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Look at the following enlightened part of a proof extracted from a paper (here $C$ is an algebraically closed field of characteristic $0$):

enter image description here

Clearly $\mathbb Q\subset C$ and $\mathbb P^1_C$ is a scheme over $\mathbb Q$ thanks to the fibered product: $$\mathbb P^1_C=\mathbb P^1_\mathbb Q\times_{\operatorname{Spec \mathbb Q}}\operatorname{Spec} C$$ but I don't understand why there exists a $\mathbb Q$-rational point on $\mathbb P^1_C$. We should have a morphism $\operatorname{Spec }\mathbb Q\longrightarrow \mathbb P^1_C$ over $\mathbb Q$, but who does ensure that it really exists?

A similar problem arises here (now note that $C=\mathbb C$):

enter image description here

In this case the author wants $K$-rational points in $\mathbb P^1_\mathbb C$!

Here the original paper: http://arxiv.org/abs/arXiv:math/0108222

Probable solution: Maybe we are in this case.

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Dubious
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  • What is the morphism $t$? – Álvaro Lozano-Robledo Mar 24 '14 at 17:28
  • $t:X\longrightarrow \mathbb P^1_C $ is a finite morphism. However I will add the intere paper to my question, I notice that is difficult to give an answer without other informations. – Dubious Mar 24 '14 at 17:31
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    This doesn't make sense. A scheme over $C$ has no $\mathbf{Q}$-rational points. On the other hand, you could choose a $\mathbf{Q}$-rational point of $\mathbf{P}_\mathbf{Q}^1$, which can then be viewed as a $C$-rational point of $\mathbf{P}_C^1$, and do something with that. – Keenan Kidwell Mar 24 '14 at 17:52
  • Probably the author uses another definition of $K$-rational point. It can't be a typo because this notation is ubiquitous in the paper. – Dubious Mar 24 '14 at 18:35
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    In classical language, $\mathbf{P}_C^n$ might be taken to mean what would now be called that set of $C$-valued points of $\mathbf{P}_C^n$, and then it makes sense to ask whether a point is a $\mathbf{Q}$-point, but the author says he's using scheme language, so I'm not sure. The only way to make sense of $\mathbf{Q}$-points of $\mathbf{P}_C^n$ in this case is the way I describe in the comments to the linked question (see also Emerton's answer). In this case $\mathbf{P}_C^n$ has a canonical $\mathbf{Q}$-model, and the set of $\mathbf{Q}$-points of that naturally embed into the set of $C$-points – Keenan Kidwell Mar 24 '14 at 18:40
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    of $\mathbf{P}_C^n$. But $\mathbf{P}_C^n$ does not have any literal (scheme-theoretic) $\mathbf{Q}$-valued points, i.e., there can be no morphism $\mathrm{Spec}(\mathbf{Q})\to\mathbf{P}_C^n$, because such a thing is equivalent to the data of a point $x$ of the target and an embedding of fields $k(x)\hookrightarrow\mathbf{Q}$. But $k(x)$ contains $C$, and $C$, being algebraically closed, cannot embed into $\mathbf{Q}$. – Keenan Kidwell Mar 24 '14 at 18:42
  • so most probably we are choosing $\mathbb Q$-rational points in $\mathbb P^1_{\mathbb Q}$ as you said in your first comment. – Dubious Mar 24 '14 at 18:50

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My guess is that the author means the following: choose a $\mathbf{Q}$-point of $\mathbf{P}_\mathbf{Q}^1$, and look at the induced $C$-point of $\mathbf{P}_C^1$.

To elaborate, the set of $\mathbf{Q}$-points of $\mathbf{P}_\mathbf{Q}^1$ is in natural bijection with the set $\mathbf{Q}^{2}\setminus\{0\}/\mathbf{Q}^\times$ (this follows e.g. from the classification of morphisms into projective space in terms of line bundles, or in more elementary ways). Likewise the set of $C$-points of $\mathbf{P}_C^1$ is in natural bijection with $C^{2}\setminus\{0\}/C^\times$. This contains the former set via $\mathbf{Q}\hookrightarrow C$ (the embedding of the $\mathbf{Q}$-points into the $C$-points via these descriptions can be identified with the natural base change map $\mathbf{P}_\mathbf{Q}^1(\mathbf{Q})\hookrightarrow\mathbf{P}_\mathbf{Q}^1(C)=\mathbf{P}_C^1(C)$). So a $\mathbf{Q}$-point of $\mathbf{P}_\mathbf{Q}^1$ naturally gives rise to a $C$-point of $\mathbf{P}_C^1$.

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  • Yes this makes sense, and very informally I think it means "pick the points $(1:q)\in\mathbb P^1(\mathbb C)$ with $q\in\mathbb Q$". Many thanks. – Dubious Mar 24 '14 at 19:13