Is that true that there is no rational curves contained in an Abelian variety? If it's true, is that because abelian varieties are not uniruled? How do I know whether an abelian variety is not uniruled?
There are no rational curves in an abelian variety, this is much stronger than not being uniruled. If there is a map $P^1 \to A$, $A$ abelian, the map would factor through the Albanese variety of $P^1$, by definition. However, for curves, the Albanese is the Jacobian (from general theory of the Jacobian) and the Jacobian of $P^1$ is a point.

$\begingroup$ Fabulous answer. Could you please provide some reference in which the Albanese and Picard functorialities of Jacobian are compared with rigor and full proof? $\endgroup$ – Anweshi Jan 10 '10 at 15:18

$\begingroup$ @Anweshi Reference: Lang, Abelian Varieties. Probably in Mumford too. $\endgroup$ – Felipe Voloch Jan 10 '10 at 16:11

$\begingroup$ Mumford avoids the Jacobian altogether. Lang's book is not very organized and uses the old language. But I suppose I will indeed find Albanese and Picard there. Thanks. $\endgroup$ – Anweshi Jan 10 '10 at 16:27

1$\begingroup$ @Anweshi Ha! Here is a short proof that, for curves, the Jacobian is the Albanese, using that maps from $P^1$ to an abelian variety are constant. A map from a curve to an abelian variety extends to divisors by linearity and is constant on linear equivalence classes because it's constant on $P^1$'s, so the map gives a welldefined map from the jacobian to the abelian variety. If you want to develop things this way from first principles, you have to avoid my answer to the original question and use one of the other answers. $\endgroup$ – Felipe Voloch Jan 10 '10 at 17:34

1$\begingroup$ @Felipe Voloch: but then you end up using the fact that you wanted to prove (that a map from $P^1$ to an ab. var. is constant). $\endgroup$ – Matthieu Romagny Mar 21 '14 at 17:32
Over $\mathbb C$ you can argue as follows. Suppose you have a morphism $\mathbb P^1(\mathbb C) \to A $ ($A$= abelian variety ). Since $\mathbb P^1(\mathbb C) $ is simply connected , the morphism lifts to the universal cover of $A$, affine space $\mathbb C^n$. But since $\mathbb P^1(\mathbb C)$ is complete and connected, the lift to affine space must be constant and hence the original morphism is constant too.
The answers by Charles, Felipe and jvp are better because they work over arbitrary fields, but since the argument just given is so ridiculously elementary (introductory topology), I thought it might still be of some interest ( also it works in the holomorphic category if $A$ is a complex torus, maybe not algebraic).

$\begingroup$ This should work over any field contained in $\mathbb C$. $\endgroup$ – Anweshi Jan 10 '10 at 15:19
Yes, an abelian variety $A$ contains no rational curves.
Suppose not and let $f: \mathbb P^1 \to A$ be a nonconstant morphism.
If $f$ is inseparable then it must be the composition of some power of Frobenius of $\mathbb P^1$ with a nonconstant separable map $g: \mathbb P^1 \to A$. Thus we may assume that $f$ is separable, i.e., $df : T \mathbb P^1 \to f^{\ast} T A$ is not the zero morphism. Therefore the general $1$form $\omega \in H^0(A,\Omega^1)$ will give rise to a nonzero $1$form $f^{\ast} \omega$ on $\mathbb P^1$. Contradiction.
Remarks:
 Above, I have expanded the original answer "If $C$ is a curve on an abelian variety then $C$ has regular $1$forms coming from $1$forms on $A$" in order to incorporate Voloch's comment about positive characteristic.
 The same argument shows that nonalgebraic compact complex tori contain no rational curves.
 Since over $\mathbb C$ the Albanese variety of a compact Kahler manifold $X$ is usually defined as $H^0(X, Omega^1)^{\ast} / H_1(X, \mathbb Z)$, the argument above is essentially the same as Voloch's when the characteristic zero.
 Let $X$ be a smooth projective variety and $f:X \to A$ be a morphism. If $df$ has maximal rank then $H^0(X,{\Omega^i}) \neq 0$ for every $i \le \dim X$. Thus an abelian variety contains no subvarieties without regular forms in any particular degree.

$\begingroup$ This is a nice and direct way to see it in characteristic zero but, in positive characteristics, you need to worry about inseparability. Your argument still works once you factor the map through a power of Frobenius. $\endgroup$ – Felipe Voloch Dec 16 '09 at 4:35

$\begingroup$ Thanks for bringing that up. I have edited my answer accordingly. $\endgroup$ – Jorge Vitório Pereira Dec 16 '09 at 11:06

$\begingroup$ The decomposition of inseparable map looks like the Zariski Main theorem which says, any morphism can be decomposed into a morphism with connected fibers and a finite morphism. Are those the same thing? Is there any relations between them? $\endgroup$ – Fei YE Dec 16 '09 at 19:25
There is not one. Reference is Milne's notes, Prop 3.9. More is true, Prop 3.10 in the same notes is that any rational map from a unirational variety to an abelian variety is constant.
Assume there is a rational curve $C$ in an Abelian variety $A$. Let $p\in C$ be a point and $q\in A$ another point. There exists an automorphism $\sigma:A\rightarrow A$ such that $\sigma(p) = q$. Then $\Gamma = \sigma(C)$ is a rational curve through $q$. In this way we see that $A$ is covered by rational curves and its Kodaira dimension is negative. A contradiction because $k(A) = 0$. We conclude that there are not rational curves on an Abelian variety.

$\begingroup$ In positive characteristic there are uniruled varieties with nonnegative Kodaira dimension, e.g., the Fermat quartic surface in characteristic congruent to $1$ modulo $4$. $\endgroup$ – Jason Starr Mar 21 '14 at 1:48
How about when the rational curve is singular?

3$\begingroup$ None of the answers above assume the rational curve is smooth. All of them consider maps from P^1 to A without assuming that the image is smooth $\endgroup$ – Jorge Vitório Pereira Dec 16 '09 at 19:00