If $X$ is a projective variety, then $O_X(n)$ has no global sections

algebraic-geometry

I have a question about Example 7.1.2 in the link https://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2002/alggeom-2002.pdf. In (ii), it is written that there are no global sections of $O_X(n)$ for $n<0$. Here $X$ is a projective variety over an algebraically closed field $k$. But why is this true? By definition, a section of $O_X(n)$ is of the form $f/g$ with $f,g$ homogeneous and $\deg(f)=\deg(g)+n$.

Edit: I see this when $X=\Bbb P^n$, because every nonconstant homogeneous polynomial should have a zero. But I can't see this in the general case.

Best Answer

Let $i : X \to \Bbb P^r$ an embedding, then we can define $\mathcal O_X(n) = i^* \mathcal O(n)$. As you said, section of $\mathcal O_X(n)$ are quotients of homogeneous polynomials of appropriate degree, restricted to $X$.

In particular, any $f/g$ (as a function on $\Bbb P^r$) has poles on an hypersurface $Y \subset \Bbb P^r$.

If $X$ has positive dimension, then $Y \cap X \neq \emptyset$ and hence $f/g$ also has poles on $X$. It follows that $\mathcal O_X(n)$ has no global sections when $n < 0$, assuming $\dim X > 0$.