First Chern class and field extensions

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.

MathOverflow Asked by manifold on January 28, 2021

1 Answers

One Answer

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 $p^{m}D$.

Correct answer by anon on January 28, 2021

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