Let $X$ be a projective variety over a field. Is there a direct way of seeing why every pair of linearly equivalent divisors $D_1$ and $D_2$ is numerically equivalent? I simply found myself unable to pass from the first definition below to the other.
Recall that $D_1 \sim D_2$ linearly if they differ by a prime divisor, and $D_1 \sim D_2$ numerically if they have the same intersection number against every curve in $X$.
Best Answer
Since $D_1\sim D_2$ is equivalent to $D_1-D_2\sim 0$, and the intersection number is additive, we can focus in the case $D\sim 0$, i.e., $D=\operatorname{div}(f)$.
If, $X$ is a curve, it it true since $\deg(\operatorname{div}(f))=0$. Now, if $X$ is a surface, we have to show that $D\cdot\operatorname{div}(f)=0$ for any divisor $D$ on $X$, but it is enogh to show it in the case where $D=C$ is a curve. In this case, we have that $C\cdot \operatorname{div}(f)=\deg(\operatorname{div}(f)|_C)=0$ since $\operatorname{div}(f)|_C$ is again a principal divisor. For the general case $\dim(X)\geq 3$, we can reduce it to the surfaces case. For example, if $\dim(X)=3$, the statement to show is that if $D$ is principal, then for any pair of divisor $D_1$, and $D_2$, $$D_1\cdot D_2\cdot D=0.$$ Here, we can note that $D_1\cdot D_2\cdot D=D_1|_{D_2}\cdot D|_{D_2}=0$ (is the surface case). Clearly, we assume that $D_1$ and $D_2$ are prime divisors.