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Let $S$ be a graded ring with $A := S_0$. Set $X := \textrm{Proj} (S)$. Then the projective coordinate ring of $X \times_A X$ is the graded ring $ \bigoplus_{n \geq 0} S_n \otimes_A S_n $, cf Hartshorne, Exercise II.5.11. (This matches with geometric intuition because this ring corresponds [at least when $S$ is nice enough, eg if $X$ is projective over $A$ and $S$ is the homogeneous coordinate ring of an embedding into $\mathbf P^m_A$] to the section ring of a very ample sheaf on $X \times_A X$ coming from the exterior product of very ample sheaves on $X$.)

However, there is another natural graded ring that we can construct here, namely $S \otimes_A S$ with the obvious grading coming from total degree. This ring is clearly not the same as the graded ring in the first paragraph above. Naively, though, by analogy to the affine case one might still expect that $\textrm{Proj}(S \otimes_A S) \cong X \times_A X$. Is this true? Probably this is not true, although I don't know how to show it; so a followup question is: what scheme does this produce?

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For one thing, the dimension is wrong. And, have you tried computing an example? –  Angelo Apr 6 '13 at 18:17
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It's helpful to remember that a graded ring S gives not only a projective scheme but also an embedding (via S(1)), and lots of rings can give the same projective scheme. If S is the ring you cite from Hartshorne, then S(1) gives the Segre embedding of your projective variety. I'm also interested in knowing if there are other easy rings to write down whose Proj gives you X x X. In the case when S is the polynomial ring k[x,y], then the total tensor product just gives you k[x,y,z,w], so you just get P^3 which isn't what we want. –  David Dynerman Apr 6 '13 at 18:24
    
Ah yes, that makes sense. So I guess this boils down to the obvious fact that a product of affine cones over a projective scheme will never be an affine cone over the product (eg by dimension considerations), which immediately shows that the naive hypothesis is false. –  Chuck Hague Apr 6 '13 at 18:36
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There's a construction called "multiproj" for rings with several $\mathbb N$ gradings. If you think of $S\otimes S$ as doubly graded and take multiproj, then you'll get the product. You can pass from a multiproj to a project by restricting to a generic ray (if you're lucky), which is what you did. –  Ben Webster Apr 6 '13 at 18:51
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It is maybe worth mentioning that for projective varieties, the full tensor product corresponds to the JOIN of X with another copy of itself after appropriate embeddings. Eisenbud's "Commutative Algebra: With a View Toward Algebraic Geometry" has a helpful discussion on page 304: books.google.ie/… –  Oliver Nash Apr 10 '13 at 18:16
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1 Answer

up vote 11 down vote accepted

A ($\mathbb{Z}$-)grading on a ring $S$ is equivalent to an action of the group $\mathbb{G}_m$ on $U=\operatorname{Spec} S$. This is an exercise in algebra; the statement is that the symmetric monoidal category of comodules over the Hopf algebra $A[t,t^{-1}]$ is equivalent to the category of graded $A$-modules. The construction $\operatorname{Proj} S$ then corresponds to taking the quotient of $U\setminus U_0$ by the $\mathbb{G}_m$ action, where $U_0$ is the subscheme corresponding to the "irrelevant ideal". This is really just the familiar fact that $\mathbb{P}^n$ is the quotient of $\mathbb{A}^{n+1}\setminus 0$ by rescaling.

What happens when we take a product of two such schemes $U$ and $V$ with corresponding graded rings $S$ and $T$? I'm going to ignore $U_0$ and $V_0$ in this discussion to simplify notation (i.e., I'm going to pretend that $\operatorname{Proj} S$ is just $U/\mathbb{G_m}$). Well, $U\times V$ will naturally have an action of $\mathbb{G}_m\times \mathbb{G}_m$; this corresponds to the natural bigrading on $S\otimes T$. By taking the diagonal $\mathbb{G}_m\to\mathbb{G}_m\times\mathbb{G}_m$, we get an action of $\mathbb{G}_m$ on the product, and this corresponds to the "total" grading on $S\otimes T$. But if we mod out $U\times V$ by the diagonal action of $\mathbb{G}_m$, we don't get $U/\mathbb{G}_m\times V/\mathbb{G}_m$; to get that, we would need to mod out the entire action of $\mathbb{G}_m\times \mathbb{G}_m$. Alternatively, instead of modding out the whole action of $\mathbb{G}_m\times\mathbb{G}_m$ at once, we could first mod out the action of the antidiagonal $\alpha:\mathbb{G}_m\to\mathbb{G}_m\times\mathbb{G}_m$ given by $\alpha(t)=(t,t^{-1})$ and then mod out the action of the diagonal; this works since the antidiagonal and the diagonal together generate the whole group. What do we get when we mod out the antidiagonal? Well, it turns out that in this case the quotient will still be affine, and its coordinate ring can be found by taking invariants of the action on $S\otimes T$. If $x\in S_n$ and $y\in T_m$, then the the antidiagonal action of $t\in\mathbb{G}_m$ sends $x\otimes y$ to $t^{n-m}x\otimes y$, so the invariants are exactly $\bigoplus S_n\otimes T_n$. Thus the product $U/\mathbb{G}_m\times V/\mathbb{G}_m$ can be obtained as $\operatorname{Proj}(\bigoplus S_n\otimes T_n)$.

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Very nice answer -- thanks! –  Chuck Hague Apr 6 '13 at 19:57
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