Timeline for In char zero $ \operatorname{Cox}(\operatorname{Bl}_{[1:1:1]}(\mathbb{P}(a:b:c))) $ is finitely generated, but not in char p. How?
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| Sep 5, 2020 at 16:58 | comment | added | Ennio Mori cone | Just because the sheaves are flat over $R$ does not mean that the Cox ring of the generic fiber is finitely generated if this is true over the special fiber. I guess the cohomology can jump over the special fiber, and the new sections make the ring finitely generated. | |
| Sep 5, 2020 at 10:05 | comment | added | Piotr Achinger | ...The point being that the multiplication maps $R_{n_1}\otimes \ldots \otimes R_{n_k} \to R_{n_1 + \cdots n_k}$ do not have constant rank. Anyway you really want the other implication. | |
| Sep 5, 2020 at 9:58 | comment | added | Piotr Achinger | I don't know if this resolves the issue, but consider the divided power algebra over $\mathbb{Z}$, i.e. the subring $R$ of $\mathbb{Q}[x]$ generated by $x^n / n!$ for all $n\geq 1$. Then $R \otimes \mathbb{Q} = \mathbb{Q}[x]$ is finitely generated, but the $R\otimes \mathbb{F}_p$ are not. However, $R$ is a free $\mathbb{Z}$-module, in fact it is graded with each grading isomorphic to $\mathbb{Z}$. So the "if and only if" claim in your last paragraph is incorrect. | |
| Sep 5, 2020 at 9:07 | history | asked | schemer | CC BY-SA 4.0 |