Let $X$ be a smooth, complex projective algebraic variety defined over a number field $K$.
Let $D$ be a divisor of $X$ defined over $K$ with the following property:



> 
> For any curve $C$ defined over $K$, we have $\operatorname{deg (D\_{|C})=0}$
> 
> 
> 


Is it then true that $c\_1(D)=0$?


In general, in order to have $c\_1(D)=0$, I should check that $\operatorname{deg (D\_{|C})=0}$
for *any curve* (not just the ones defined over $K$). I'm asking if in this particular setting, the curves defined over $K$ are enough.



First Chern class and field extensions
Every curve on $X$ is algebraically equivalent to a curve defined over a finite extension of $K$, and then a union of Galois conjugates will be defined over $K$. So, if you allow reducible curves, then the answer is yes.


Added: The intersection product is Galois invariant.


For a nonperfect field $k$ and a divisor $D$ defined over a purely inseparable extension of $k$ of degree $p^m$, the divisor $p^m D$ is defined over $k$.


Regard $D$ as the
Cartier divisor defined by a family of pairs $(f\_{i},U\_{i}^{\prime})$,
$f\_{i}\in k^{\prime}(X)$, and let $U\_{i}$ be the image of $U\_{i}^{\prime}$ in
$X$; then $k^{\prime}(X)^{p^{m}}\subset k(X)$, and so the pairs $(f\_{i}%
^{p^{m}},U\_{i})$ define a divisor on $X$ whose inverse image on $X\_{k^{\prime}}$ is $p^{m}D$.


