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. 


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). 


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:



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. 


How about when the rational curve is singular? 


See also CornellSilverman, p. 107. 

