Timeline for Is having no rational point always witnessed over a place?

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Mar 23, 2020 at 17:55	comment	added	Fabian Meumertzheim		I see now that this trick will only work in rather special situations. Do you know whether there is anything else one can do?
Mar 22, 2020 at 20:02	comment	added	François Brunault		The trick will work as long as there exists a residue field that embeds into $K$. Say that $K$ is the function field of a curve $C$ over a number field $k$. Usually $C$ is geometrically connected, which means that $k$ is algebraically closed in $K$. This implies that any residue field embedding into $K$ must be equal to $k$. If $C$ has a $k$-rational place, then the answer to your question is yes. But in general the $k$-rational places correspond to the $k$-rational points of the normalisation of $C$, and this set can be empty.
Mar 21, 2020 at 17:09	comment	added	Fabian Meumertzheim		@GTA Thanks for your comment! If I understood you correctly, this does seem to resolve my question in the special case where K can be written as a purely transcendental extension of an extension of $\mathbb{Q}$. But if we choose $K$ to be e.g. $\mathbb{Q}(t, \sqrt{t^3-t})$, this trick does not seem to apply anymore. Do you maybe have an idea how to handle the general case?
Mar 20, 2020 at 21:00	comment	added	GTA		If $K=L(t)$ for $t$ trascendental over $L$, $L$ is a residue field but also $L$ embeds into $K$, so any $L$-rational point can be made a $K$-rational point?
Mar 20, 2020 at 20:38	history	asked	Fabian Meumertzheim	CC BY-SA 4.0