I heard that there is a theorem due to Rosenlicht which says the following:



> 
> **Theorem.** Let $X$ be a complex projective manifold and $V$ a non-trivial holomorphic vector field on $X$. Then $X$ is uniruled,ie,can be covered by rational curves,if $V$ has a zero.
> 
> 
> 


I have thought for a few days and failed to give myself a proof. Can somebody give me the reference or say something about the idea of proof?


Thanks in advance.



Rosenlicht theorem about uniruledeness and zeroes of holomorphic vector field on complex projective manifold
You can look at Lieberman's paper [Holomorphic Vector Fields on Projective Manifolds](http://books.google.com.br/books?id=pa1I1gG3eyUC&pg=PA273&lpg=PA273&dq=Holomorphic+vector+fields+on+projective+varieties.&source=bl&ots=QxbsauUatN&sig=8gFvKiAuur3UwDrP0eGPu94DLH4&hl=pt-BR&ei=KhIVTrmQKOPx0gGd-pRT&sa=X&oi=book_result&ct=result&resnum=4&ved=0CDgQ6AEwAw#v=onepage&q=Holomorphic%2520vector%2520fields%2520on%2520projective%2520varieties.&f=false).


His proof is more or less as follows. A result of Grothendieck asserts that $\mathrm{Aut}^0(X)$, the connected component 
of the identity of the automorphism group of $X$, is an algebraic group which acts
algebraically on $X$.


Look at the (analytic) subgroup generated by your vector field and let $G$ be its Zariski closure in $\mathrm{Aut}^0(X)$. Notice that $G$ is abelian.


If $p \in X$ is a zero of your vector field then 
$p$ is fixed by the action of $G$ on $X$. Thus for $k \in \mathbb N$, $G$ acts on $$\frac{\mathcal O\_{X,p}}{\mathfrak{m}\_p^k}, $$ where $\mathfrak m\_p$ is the maximal ideal of $\mathcal O\_{X,p}$. Moreover, if $k \gg 0$ then the action is faithfull.
Thus $G$ is isomorphic to a linear algebraic group and [yet another result of Rosenlicht](http://www.jstor.org/pss/2372523) says that a Zariski-closed abelian 
subgroup of a linear algebraic group is of the form $(\mathbb Cˆ\*, \cdot)^r \times (\mathbb C,+)^s$. The action of the factors of this decomposition generate the sought rational curves.




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**Added later:** For an alternative proof see Theorem 6.4 of this [paper](http://arxiv.org/abs/1107.1538). There it is proved that the existence of a non-zero section of $\bigwedge^q TX$ vanishing at a point suffices to ensure that $X$ is uniruled.


