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Let $ X(a_{1}:a_{2}:a_{3}) $ be the blow-up of $ \mathbb{P}(a_{1}:a_{2}:a_{3}) $ at $ [1:1:1] $, the identity of the torus. In Steven Dale Cutkosky's paper Symbolic Algebras of Monomial Primes Cutkosky gives an example of what appears to be weights $ a_{i}, i \in \{1,2,3\} $ such that $ \operatorname{Cox}(X(a_{1}:a_{2}:a_{3})) $ is not finitely generated in characteristic zero, but is in positive characteristic. This does not make sense to me for the following reason.

Let $ R $ be a ring of mixed characteristic, and let us denote the affine coordinate ring of $ \mathbb{G}_{m}(R) $ by $ R[t]_{t} $. If $ \mathbb{A}^{3}_{R} $ has affine coordinate ring $ R[x_{1},x_{2},x_{3}] $, then the ring homomorphism $ \beta^{\sharp}: R[x_{1},x_{2},x_{3}] \to R[x_{1},x_{2},x_{3}][t] $ which sends $ x_{i} $ to $ x_{i} t^{a_{i}} $ is the co-action for an action $ \beta $ of $ \mathbb{G}_{m}(R) $ on $ \mathbb{A}^{3}_{R} $. The quotient $ (\mathbb{A}^{3}_{R} \setminus (0,0,0))//\mathbb{G}_{m}(R) $ is a scheme $ \mathbb{P}(a_{1}:a_{2}:a_{3})_{R} $ whose fibres are all isomorphic to $ \mathbb{P}(a_{1}:a_{2}:a_{3}) $ over the respective fields. The Cox ring of $ \mathbb{P}(a_{1}:a_{2}:a_{3})_{R} $ is $ R[x_{1},x_{2},x_{3}] $, and is therefore flat over $ R $.

For any $ d \in \mathbb{Z} $ and invertible sheaf $ \mathcal{O}_{\mathbb{P}(a_{1}:a_{2}:a_{3})}(d) $ the number of global sections is purely determined by the weights $ a_{1},a_{2} $ and $ a_{3} $. Since the Picard group of $ \mathbb{P}(a_{1}:a_{2}:a_{3})_{R} $ is isomorphic to $ \mathbb{Z} $, any invertible sheaf $ \mathcal{O}_{\mathbb{P}(a_{1}:a_{2}:a_{3})_{R}}(d) $ is flat over $ R $ for $ d \in \mathbb{Z} $. If $ \mathcal{I} $ is the ideal sheaf of the $ R $-valued point $ [1:1:1] $, then since $ R $ is flat over $ R $, and $ R[x_{1},x_{2},x_{3}] $ is flat over $ R $, the Hilbert polynomial of the ideal sheaf $ \mathcal{I} $ is the same for all points of $ \operatorname{Spec}(R) $. Therefore $ \mathcal{I} $ is flat over $ R $.

Let $ X(a_{1}:a_{2}:a_{3})_{R} $ be the blow-up of $ \mathbb{P}(a_{1}:a_{2}:a_{3}) $ at the $ R $-valued point $ [1:1:1] $. If $ \pi: X(a_{1}:a_{2}:a_{3})_{R} \to \mathbb{P}(a_{1}:a_{2}:a_{3}) $ is the natural projection map, then the effective divisors of the Picard group of $ X(a_{1}:a_{2}:a_{3})_{R} $ are contained within the semi-group generated by divisors of the form $ \pi^{\ast}(\mathcal{O}_{\mathbb{P}(a_{1}:a_{2}:a_{3})_{R}}(d)) \otimes \pi^{\ast}(\mathcal{I}^{m}) $ for $ (d,m) \in \mathbb{N}_{0}^{2} $. Since these sheaves are flat over $ R $, this would mean that the Cox ring of $ X(a_{1}:a_{2}:a_{3})_{R} $ would be flat over $ R $.

However, if the Cox ring of $ X(a_{1}:a_{2}:a_{3})_{R} $ is flat over $ R $, then it is finitely generated over its generic fibre if and only if it is finitely generated over its special fibre. This seems to contradict Cutkosky's result in section 3 of the aforementioned paper. What is wrong with this argument, or is there something I missed regarding Cutkosky's argument?

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    $\begingroup$ 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. $\endgroup$ Commented Sep 5, 2020 at 9:58
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    $\begingroup$ ...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. $\endgroup$ Commented Sep 5, 2020 at 10:05
  • $\begingroup$ 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. $\endgroup$ Commented Sep 5, 2020 at 16:58

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