# Very special rational points on curves over number fields

For some reason, I'm convinced the answer to the following question should be (obviously) negative, but I can't come up with a good reason.

Does there exist a number field $K$, a smooth projective geometrically connected curve $X/K$ of genus $g\geq 2$ with a $K$-rational point $x$ such that, for any number field $L/K$, $x$ does not intersect $X(L)$?

Let me make the last part of the question more precise. Firstly, let $\mathcal X$ be the minimal regular model of $X$ over $O_K$. When I say that $x$ does not intersect $X(L)$ I mean that the intersection product $(x,y)_{\mathcal X}$ on $\mathcal X$ equals zero for all $y\in X(L)-\{x\}$. (Here $x$ and $y$ also denote their Zariski closures in $\mathcal X$.)

I think it could happen that some $K$-rational point does not intersect any other $K$-rational point. Take for example a curve with $X(K) = \{pt\}$. For some reason I do think that there should always be some (other) $L$-rational point which intersects this $K$-rational point.

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What do you mean by intersection product of two horizontal curves on $\mathcal X$ ? – user18119 May 29 '12 at 20:31
Let $D$ and $E$ be horizontal divisors on $\mathcal X$ over $S=\mathrm{Spec} O_K$. Then $(D,E)$ is defined to be $\sum_{s\in \vert S\vert} i_s(D,E) \log \# k(s)$, where the sum runs over the closed points of $S$ and $i_s(D,E)$ is the local intersection multiplicity of $D$ and $E$ over $s$. – Harry May 29 '12 at 20:45

The answer is no (for any genus). Fix a closed point $s\in \mathrm{Spec}(O_K)$ such that $\mathcal X_s$ is smooth. Let $x_s$ be the intersection point of $\overline{\{ x\}}$ with the fiber $\mathcal X_s$. Look (Zariski) locally at $\mathcal X$ around $x_s$. Lift a generator of the maximal ideal of $O_{\mathcal{X}_s, x_s}$ to $O_{\mathcal X, x_s}$, then you get a horizontal curve in $\mathcal X\otimes O_{S,s}$ passing through $x_s$. Taking the Zariski closure of the horizontal curve in $\mathcal X$ then gives you a horizontal curve in $\mathcal X$ passing through $x_s$. Its intersection with $\overline{\{ x\}}$ is positive, and its generic fiber is a point of $X(\bar{K})$ hence of $X(L)$ for some finite extension $L/K$.

Finally, there are infinitely many local liftings of a given generator (just perturb a lifting by adding a multiple of the uniformizing element of $O_{S,s}$), so at least one of them will gives a horizontal curve different from $\overline{\{ x\}}$.

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A "system of generators" of $m_{X_s,x_s}$ is just a "generator" in this case, right? I thought the local ring of $X_s$ at $x_s$ would be a dvr as $x_s$ is smooth on $\mathcal X_s$. As a not very important side question, I just starting wondering whether this lifting could be done in some "almost uniform" sense for all curves $X$ over $K$ of fixed genus $g$ (and maybe some "extra fixed data"). For example, could one control the degree of the extension $L/K$ in terms of this data? Do different local liftings give rise to bigger and bigger extensions? – Harry May 29 '12 at 21:21
@Harry, yes I will correct the answer. For the second part of your question, it is easy to have extension $L/K$ of big degree. I don't know how to get small extensions except for local henselian DVR. – user18119 May 30 '12 at 9:25
Yeah, I also soon realized it is easy to obtain $L/K$ of arbitrary big degree. This is already clear with the zero section of $\mathbf{P}^1_{\mathbf{Z}}$. But, I'm actually very interested in knowing how to choose a rational point with "controlled" degree even in the henselian case. Naive question: if we can control $[L:K]$ for henselian local dvr's, then wouldn't we be able to also do it for local dvr's by taking their henseliazation? – Harry May 30 '12 at 10:17
@Harry, in the henselian case, one can take $L=K$. Indeed, $\mathcal X$ is Zariski locally around $x_s$ étale over $\mathbb A^1_{O_K}$. Pull-back any section of $\mathbb A^1_{O_K}$ passing through the origin of the special fiber will give a section of $\mathcal X$ passing through $x_s$ (because $O_K$ is henselian). But this will not give control in the general case because in the henseliazation, there may be algebraic extensions of big degrees. – user18119 May 30 '12 at 20:25