A divisor on a smooth curve such that $\Omega$ minus it has no nonzero sections Let $X$ be a smooth projective curve of genus $g$. Let $\Omega$ be the sheaf of differentials. Mumford (in Abelian Varieties, sec. 2.6, in proving the theorem of the cube) asserts that there is an effective divisor of degree $g$ on $X$ such that $H^0(X, \Omega \otimes L(-D))=0$. 
This should be easy, but I'm missing the argument. Could someone explain?

Let $\mathcal L$ be an invertible sheaf on the curve $X$, and suppose
that $H^0(X,\mathcal L)$ has dimension $d$.  (In the case when $\mathcal L
= \Omega$, we have $d = g$.)
If we choose a closed point $x \in X$, then there is a map $H^0(X,\mathcal L) \to 
\mathcal L_x/\mathfrak m_x \mathcal L_x$ given by mapping a section to its fibre at $x$.
To make my life easier, let's assume that $k$ (the field over which $X$ is defined) is algebraically closed, so that $x$ is defined over $k$, and we 
may identify $\mathcal L_x/\mathfrak m_x\mathcal L_x$ with $k$ (since it one
dimensional, being the fibre of an invertible sheaf).
Evaluation is then a functional $H^0(X,\mathcal L) \to k.$  Now this functional will be identically zero if and only if every section vanishes at $x$.  But a non-zero section has only finitely many zeroes, and $X$ has an infinite number of closed points, so we may certainly choose $x$ so that this functional is not
identically zero.
The evaluation map sits in an exact sequence
$$0 \to H^0(X,L(-x)) \to H^0(X,L) \to k,$$
and so if we choose $x$ such that evaluation is surjective, we find that
$H^0(X,L(-x))$ has dimension $d - 1$.  Proceeding by induction, we find
points $x_1,\ldots,x_d$ such that $H^0(X,L(-x_1-\cdots - x_d)) = 0$.
In summary: we have shown that we may find an effective degree $d$ divisor $D$
such that $H^0(X,L(-D))$ vanishes.  (And in fact, looking at the proof, we see
that this vanishing will hold for a generic choice of $D$.)
