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Let $S$ and $T$ be graded rings, finitely generated in degree 1 over degree 0. Suppose furthermore that $S_0$ and $T_0$ are isomorphic (call it $R$). I want to determine which kinds of ideal on $S \otimes_R T$ correspond to closed subschemes of $Proj(S) \times_{Spec(R)} Proj(T)$, and how to determine if two ideals give the same closed subscheme.

I know this should be possible, since there is the Segre embedding $\mathbb{P}^m_R \times_R \mathbb{P}^n_R \rightarrow \mathbb{P}^N_R$, so any closed subscheme $Z \rightarrow Proj(S) \times_{Spec(R)} Proj(T)$ is also a closed subscheme of $\mathbb{P}^N_R$.

If $F$ is a quasicoherent sheaf of ideals on $Proj(S) \times_{Spec(R)} Proj(T)$, then on $U = D(s) \times D(t)$ (where $s \in S, t \in T$ are homogeneous of degree 1), $F(U)$ is an ideal of $S_{s : 0} \otimes_R T_{t : 0}$, and this ideal must agree on overlaps. Thus, I think that the ideals of $S \otimes_R T$ corresponding to closed subschemes of $Proj(S) \times_{Spec(R)} Proj(T)$ should be the "doubly homogeneous ideals" : if $x = \sum_{i, j} s_i \otimes t_j \in I$, where each $s_i \in S, t_j \in T$ are homogeneous of degrees $i$ and $j$, then $s_i \otimes t_i \in I$. But how do I prove this?

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Let me assume that either $\mathrm{Proj}(S)$ or $\mathrm{Proj}(T)$ is flat over $R$ (more generally, it is enough to assume that they are $\mathrm{Tor}$-independent). Then the Künneth formula (applied to the Segre embedding) implies that $$ \mathrm{Proj}(S) \times_{\mathrm{Spec}(R)} \mathrm{Proj}(T) \cong \mathrm{Proj} \left( \bigoplus_{k=0}^\infty S_k \otimes_{R} T_k \right). $$ The graded algebra in the right side is generated by its first component, so you can associate closed subschemes to ideals in it in the usual way.

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  • $\begingroup$ What version of the Kunneth formula are you using? I googled and got some results about cohomology of product spaces, which I'm not sure how to apply here. Do you have a link or reference to the precise statement of the version of Kunneth formula? $\endgroup$
    – David Lui
    Commented Dec 18, 2023 at 13:43
  • $\begingroup$ See, e.g., Section 3.10 in [Joseph Lipman, Notes on derived functors and Grothendieck duality, in: Foundations of Grothendieck Duality for Diagrams of Schemes, in: Lecture Notes in Math., vol. 1960, Springer, Berlin, 2009, pp. 1–259]. $\endgroup$
    – Sasha
    Commented Dec 18, 2023 at 14:25

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