The projective functor $\mathbb{P}^n: \operatorname{CRing} \to \operatorname{Set}$ is not representable: categorical argument

Question

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?

Answer 1

$\def\PP{\mathbb{P}}\def\AA{\mathbb{A}}\def\GG{\mathbb{G}}\def\Spec{\mathrm{Spec}}$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.

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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.