding projective spaces, but the first thing I really understood and liked was the projective line, patched together by pieces of $k[t]$. It makes it clear to me why a meromorphic function on $X$ is the same thing as a morphism $X\to\mathbb{P}^1$. This complements nicely the notion of a regular function as a morphism $X\to\mathbb{A}^1$. If I was able to see from the definition of $\mathbb{P}^I$, that this was the case I would be every happy. At least you've told me what I need to stare at until it makes sense :-) – Eivind Dahl Mar 16 '12 at 23:02
Charts can be confusing, but they are useful for doing local calculations. It's important to understand both approaches (as well as the homogeneous space point of view). – Michael Joyce Mar 17 '12 at 0:11
Michael: I agree, but often computations with charts are made without any global data. This is always confusing and clumsy. @Eivind: You can derive this from the functorial definition. – Martin Brandenburg Mar 18 '12 at 1:09
This got way clearer with time. Thanks again :-) – Eivind Dahl Apr 14 '12 at 21:19
@Eivind: I am very glad to hear that. – Martin Brandenburg Mar 26 '13 at 0:14

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