Is there any rational curve on an Abelian variety? 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?
 A: 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.
A: 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.
A: 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).  
A: 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.
A: Yes, an abelian variety $A$ contains no rational curves.
Suppose not and let $f: \mathbb P^1 \to A$ be a non-constant morphism. 
If $f$ is inseparable then it must be the composition of  some power of Frobenius of $\mathbb P^1$ with a non-constant 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 non-zero $1$-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  non-algebraic 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.

A: How about when the rational curve is singular?
A: See also Cornell-Silverman,  p. 107.
