Suppose $X$ is a scheme, the structure morphism $X \rightarrow \mathrm{Spec}\mathbb{Z}$ smooth and surjective. Assume further that $H^0(X \times \mathrm{Spec} \mathbb{C}, \mathcal{O}^\times) = \mathbb{C}^\times$. Then is it necessary that $H^0(X, \mathcal{O}^\times) = \pm 1$ ?
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$\begingroup$ I don't understand the title. $\endgroup$– Martin BrandenburgMay 25, 2012 at 6:21
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2$\begingroup$ Martin, I agree that it is a bit of a riddle. But the automorphism group of a line bundle can be identified with $H^0(\mathcal{O}^*)$ if you think about it. $\endgroup$– Donu ArapuraMay 25, 2012 at 11:32
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$H^0(X,\mathcal O_X^\times)=\Gamma(X,\mathcal O_X)^{\times}$ since the first consists of regular functions that are everywhere locally invertible and the second consists of regular functions that are globally invertible, but these are the same thing as local inverses can be patched to global inverses.
Assume the statement is false. Then $\Gamma(X,\mathcal O_X)$ must contain some unit $a$ not in $\pm 1$. That unit cannot be in $\mathbb Q$. If it were, the ring would contain $1/p$ for some $p$, which would imply that every local ring of $X$ would also contain $1/p$, so $X$ does not map surjectively onto $\textrm {Spec} \mathbb Z$ since there are no points lying over $p$. So it contains a unit not in $\mathbb Q$.
Since $\mathcal O_X$, being flat, is non-torsion $\mathbb Z$-module, so the map $\mathcal O_X \to \mathcal O_X\otimes \mathbb Q$ is an injection, so by left exactnes $\Gamma(X,\mathcal O_X)\in \mathbb \Gamma(X,\mathcal O_X\otimes \mathbb Q)=\Gamma(X\otimes \mathbb Q, \mathcal O_{X\otimes \mathbb Q})$. Therefore this extra unit of $\Gamma(X,\mathcal O_X)$ also lives in $X \otimes \mathbb Q$ and is not in $\mathbb Q$.
By the flat base change theorem, $\Gamma (X\otimes \mathbb C, \mathcal O_{X\otimes \mathbb C})=\Gamma(X\otimes \mathbb Q, \mathcal O_{X\otimes \mathbb Q})\otimes \mathbb C$. If you take a ring that contains $\mathbb Q[\alpha]$, with $\alpha$ a unit not in $\mathbb Q$, and tensor with $\mathbb C$ you will get a vector space over $\mathbb C$ of positive dimension, hence not $\mathbb C$.
answered May 25, 2012 at 6:02
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2$\begingroup$ "If a ring over $\mathbb C$ has group of units $\mathbb C^{\times}$ then it is $\mathbb C$.": what about $R=\mathbb{C}[x]$? $\endgroup$ May 25, 2012 at 7:46
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$\begingroup$ Good point. I believe the argument can be fixed, but I will have to think about it. $\endgroup$ May 25, 2012 at 8:06
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$\begingroup$ "If you take a ring that contains $\mathbf{Q}[\alpha]$, with $\alpha$ a unit not in $\mathbf{Q}$, and tensor with $\mathbf{C}$ you will get $\mathbf{C}[\alpha]$." What if I tensor $\mathbf{Q}[\sqrt{2}]$ with $\mathbf{C}$? Then I get $\mathbf{C}\otimes \mathbf{C}$. This is not $\mathbf{C}[\sqrt{2}] = \mathbf{C}$. Moreover, why "with $\alpha$ not a unit in $\mathbf{C}$"? What am I missing here? I shouldn't be able to take $\alpha = \sqrt{2}$? $\endgroup$– HarryMay 27, 2012 at 23:28
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$\begingroup$ I think the last sentence would be slightly improved if it were replaced with something like "you will get a commutative $\mathbb{C}$-algebra, where the image of $\alpha$ is a unit that is not in the $\mathbb{C}$-span of $1$. This violates the initial assumption." $\endgroup$– S. Carnahan ♦Jun 13, 2018 at 4:20