 Rigidity of a negative curve

If an effective divisor $D$ is linearly equivalent to a negative curve $C$, then $D=C$. Otherwise $D\cdot C=C^2<0$ forces $C$ to be a component of $D$, since distinct irreducible curves have nonnegative intersection multiplicities. Write $D=C+D^{\prime }$ with $D^{\prime }$ effective and linearly equivalent to zero. An ample divisor intersects $D^{\prime }$ positively unless $D^{\prime }=0$, while linear equivalence forces that intersection to vanish.