Generalize the Proj construction?

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I'm wondering if there is a generalization of the Proj construction used in algebraic geometry. Given a graded ring R, which is a monoid homomorphism $R\to \mathbb{N}$, we can form the scheme Proj(R), which is the union of Spec$(R_f)_0$.

I'm wondering if there is a way to extend this construction to general homomorphisms from a ring to a (sharp) monoid.

In particular, I want to see if there is a Proj'' construction which gives the product $\mathbb{P}^m\times\mathbb{P}^n$ from the bidegree $k[x_0,\ldots,x_m,y_0,\ldots,y_n]\to\mathbb{N}^2$.

(I know this can be seen via two consecutive Proj, but want to see if this generalizes, as I want to see what happens if the degree map take value in a general toric monoid.)

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What does "sharp" mean? That is has an identity and embeds into its Grothendieck group? Googling produces many instances of the phrase "sharp monoid", but not a definition. – Anton Geraschenko♦ Dec 3 2010 at 7:39

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Sharp means no (non-trivial) units. – 36min Dec 3 2010 at 7:43

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This is called "multi-proj" for submonoids of Z^n and is annoyingly hard to google due to google thinking you really meant "multi-project". Look in Miller and Sturmfels. – Ben Webster♦ Dec 3 2010 at 8:31
@Ben: I found the book at books.google.com/…, but can you give a more detail reference on sections, I can't search Multi-Proj in it either. – Yuhao Huang Dec 3 2010 at 8:40
By the way, the only varieties you'll ever get are the usual projs of taking the graded pieces for multiples of a single generic vector in your monoid (maybe after saturation). You might also want to look into geometric invariant theory. – Ben Webster♦ Dec 3 2010 at 8:45
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2 Answers

My naive guess from the examples is the following:

Spec$k[P^{gp}]$ (which is just a product of $\mathbb{G}_m$'s) (somehow) acts on Spec$R$, and a GIT-quotient gives the construction you need, because the Proj construction is just a GIT-quotient of Spec$R$ by $\mathbb{G}_m$ w.r.t a certain linearization.

I'm not sure if the above construction generalizes directly, maybe some extra data is necessary.

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The monoid and the sharpness condition feel misplaced to me. After all, there's no reason a ring must be positively graded in order to take Proj of it.

A grading on $R$ is "really" a $\mathbb G_m$-action on $R$, which happens to be the same as a monoid homomorphism from $R$ to the character group of $\mathbb G_m$. $Proj(R)$ is obtained by cutting the $\mathbb G_m$-fixed locus out of $Spec(R)$ and quotienting what's left by the action of $\mathbb G_m$. So it feels like the more natural generalization to shoot for is "Proj" of a ring with group action.

But maybe not. I think a productive case to think about is $R=k[x,xy,x^2y]$ with the bi-grading inherited from $k[x,y]$. It seems to me unlikely that any reasonable generalized $Proj(R)$ would depend on which specific monoid you take, whether its $\mathbb N\times \mathbb N$, or something bigger, or something smaller. It may as well be $\mathbb Z\times \mathbb Z$.

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Anton- You should know better than this.  By your recipe, what is the Proj of k[x,y] with x of weight 1 and y of weight -1?  This is one of the standard examples for why geometric invariant theory is a good reason and it involves (surprise, surprise) picking a positive direction.  The monoid you use for your proj should be a submonoid of the ample cone on the result;  it doesn't sound like you really want to have lots of units.  Such line bundles can only see the affinization of your variety. – Ben Webster♦ Dec 3 2010 at 8:42
@Ben: I don't understand your objection. The recipe produces something over Spec of the zero-graded (invariant) subring. In your example, you get the non-separated line over the line Spec(k[xy]). It's weird, but what's so unreasonable about it? You're right that cutting out the fixed locus is probably the wrong thing to do, since you could artificially change this locus--the right thing is to remove points with large stabilizers. But once you do that, this recipe is basically the quotient of the pre-stable locus. It seems like I don't have to pick line bundle to construct things like ℙ^n. – Anton Geraschenko♦ Dec 3 2010 at 15:25
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