What the title says.

Let $S$ be a graded $k$-algebra, generated in degree $1$, and the same for $R$. Then $S \otimes_k R$ is graded as well, with $k$'th graded piece $\bigoplus_{j+l=k} S_j \otimes_k R_l$. Then $S \otimes_k R$ is generated in degree $1$ as well, and we can form $\mathrm{Proj} \left( S \otimes_k R \right)$.

Here's the concrete case I'm considering: let $S=k[x_1,\ldots,x_n]/I$ for some homogeneous ideal $I$, and same for $R$. I then want to know, what is $\mathrm{Proj}(S \otimes_k R)$ geometrically?

Even more concrete: Let $R=S = k[x_0,x_1,x_2]$ and $I=\langle (y^2=g(x)\rangle$ where $g(x)$ is of degree $2$. Then $E=\mathrm{Proj}(R/I),E'=\mathrm{Proj}(S/I)$ are elliptic curves, and $\mathrm{Proj}(R/I \otimes_k S/I)$ is some (singular) threefold. What can I say about this threefold? What are its cohomology groups for examle?


It might help to remember that in these cases, $\mathrm{Spec}(R)$ is the affine cone over $\mathrm{Proj}(R)$. So $\mathrm{Spec}(R\otimes S)$ is the product of the affine cones over $\mathrm{Proj}(R)$ and $\mathrm{Proj}(S)$, which is again a cone. Then we take $\mathrm{Proj}$ of it.

In terms of the ambient affine and projective spaces, this is like going from $\mathbb{P}(V)$ and $\mathbb{P}(W)$ to $\mathbb{P}(V \oplus W)$.

I agree that the picture is a little weird. For example, $\mathrm{Proj}(R\otimes S)$ contains disjoint copies of $\mathrm{Proj}(R)$ and $\mathrm{Proj}(S)$ (which are induced by the disjoint inclusions $\mathbb{P}(V),\mathbb{P}(W) \hookrightarrow \mathbb{P}(V \oplus W)$ above).

What we get n your example is the following: in $\mathbb{P}^5$, take two disjoint copies of $\mathbb{P}^2$, say the planes $X_1 = (*:*:*:0:0:0)$ and $X_2=(0:0:0:*:*:*)$, and embed $E_1$ in $X_1$ and $E_2$ in $X_2$. Then

$$\mathrm{Proj}(R\otimes S) = \bigcup_{(p_1,p_2) \in E_1 \times E_2} \overline{p_1p_2},$$

the union of all lines from points on $E_1$ to points on $E_2$. Note that this is indeed 3-dimensional, and singular along both $E_i$'s. I don't know what its cohomology is, but maybe this description can lead to it.

In general, that's what you get -- maybe this could be called "the suspension" of the two projective varieties we started with, the union of all lines connecting them in an appropriately-sized projective space: namely, the projective space whose homogeneous coordinates are the union of the homogeneous coordinates for $\mathrm{Proj}(R)$ and $\mathrm{Proj}(S)$.

EDIT: Thanks Asal, this is called the join of the varieties, not the suspension.

It's a little more natural to think of $R \otimes S$ as bigraded and to take the (bi-)$\mathrm{Proj}$ of it, in which case we actually do get the product $\mathrm{Proj}(R) \times \mathrm{Proj}(S)$ inside the corresponding product of projective spaces, which in this case would be a smooth surface in $\mathbb{P}^2 \times \mathbb{P}^2$.

  • $\begingroup$ I think the construction you describe is called the join of the two varieties (at least in Harris' book, which I don't have with me at the moment). $\endgroup$ – user64687 Oct 23 '14 at 11:03
  • $\begingroup$ @AsalBeagDubh Thanks -- edited that in. $\endgroup$ – Jake Levinson Oct 23 '14 at 13:19

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