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I know that there are isomorphic projective varieties which have nonisomorphic coordinate rings, but I'm a little mystified as to "why" this is the case. Why doesn't a usual functoriality proof go through to prove this, and is there any insight into this other than that the definitions just work out this way?

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consider scaling the grading for a graded ring, and taking the proj construction. Details can be found in 7.4 of Ravi Vakil's notes: math.stanford.edu/~vakil/216blog/FOAGmar1011public.pdf –  BBischof Mar 15 '11 at 5:36
Wow, I am surprised to learn this as well. I always thought of the coordinate ring and the variety as "two sides of the same coin", so to speak. –  Alex Becker Mar 15 '11 at 6:21
@Alex: this is true for affine varieties, but for projective varieties to get something completely invariant one really has to work with the scheme. –  Qiaochu Yuan Mar 15 '11 at 10:06
@Qiaochu: ah, thank you for the clarification. I've never worked with schemes before, but I hope to study them soon! –  Alex Becker Mar 16 '11 at 0:24
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3 Answers

up vote 11 down vote accepted

The homogeneous coordinate ring of a projective variety embedded in $\mathbb{P}^n$ is not the "ring of functions" on the variety but actually the direct sum of sections of symmetric powers of some line bundle. The point is that this construction actually depends on the line bundle, and the same variety can have many line bundles which give rise to non-isomorphic rings.

I'm not sure what "a usual functoriality proof" is, but you should find either that you end up proving something about the category of projective varieties together with an embedding or that you can't define the homogeneous coordinate ring at all without a choice of embedding.

I'm not totally comfortable with this answer since I can't claim to have much experience with line bundles, so here's a more straightforward version. First, the reason this issue doesn't come up for affine varieties is that once you have a definition of a morphism of affine varieties, the ring of functions $k[V]$ on an affine variety $V$ really is just the set of morphisms $V \to \mathbb{A}^1(k)$, and this definition is clearly functorial.

The same cannot be said for the homogeneous coordinate ring of a projective variety $V \subset \mathbb{P}^n$. First of all, its elements are not functions on the variety. Quotients of two elements of the same degree do define morphisms rational maps $V \dashrightarrow \mathbb{P}^1$, but you're considering much more than just such quotients. When $V$ is irreducible you might be tempted to replace the ring of functions with the field $k(V)$ of rational functions, but $k(V)$ does not completely capture $V$ (for example it ignores the removal of finitely many points).

After you think about this problem for awhile you should realize that the problem here is that the homogeneous coordinate ring really is defined in terms of the embedding, and so there's no reason to expect it to be independent of the choice of embedding. So next you might look for a definition of projective variety that is independent of the choice of embedding, and then sooner or later you have to start studying schemes.

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Dear Qiaochu, A quotient of two elements of the same degree does not necessarily define a map to $\mathbb P^1$. If $F$ and $G$ are homogeneous polys. of degree $d$, cutting out divisors $D_1$ and $D_2$ on $X$, then the map to $\mathbb P^1$ induced by $F/G$ will not necessarily be well-defined at the intersection of $D_1$ and $D_2$ (which will be non-empty once $X$ has dimension $> 1$). Thus we get a rational map from $X$ to $\mathbb P^1$, but to make it an honest morphism you will need to blow up $X$ along $D_1\cap D_2$. (As an example of this, you are probably already familiar with the ... –  Matt E Mar 16 '11 at 2:45
... fact that there are no non-trivial morphism $\mathbb P^n \to \mathbb P^1$ if $n > 1$. More generally, there are no morphisms $\mathbb P^n \to \mathbb P^m$ if $m < n$.) Regards, –  Matt E Mar 16 '11 at 2:45
Right, right. I've been spending too much time working with smooth curves... –  Qiaochu Yuan Mar 16 '11 at 12:47
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Just to elaborate slightly on Qiaochu's answer:

If $X \subset \mathbb P^n$ is a projective variety, then the given embedding of $X$ in $\mathbb P^n$ equips $X$ with extra structure: a line bundle $\mathcal L$ (the restriction of $\mathcal O(1)$), and sections $x_0,\ldots,x_n \in \Gamma( X,\mathcal L)$ (the restriction of the sections $x_i$ of $\mathcal O(1)$).

Conversely, given an abstract variety $X$, equipped with a line bundle $\mathcal L$ and sections $x_0,\ldots,x_n$ of $\Gamma(X,\mathcal L)$ satisfying appropriate axioms (see e.g. Chapter II, Section 7 of Hartshorne), one obtains a projective embedding.

The homogeneous coordinate ring of $X$ depends functorially on the data $\bigr(X,\mathcal L, (x_i)_{i= 0,\ldots,n}\bigr).$ But of course, this is more data than just $X$ alone.

As a concrete example, think about the difference between the homogeneous coordiante ring of $\mathbb P^1$ embedded via the identity map into itself, and embedded as a smooth conic in $\mathbb P^2$.

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Here's a slightly tangential elaboration on the above two answers.

Let $X = \mathrm{Proj} A$ for $A$ a noetherian ring. So here $A$ should fade into the background, but $X$ is the important thing; it turns out that the $\mathrm{Proj}$ depends only on the asymptotic part of $A$ in high dimensions. Then we can consider the category of coherent sheaves on $\mathrm{Proj} A$. The sheaf of regular functions is the canonical example.

There is a functor from finitely graded $A$-modules to coherent sheaves on $\mathrm{Proj} A$, denoted by a tilde. (This is as in Hartshorne II.5.) Unlike the case for an affine scheme, however, it is not an equivalence of categories. It is not too hard to show that the functor is essentially surjective, since given a coherent sheaf $\mathcal{F}$ on $\mathrm{Proj} A$ you can construct the graded $A$-module $\bigoplus \Gamma(X, \mathcal{F}(n))$ of the global sections of all the twists. Then one can show using elementary methods that this recovers $\mathcal{F}$. (In fact, this is true even if $A$ is not noetherian and $\mathcal{F}$ is simply assumed quasi-coherent.)

However, if we have a graded $A$-module $M$, we cannot recover $M$ from the associated sheaf $\tilde{M}$. For instance, if $M$ has finitely many nonzero components, then the homogeneous localizations $M_{(f)}$ vanish (for any homogeneous element $f$), so the associated sheaf is zero. But $M$ may not be zero.

This suggests that if we consider the abelian category of finitely generated graded $A$-modules, and quotient out by the Serre class of modules that are "asymptotically zero," then we will get the category of coherent sheaves on $\mathrm{Proj} A$. This is in fact true, though the proof uses a bit of cohomological machinery (see EGA III.2).

So the moral of the story is that the $\mathrm{Proj}$ sees only asymptotic data. If we have a morphism $A \to A'$ that induces an isomorphism in large enough degrees, then $\mathrm{Proj} A, \mathrm{Proj} A'$ will be isomorphic (as one example).

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