$\mathbb{Z}$ has no torsion? What does is mean to say that $\mathbb{Z}$ has no torsion? 
This is an important fact for any course?
Thanks, I heard that in my field theory course, but I don't know what it is. 
 A: It means that 
$$\forall\,n,z\in\Bbb Z\,\,,\,n\neq 0\,\,,\,nz=0\Longrightarrow z=0$$
And yes, it is a rather important notion in group theory in general.
A: A group $G$ has torsion if there are elements $g\neq0\in G$ such that $ng=0$ for some $n\neq 0$ (depending on $g$).  In $\mathbb Z$, we have that $ng\neq 0$ for all $n,g\neq 0$ and so $\mathbb Z$ is torsion free.
A: @d555, you might want to know that the notion of torison is extremely important, for example if $A$ is finitely generated and abelian group, then it can be written as the direct sum of its torsion subgroup $T(A)$ and a torsion-free subgroup (but this is not true for all infinitely generated abelian groups). $T(A)$ is uniquely determined.
A: In general, given an $R$-module $M$, an element $m \in M$ is called a torsion element if there exists some non zero $r \in R$ such that $rm=0$. Here I denote the set of torsion elements of $M$ is denoted $\text{Tor}(M)$, although I have also seen it denoted by $T(M)$. If $\text{Tor}(M)=0$, then $M$ is said to be torsion free. 
For any ring $R$, one can think of $R$ as an $R$-module over itself, where scalar multiplication is simply the ring multiplication. In your case you are looking at the ring $\mathbb{Z}$, which is certainly a $\mathbb{Z}$-module. So the torsion elements of $\mathbb{Z}$ would be the set $\text{Tor}(\mathbb{Z})$. If $a \in \text{Tor}(\mathbb{Z})$, then there exists some $n\in \mathbb{Z}$ such that $na=0$. Since $\mathbb{Z}$ is an integral domain (the prototypical one in fact), the equation $na=0$ forces either $n=0$ or $a=0$. Thus $\text{Tor}(\mathbb{Z})=\{0\}$.
