I have been working on Exercise IV.1.5 in Hartshorne's Algebraic Geometry, which says
For an effective divisor $D$ on a curve $X$ of genus $g$, show that $\operatorname{dim} |D| \le \operatorname{deg} D$. Furthermore, equality holds if and only if $D = 0$ or $g = 0$.
Note that by "curve" Hartshorne means complete, nonsingular curve over algebraically closed field. I was able to show the inquality using Riemann-Roch, Serre duality, and the long exact sequence of cohomology.
Regarding the equality statement, it's easy to show that $D = 0$ or $g = 0$ forces equality. However, I concluded something stronger than what Hartshorne claims in the reverse direction. I can show that
If equality holds, then $D= 0$.
As an immediate corollary, one obtains that every effective divisor on a curve of genus zero is the zero divisor, because $g = 0 \implies \deg |D| = \dim D \implies D = 0$. Have I made a mistake and concluded something false?