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?



Why does the naive choice of homogeneous coordinate ring of a product of projective schemes not work?
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 $\operatorname{Proj}(\bigoplus S\_n\otimes T\_n)$.


