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I'm having trouble reconciling the following claims (everything over $\mathbb{C}$ for simplicity):

  1. [GS II, Remark 2.14] Every toric stack embeds as an open substack of some $[X_0 / G]$ with $X_0$ affine and $G$ a commutative reductive group.
  2. [CLS, Example 4.2.13] There exist positive-dimensional complete toric varieties with no nontrivial line bundles.

More precisely, it seems to me that (1) implies that every complete toric variety with no nontrivial line bundles is a point, by the following argument.

Suppose $X$ is such a toric variety. By (1), we may view $X$ as an open substack of some $[X_0 / G]$. Because $G$ is commutative and reductive, the category $\mathsf{QCoh}(BG) = \mathsf{Rep}(G)$ is generated by line bundles, i.e. every object admits a surjection from a direct sum of line bundles. Thus $\mathsf{QCoh}([X_0 / G])$ is also generated by line bundles, since it's a category of modules over an algebra object of $\mathsf{QCoh}(BG)$. It follows that the category $\mathsf{QCoh}(X)$, being a Serre quotient of $\mathsf{QCoh}([X_0 / G])$, is also generated by line bundles. By our hypotheses on $X$ this means $\mathsf{QCoh}(X)$ is generated by the structure sheaf $\mathscr{O}_X$. But then $X$ is quasiaffine by [EGA II, Proposition 5.1.2], and any toric variety which is both complete and quasi-affine is just a point.

So... what gives? Is (1) incorrect, or is there some key mistake I'm making?

References:

  • [GS II] Geraschenko and Satriano, "Toric Stacks II: Intrinsic Characterization of Stacky Fans".
  • [CLS] Cox, Little, and Schenck, "Toric Varieties".
  • [EGA II] Alexander Grothendieck, "Éléments de géométrie algébrique: II. Étude globale élémentaire de quelques classes de morphismes."
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If I am not mistaken, there seems to be a mistake in the proof of Lemma 2.10 of [GS II], as I will explain below.

Diving into Example 4.2.13 of [CLS] shows a great similarity with the arguments in [GS II]. Indeed, the idea is to compute $\operatorname{Pic}(X_\Sigma)$ by computing the free abelian group $\operatorname{SF}(\Sigma,N)$ of support functions on $\Sigma$.

On the other hand, the proof of Theorem 2.13 in [GS II], when specialised to toric varieties, goes by computing the colimit $\operatorname{colim}_{\sigma \in \Sigma} \sigma \cap N$. Unwinding the definitions, we see that $\operatorname{SF}(\Sigma,N)$ coincides with $\operatorname{Hom}(\operatorname{colim}_{\sigma \in \Sigma} \sigma \cap N,\mathbf Z)$. Equivalently, by [GS II, Rmk. 2.3] and the universal property of group completion, this is $\operatorname{Hom}(\operatorname{colim}_{\sigma \in \Sigma} (\sigma \cap N)^\text{gp},\mathbf Z)$.

Now the computation of [CLS] is exactly that the map $\operatorname{colim}_{\sigma \in \Sigma}(\sigma \cap N)^{\text{gp}} \to N$ is an isomorphism in Example 4.2.13 (this is equivalent to the statement that $M \to \operatorname{SF}(\Sigma,N)$ is an isomorphism by duality). However, the proof of [GS II, Thm. 2.13] seems to say that the image of each $\sigma \cap N$ is a face in the colimit, which is clearly not true in this example.

Backtracking through the proofs of Corollary 2.12 and Lemma 2.10 with the example of [CLS] in mind reveals a mistake in the proof of Lemma 2.10. It is claimed that $D^1$ is a join-closed subdiagram since the newly added face is maximal. However, running this argument for the example of [CLS] shows that this is not true: if we start with the diagram $D^0$ of faces of $\sigma_1$, and adjoin all faces of $\sigma_2$, we see that the join of the 1-dimensional faces spanned by $u_2$ and $u_4$ is the 3-dimensional face $\sigma_3$, which is not contained in $D^1$. (If we use $\sigma_1$ and 'back' instead, we can also get a 2-dimenional join.)

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    $\begingroup$ Great answer, thanks! Looking at the proof more, it also seems that Corollary 2.12 is not the "right statement" to try to prove - even if we could show that all of the colimit structure maps were inclusions of faces, that would be effectively showing that the map $X \to [X_0 / G]$ is an open immersion locally on the source, which isn't enough to show that it's an open immersion (consider the forgetful map from $\mathbb{A}^1$ with doubled origin to $\mathbb{A}^1$). I wonder if / how this is fixable... $\endgroup$ Commented Apr 15 at 7:45

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