TITLE: Which functor does the projective space represent?
QUESTION [17 upvotes]: 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.

REPLY [17 votes]: 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 $R^{n+1}\to Q$.