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How far can one go in proving facts about projective space using just its universal property?

Can one prove Serre's theorem on generation by global sections, calculate cohomology, classify all invertible line bundles on projective space?

I don't like many proofs of some basic technical facts very aesthetic because one has to consider homogeneous prime ideals, homogeneous localizations, etc. Do there exist nice clean conceptual proofs which avoid the above unpleasantries?

If you include references in your answer it would be very helpful, thanks.

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Hmmmm... abstract nonsense can do quite a few things, but I doubt it can go that far. Still modding the question up since I'd gladly be proven wrong :-) –  Julien Puydt Oct 18 '11 at 6:04
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2 Answers

up vote 22 down vote accepted

I think the answer is "probably not." The reason is that projective space has two universal properties which are used to prove different kinds of things about it. One of these is the slick universal property you like, and the other is the clunky one which results in unpleasantries.

Though each universal property implies the other (since it uniquely identifies projective space), it seems unlikely to me that you can effectively do anything if you try to avoid one of them altogether.

One universal property makes it easy to understand maps to projective space:

$$ Hom(T,\mathbb P^n) = \{\mathcal O_T^{n+1}\twoheadrightarrow \mathcal L| \mathcal L\text{ a line bundle}\} $$

Without bending over backwards (i.e.~reproducing the usual theory), I'd be surprised if you could use this universal property to even prove that there are no non-constant regular functions on $\mathbb P^n$.

I expect constructions that naturally pull back along morphisms (e.g. line bundles, regular functions) to behave like morphisms from projective space, so it would be strange if you could attack such constructions with this universal property.

Another universal property makes it easy to understand maps from projective space: $Hom(\mathbb P^n,T)$ is the equalizer of the two restriction maps $Hom(\coprod_{i=0}^n \mathbb A^n,T)\rightrightarrows Hom(\coprod_{i,j}\mathbb A^{n-1}\times (\mathbb A-0),T)$.

I guess this is the one that you don't like, but we're lucky to have it since it actually makes it possible to make sense of projective space having Zariski local properties (e.g. being smooth, $n$-dimensional, etc.), and thereby makes it possible to do geometry on it.

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+1, very concise answer. –  Martin Brandenburg Oct 18 '11 at 9:01
my personal taste is to view the first universal property you mention as defining projective space. After all I think of it as a moduli space of lines in affine space. But once you've defined it you'd like to prove (for example) that it is a scheme! For this, you need to construct an open atlas. The first one which comes to mind shows that projective space is a coequalizer of the spaces Anton uses in the second universal property. But I still hate the Proj construction... –  Yosemite Sam Oct 18 '11 at 23:25
@konb: +1 - I also hate Proj construction :) –  Mikhail Gudim Oct 22 '11 at 3:43
It depends what you mean by "the Proj construction". It also has a universal property (which is nice-ish to work with) and an open cover which shows it's a scheme (which is not so nice to work with). –  Anton Geraschenko Nov 3 '11 at 3:40
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I agree with Anton that it would be too much to hope for to get serious results (e.g. cohomology of line bundles) from the "nice" universal property of projective space, but one can indeed prove that there are no non-constant regular functions on $\mathbb{P}^n$ using only the universal property.

Namely, it suffices to check that $\mathbb{P}^n$ is proper and connected. For properness, one may use the valuative criterion. Namely, let $R$ be a valuation ring and $K$ its fraction field. Then a map from $\operatorname{Spec}(K)\to \mathbb{P}^n$ is a surjection $K^{n+1}\to K$; then the image of $R^n\hookrightarrow K^n\to K$, where the inclusion is the obvious one, is isomorphic to $R$. In particular, the map $K^{n+1}\to K$ lifts to a surjective map $R^{n+1}\to R$, which is a map $\operatorname{Spec}(R)\to \mathbb{P}^n$ fullfilling the valuative criterion (its uniqueness up to automorphisms of the given diagram is also clear).

As for connectedness, it suffices to check that $\mathbb{P}^n$ is path-connected in the following sense--for any two geometric points, represented by surjections $x_0: \bar k^{n+1}\to\bar k, x_1: \bar k^{n+1}\to\bar k$, there is a ``path" connecting them; namely a map $f: \mathbb{A}^1\to \mathbb{P}^n$ with $f(0)=x_0, f(1)=x_1$. This is a surjection $\bar k[t]^{n+1}\to \bar k[t]$ such that reducing mod $(t), (t-1)$ gives the desired maps. Translating, we must choose polynomials $f_1, ..., f_{n+1}$ such that $f_i(0)=x_0(e_i), f_i(1)=x_1(e_i)$, where $e_i$ are the standard basis of $\bar k^{n+1}$, and where all the $f_i$ do not vanish simultaneously.

But one can do this by Lagrange interpolation; choose any $f_1$ with the desired values at $0,1$, then any $f_2$ with the desired values at $0, 1$ and not vanishing at the other zeros of $f_1$, then any $f_3$ analogously, and so on.

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I like it, but I'm skeptical. To prove that proper connected things have no global sections, don't you use Chow's lemma to reduce to the case of projective space? –  Anton Geraschenko Oct 19 '11 at 1:13
Now that I think about it, you're right, that's essentially the argument I know. Furthermore, I think I'd actually need that $\mathbb{P}^n$ was reduced (which I think can be done from the universal property but I haven't worked out). In any case, one can at least check properness and connectedness :). –  Daniel Litt Oct 19 '11 at 1:28
@Anton: For a reduced algebraic variety over a field one can show by hand that its global sections are a finite dimensional vector space, Liu does it this way in his book. Then the fact that P^n is geometrically integral shows that its global sections are a one-dimensional. –  Lennart Galinat Oct 19 '11 at 4:58
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