TITLE: Linear equivalence of Divisors on a Surface
QUESTION [6 upvotes]: Studying algebraic geometry, while the abstract theory is pretty clear to me I often feel puzzled in practice. Here I am trying to understand linear equivalence of divisors in some practical situations. Consider the following example.
Let $S\subset\Bbb{P}^3$ be a smooth surface of degree $d\geq3$ and let $\ell\subset S$ be a line. A plane containing $\ell$ cuts on $S$ the divisor 
$$ H_0=\ell+C$$
where $C\subset S$ is a curve of degree $d-1$. Let $H$ be the divisor cut on $S$ by a generic plane. So $H$ cuts $\ell$ in one point. I am not sure about the following:
Question: why can we say that $H_0\sim H \ $ (linear equivalence) ?
This is then a nice example for showing that even the most simple divisor like $\ell$ can have negative self-intersection: if $H_0\sim H$ then
$$1=H\cdot\ell=H_0\cdot\ell=(\ell+C)\cdot\ell=\ell^2+C\cdot\ell=\ell^2+d-1 $$
Hence $\ell^2=2-d<0$.

REPLY [2 votes]: (Turning my comment into an answer...) The point is that restriction of divisors respects linear equivalence (as follows quickly from the definition). Any two planes in the ambient space $\mathbf{P}^3$ are linearly equivalent divisors (easy exercise!), so the same remains true when we restrict them to the surface $S$.