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

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?

• 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. – Neil Strickland Aug 8 at 9:42
• 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. – David E Speyer Aug 8 at 13:07
• I now see that I wrote the same thing as Neil Strickland, sorry. – David E Speyer Aug 8 at 13:09

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