Lazarsfeld said in his book (Positivity in AG 1,example 2.1.12) that if $f:X\rightarrow Y$ is a projective surjective morphism of normal (complex)varieties, and $\mathbb{C}(Y)\subset\mathbb{C}(X)$ the corresponding extension of function fields. Then $f$ is a fibre space if and only if $\mathbb{C}(Y)$ is algebraically closed in $\mathbb{C}(X)$.
Here $f:X\rightarrow Y$ is a fibre space means that $f$ is a projective surjective morphism s.t. $f_{*}\mathcal{O}_X=\mathcal{O}_Y$.
I can not understand the "only if" part, the following is the method in his book:
Suppose $\mathbb{C}(Y)$ is not algebraically closed in $\mathbb{C}(X)$, then $f$ facors as a composition $X\stackrel{u}\dashrightarrow Y'\stackrel{v}\dashrightarrow Y$ of rational maps where $Y'$ is generically finite of degree >1 over Y. Replacing $Y'$ and $X$ by suitable birational modifications(projective birational morphism to $X$ and $Y'$), since $X$ is normal, it doesn't affect the fibre space hypothesis, one can suppose that $u$ and $v$ are in fact morphisms, and then f fails to be a fibre space since some fibres may not be connected.
I don't know how to find a "suitable modification" of $X$ so that one can assume $u$ a morphism, can we take the graph of the rational map $u$?, but then how can we guarantee that the morphism $\Gamma_u\rightarrow X$ is a projective morphism?
I hope I can get some help, thank you!
I think I've found the rihgt answer. We can factor $f$ as a composition of rational maps $u$ and $v$ as Lazarsfeld does, so that $Y'$ be a projective variety. Then denote the graph of $u$ by $\Gamma_u$ we have the natural morphism: $\Gamma_u\rightarrow X$ is a birational projective morphism.(actually being proper is enough).
So the composition $\Gamma_u\rightarrow X\rightarrow Y$ is still a fibre space ( X is normal thus the pushforward of $\mathcal{O}_{\Gamma_u}$ is equal to $\mathcal{O}_X$). So the morphism $\Gamma_u\rightarrow X$ is the desired modification of X and we can assume the rational map $u$ to be regular.