Look at the following enlightened part of a proof extracted from a paper (here $C$ is an algebraically closed field of characteristic $0$):
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 $C=\mathbb C$):
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.
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 $\mathbf{P}_C^1$.