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Using a "geometrical" argument of dimension, like the one here, one can show that the projective space is not affine.

I am interested in showing that, but using a categorical argument, i.e. I want to show that $\mathbb{P}^n:\operatorname{CRing} \to \operatorname{Set}$ which sends a ring $R$ to the set of equivalence classes $\mathbb{P}^n(R):= R^{n+1}/R^{\times}$ is not representable.

Similar to the example of the $\operatorname{Nil}$ functor, this could be done either by showing that its category of elements has no initial object, or by showing that it does not preserve limits.

Any ideas what could work?

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    $\begingroup$ As you have defined it the functor need not preserve pullback squares of the form $(R,R[a^{-1}],R[(1-a)^{-1}],R[(a(1-a))^{-1}])$, and so is not representable. You can see this by finding an example where there is a non-free projective submodule $L<R^{n+1}$ of rank one, such that $L[a^{-1}]$ and $L[(1-a)^{-1}]$ are free over $R[a^{-1}]$ and $R[(1-a)^{-1}]$. But this just shows that your definition is wrong: $\mathbb{P}^n(R)$ should be defined as the set of rank-one projective submodules of $R^{n+1}$. I'm not sure what's the simplest proof that that is not representable. $\endgroup$ – Neil Strickland Aug 8 at 9:42
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    $\begingroup$ The functor which is usually called $\mathbb{P}^n$ (corresponding to projective $n$-space as a variety) is not $R^{n+1}/R^{\times}$. The right statement is that $\mathbb{P}^{n+1}(R)$ is rank one direct summands of $R^{n+1}$, see math.stackexchange.com/questions/121105 . Your functor is both too large and too small -- Given $(r_0, r_1, \ldots, r_{n+1})$ in $R^{n+1}$, we can form the rank one module it spans, but it is neither true that this need be a summand, nor that all rank one summands are of this form. $\endgroup$ – David E Speyer Aug 8 at 13:07
  • $\begingroup$ I now see that I wrote the same thing as Neil Strickland, sorry. $\endgroup$ – David E Speyer Aug 8 at 13:09
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$\def\PP{\mathbb{P}}\def\AA{\mathbb{A}}\def\GG{\mathbb{G}}\def\Spec{\mathrm{Spec}}$This is probably going to sound too classical to satisfy, but it seems straightforward to me. Let $\PP^{n}_{charts}$ be the functor represented by the scheme which is normally called projective $n$-space. In other words, $\PP^{n}_{charts}$ is the co-equalizer of a certain diagram $(\AA^{n-1} \times \GG_m)^{\binom{n+1}{2}} \rightrightarrows (\AA^n)^{n+1}$.

As discussed in comments, the correct definition of $\PP^n(R)$ is that $\PP^n(R)$ is the set of rank one direct summands of $R^{n+1}$ (see here). Grothendieck preferred to dualize and work with rank one projective quotients of $R^{n+1}$. I'm not sure if there is a deep reason which this is better; from a shallow perspective, it seem to me to introduce unnecessary duals in the notation. I'll work with the summand version.

For $0 \leq j \leq n$, let $X_j$ be the submodule $(r_0, r_1, \ldots, r_{j-1}, 0 , r_{j+1}, \ldots, r_n)$ of $R^{n+1}$. Let $U_n$ be the subfunctor of $\PP^n$ where $U_j(R) = \{ L \subset R^{n+1} : L + X_j = R^{n+1} \}$. Every submodule in $U_n(R)$ is uniquely of the form $R(u_0, u_1, \ldots, u_{j-1}, 1, u_{j+1}, \ldots, u_n)$; the coordinates $(u_0, \ldots, u_{j-1}, u_{j+1}, \ldots, u_n)$ give an isomorphism $U_j \cong \AA^n$. The overlap $U_i \cap U_j$ is the chart $u_i \in R^{\times}$ in $U_j$, so $U_i \cap U_j \cong \AA^{n-1} \times \GG_m$, and the gluing is precisely the classical chart formula. So, by the universal property of co-equalizers, we get a map $\PP^{n}_{charts} \to \PP^{n}$.

It shouldn't be hard to show that this is an isomorphism, but we don't need to work that hard to show that $\PP^{n}$ isn't affine. For a field $k$, the map $\PP^{n}_{charts}(k) \to \PP^{n}(k)$ is definitely a bijection. If $\PP^{n}$ were affine, then global functions on $\PP^{n}$ would separate $k$-points, so such functions pulled back to $\PP^{n}_{charts}$ would separate $k$-points. But global functions on $\PP^{n}_{charts}$ don't separate $k$-points, a contradiction.


Or, briefer but less intuitively: The module $k[t] (1,t) \subset k[t]^2$ corresponds to a map $\Spec\ k[t] \to \PP^1$, so if $\PP^1$ were $\Spec\ S$, there would be a corresponding map $S \to k[t]$. The module $k[t^{-1}] (t^{-1},1) \subset k[t^{-1}]^2$ would similarly give a map $S \to k[t^{-1}]$. The inclusions of $k[t]$, $k[t^{-1}]$ into $k[t, t^{-1}]$ give us the same module, $k[t,t^{-1}] (1,t) = k[t,t^{-1}](t^{-1},1)$ inside $k[t,t^{-1}]^2$, so the maps $S \to k[t]$ and $S \to k[t^{-1}]$ must become the same after further composing to $k[t,t^{-1}]$. So the image of $S$ would have to lie in $k[t] \cap k[t^{-1}] = k$. But then all the different maps $k[t] \to k$ would have to give to the same $k$-submodule of $k^2$, and they don't.

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    $\begingroup$ The reason why Grothendieck works with quotients is that this gives a representable functor for any module, not just finitely generated projective modules. $\endgroup$ – Angelo Aug 11 at 11:38

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