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On page 10 of the article https://arxiv.org/abs/2012.04806# I found morphismp which behaves on smooth projective varieties over $\mathbb{R}$ like $X \mapsto \mathrm{Spec}(H^0(X, \mathcal{O}_X))$. In this context this morphism send to $\mathrm{Spec}(H^0(X, \mathcal{O}_X))$ only projective varieties over $\mathbb{R}$, on other types it behaves differently. So I need to understand, how it behaives on base changes.

More generaly, consider $X$ to be a smooth projective variety over $\mathbb{R}$. Under what condition $X_{\mathbb{C}}$ is still smooth projective variety over $\mathbb{R}$?

To be more precise, does the variety remain projective after the change of base?

As example, is base change $E_{\mathbb{C}}$ of elliptic curve $E$ over $\mathbb{R}$ a projective variety over $\mathbb{R}$?

zieonasta
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Apr 02 '23 at 12:56
  • This situation you're asking about is a little unusual - can you explain why you want to consider the base change as a scheme over $\Bbb R$? – KReiser Apr 02 '23 at 14:28
  • I need to understand how the map $X\mapsto Spec H^0(X, \mathcal{O}_X)$ behaves on smooth projective varieties over $\mathbb{R}$. This map is defined only on smooth projective varieties over $\mathbb{R}$ – zieonasta Apr 02 '23 at 14:43
  • No, that map is defined on all schemes due to the adjunction between $\operatorname{Spec}$ and global sections; see here for more info. (Also, when responding specifically to another user, @replies are your friend.) – KReiser Apr 02 '23 at 14:56
  • @KReiser No, that's not exactly what I meant (sorry for bad English). Slightly different context. I found this morphism in this article: https://arxiv.org/abs/2012.04806, page 10. In short, this morphism acts on generators that are smooth projective over $\mathbb{R}$, I need to understand whether this morphism works the same way on base changes. – zieonasta Apr 02 '23 at 15:09
  • Ah, that's valuable context for your question - please add that to your post with an [edit]. – KReiser Apr 02 '23 at 16:28
  • Do you require a "variety" to be geometrically connected? If so, this fails even for $X = \mathrm{Spec} \mathbb R$. – Ravi Fernando Apr 02 '23 at 18:30

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Since $X\to \operatorname{Spec}(\mathbb{R})$ is smooth, $X_{\mathbb{C}}\to \operatorname{Spec}(\mathbb{C})$ is smooth.

Since $\mathbb{C}/\mathbb{R}$ is separable, $\operatorname{Spec}(\mathbb{C})\to \operatorname{Spec}(\mathbb{R})$ is smooth.

In the end, the composition $X_{\mathbb{C}}\to \operatorname{Spec}(\mathbb{C})\to \operatorname{Spec}(\mathbb{R})$ is smooth.

Captain Lama
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