Effective divisors $D_1\leq D_2$ such that $h^0(D_1)=h^0(D_2)$, then $D_1=D_2?$

Question

Let $X$ be a smooth algebraic surface over $\Bbb{C}$. I'll use the following notation: \begin{align*} H^0(D)&:=H^0(X,\mathcal{O}_X(D))\\ h^0(D)&:=\dim_\Bbb{C}H^0(D)\\ |D|&:=\{\text{effective divisors linearly equivalent to }D\} \end{align*}

Suppose $D_1, D_2$ are effective divisors such that $D_1\leq D_2$ (meaning the multiplicity of each component in $D_1$ is less or equal the multiplicity for the same component in $D_2$). In this case, $H^0(D_1)$ is a subspace of $H^0(D_2)$.

If we have $h^0(D_1)=h^0(D_2)$, the vector spaces are equal, so $H^0(D_1)=H^0(D_2)$.

My question is: in this case, can we conclude $D_1=D_2$? Or at least that $D_1,D_2$ are linealy equivalent?

Answer 1

The answer is no. For an example, let $X$ be a blow up of a smooth surface at a point and let $E$ be the exceptional divisor. Let $D$ be any curve not meeting $E$, for example, a curve in the original surface not passing through the blown up point, pulled back to $X$. Then, $H^0(D)=H^0(D+E)$, $D\leq D+E$, but $D$ is not linearly equivalent to $D+E$.

Answer 2

Here's a simpler example. Take a non-hyperelliptic curve (Riemann surface) $C$, and let $P,Q\in C$ be distinct points. Then take $D_1=P$, $D_2=P+Q$. (Clearly $D_1\le D_2$, but they are not linearly equivalent.) We have $h^0(C,\mathscr O(P)) = 1 = h^0(C,\mathscr O(P+Q))$. Note that if we had $h^0(C,\mathscr O(P+Q)) = 2$, this would make $C$ hyperelliptic: The linear system $|P+Q|$ gives the branched double cover of $\Bbb P^1$.