Question about a functor related to projective space If $V$ is a vector space over $k$, then the classical way of defining $\mathbb{P}(V)$ is as $(V\backslash\{0\})/k^\times$. When generalizing this notion to schemes, and in particular if we wish to view projective space as a functor, then this point of view is somewhat abandoned (although of course is always "in the background").
I was wondering, if $R$ is a fixed ring and $M$ is a fixed (finitely generated) $R$-module, what can be said about the functor that takes an $R$-algebra $S$ and sends it to $((M\otimes_{R}S)\backslash\{0\})/S^\times$?
Here the quotient is taken to mean the orbit space of the natural action of $S^\times$ and so the functor can be seen as going from the category of $R$-algebras to the category of sets.
 A: (This is mostly just elaborating on what Alex said in the comments.)
This question gets at one of the more subtle things about algebraic geometry if you start (like I do) from the functor-of-points POV, which is that "deleting a point" (or more generally a closed subscheme) from a scheme is a really wacky operation to perform and it's surprising that we can perform it! (Edit: But see the discussion in the comments, I'm happier now.)
As Alex says in the comments, this issue already appears for $\mathbb{A}^1$ when we ask, functorially, what "$\mathbb{A}^1 \setminus \{ 0 \}$" could possibly actually mean. $\mathbb{A}^1$ itself is very easy to define: its functor of points is just $\mathbb{A}^1(R) = R$. The most naive definition of $\mathbb{A}^1 \setminus \{ 0 \}$ would then be that its functor of points sends $R$ to $R \setminus \{ 0 \}$, and as Alex says in the comments this fails to even be a functor because morphisms $f : R \to S$ can send nonzero elements to zero elements. Now one can try to rescue the definition by asking: what can we do that's like taking "nonzero elements" but is guaranteed to be functorial?
The answer is the following: an element $r \in R$ is guaranteed to remain nonzero after applying every nonzero morphism $f : R \to S$ iff it avoids every proper ideal iff it's a unit. So we define $\mathbb{A}^1 \setminus \{ 0 \}$ to be the functor sending $R$ to $R^{\times}$ and now this is even representable by affine scheme, and so a Zariski sheaf etc. and everyone is happy. But I want to emphasize that from the functor-of-points POV it's not entirely obvious that this operation should be called "deleting a point"!
Similarly we can reproduce this discussion for $\mathbb{A}^2$ and ask ourselves what "$\mathbb{A}^2 \setminus \{ 0 \}$" ought to mean. Again $R \mapsto R^2 \setminus \{ 0 \}$ is not a functor and again we can ask: which pairs of elements $(r_1, r_2) \in R^2$ are guaranteed to remain nonzero after applying every nonzero morphism? The answer is precisely the pairs which are not both contained in any proper ideal, or equivalently the pairs which generate the unit ideal. So this is how we define $\mathbb{A}^2 \setminus \{ 0 \}$, and more generally $\mathbb{A}^n \setminus \{ 0 \}$.

Now, what can we say about projective spaces from here?
We'd like the presheaf quotient of $\mathbb{A}^2 \setminus \{ 0 \}$ by scaling to have something to do with $\mathbb{P}^1$. This quotient identifies two pairs $(r_1, r_2) \sim (r_1', r_2')$ iff they generate the same $R$-submodule of $R^2$ and so can be thought of as describing lines in $R^2$; more formally you can show that the maps $R \to R^2$ arising in this way are precisely the inclusions of direct summands.
This is a perfectly fine presheaf, but it's no longer a (Zariski, for example) sheaf. It classifies trivial rank $1$ direct summands of the rank $2$ trivial vector bundle, and in general it's possible to glue these together over a Zariski cover to obtain a nontrivial line bundle. So we can sheafify, and I think it should be true that the Zariski sheafification is already the correct $\mathbb{P}^1$ (classifying rank $1$ direct summands, not necessarily trivial).
