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Let $X$ be an irreducible smooth projective variety over a field $k$ (algebraically closed and of characteristic zero if needed). Let $U \subseteq X$ an affine open such that $O_X(U)$ is factorial and $O_X(U)^* = k^*$. Let $D_1, \dots D_k$ be the irreducible closed subvareities of $X$ such that $X= U \cup D_1 \cup \dots \cup D_k$. As $U$ is affine it's pure in codimension one, hence each $D_i$ is a divisor, and uder our hypothesis we have that $$Pic(X) \simeq \bigoplus_{i = 1}^k \mathbb{Z} O(D_i).$$ I would like to undertand what can be said on the Cox RingC of $X$ under the above hypothesis. Namely can we say something about the following questions?

  1. Is it true that under the above hypothesis $H^0( X, O (\sum_{i=1}^k n_i D_i) ) \neq 0$ implpies that each $n_i$ is non-negative? Probably this is a too strong statement but is it true if we also assume that $U$ is an affine space, so $U \simeq \mathbb{A}^r$ for a certain $r$?

  2. The following question looks false to me but still I'm not able to produce a countrexample. Is it true that the $k$-algebra $$R:= \bigoplus _{\underline{n} \in \mathbb{Z}^k} H^0(X, O(\sum_{i=1}^k \underline{n}_i D_i) )$$ is generated by $\oplus H^0(X, O(D_i))$?

  3. Is question 2 true if we assume that there is a reductive algebraic group $G$ acting on $X$ such that the $D_i$ are $G$-stable and $U$ contains an open $B$-orbit where $B$ is a Borel subgroup of $G$ ?

Probably these facts are known to specialists so if a reference exists it will be largely enough. Thank you.

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    $\begingroup$ The hypotheses that you write do not imply that the Picard group is generated by the Cartier divisors $D_i$. For instance, let $X$ be a smooth projective curve of genus $g>0$ and let $U$ be the complement of one rational point. You need to assume that the Picard group of $U$ is trivial to get your conclusion. $\endgroup$ Commented Nov 7, 2022 at 12:25
  • $\begingroup$ Yes thanks I forgot to ask $O_X(U)$ factorial, which implies that the Picard group of $U$ is trivial. I edited the question. $\endgroup$ Commented Nov 8, 2022 at 9:40
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    $\begingroup$ Did Prince Albert have a Cox ring, or merely an Albert algebra? $\endgroup$
    – Ben McKay
    Commented Nov 8, 2022 at 11:12

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The answer to all these questions is no.

Let $X=F_1$, the first Hirzebruch surface. Let $D_0$ be a fibre of the projection to ${\mathbb P}^1$. Let $D_{-1}$ be the "negative" section of this projection, and let $D_1$ be a disjoint positive section. Then $(X,D_0,D_1)$ satisfies your conditions, with $U={\mathbb A}^2$ (with trivial Picard group). The torus $G=(k^*)^2$ acts on $X$, fixing (suitably chosen) $D_0$ and $D_1$, and $B=G$ has an open orbit in $U$.

We have $D_0^2=0, D_1^2=1, D_{-1}^2=-1, D_{-1} D_1=0, D_0D_1=D_0D_{-1}=1$ and in the Picard group, $[D_{-1}]=[D_1]-[D_0]$. Inside ${\rm Pic}(X)={\mathbb Z}[D_0] \oplus {\mathbb Z}[D_1]$, the nef cone is generated by $[D_0]$ and $[D_1]$: non-negative linear combinations of $[D_0]$ and $[D_1]$ give all nef classes. But there are effective classes that are not in the nef cone, in particular $[D_{-1}]$ itself. So the answer to 1. is negative. This also implies of course that the answer to 2. is also negative.

Note that we could also have chosen to look at $(X,D_{-1},D_0)$. This choice also satisfies your conditions. In this case, the answer to 1. is positive: the effective cone of $X$ is generated by $[D_{-1}]$ and $[D_0]$. However 2. still fails: sections of $[D_{-1}]$ and $[D_0]$ generate only a non-trivial subring of the whole Cox ring of $X$.

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  • $\begingroup$ Thank you! I will carefoully look at your answer! $\endgroup$ Commented Nov 8, 2022 at 9:52

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