unirational complex variety has $H^i(X,O_X) = 0$ for i > 0

Question

Let $X/\mathbf{C}$ be a smooth projective connected unirational variety. Why do we then have $H^i(X,O_X) = 0$ for $i > 0$?

Answer 1

The vector space $H^i(X,O_X)$ is complex conjugate to $H^0(X,\Omega^i)$, the space of global holomorphic $i$-forms. So it is enough to show that a unirational variety has no nonzero such forms.

Let $X$ be unirational, and $\mathbf P^n \dashrightarrow X$ be a dominant rational map. We can resolve the indeterminacy of this to get a nonsingular rational variety $Y$ with a birational morphism $\pi: Y \rightarrow \mathbf P^n$ and a surjective morphism $f: Y \rightarrow X$.

Now suppose $\omega$ was a nonzero global holomorphic $i$-form on $X$; then $f^* \omega$ would be a nonzero global holomorphic $i$-form on $Y$. (The fact that $f^*\omega$ is actually nonzero is not completely straightforward: this part can actually be false in positive characteristic, more precisely if $f$ is not separable.)

But now $Y$ is actually rational, and it is pretty straightforward to show that a smooth projective rational variety has no holomorphic forms — at least, assuming you know this for $\mathbf P^n$ itself. (See Hartshorne Exercise II.8.8.)