Abelian group is torsion if tensor product with reals is zero. I am confused about the fact that if the tensor product of an abelian group (thought of as a $\mathbb{Z}$-module) with the reals is zero, then the group has only torsion elements. I know that this question was also addressed for the rationals in An abelian group A is torsion iff A ⊗ Q = 0, but I do not understand the answer (specifically, why is $a \otimes 1$ nonzero if $na$ is nonzero for all $n$?) which is probably because I do not understand products of modules well enough yet. What is so specific about $\mathbb{Q}$ (or $\mathbb{R}$, for that matter) that makes this work? Can someone please clarify this, preferably only using elementary notions like the universal property of tensor product modules? Thanks. 
 A: Here is a fairly simple argument directly from the universal property.  Given an abelian group $A$, define a group $F$ of "fractions" $\frac{a}{n}$ where $a\in A$ and $n\in\mathbb{Z}\setminus\{0\}$.  We impose an equivalence relation on these fractions by saying $\frac{a}{n}=\frac{b}{m}$ iff there exists $k\in\mathbb{Z}\setminus\{0\}$ such that $k(ma-nb)=0$.  Using the usual formula for addition of fractions, you can check that the set $F$ of equivalence classes of fractions forms an abelian group.
There is now a bilinear map $f:A\times\mathbb{Q}\to F$ defined by $f(a,m/n)=\frac{am}{n}$.  This induces a homomorphism $g:A\otimes\mathbb{Q}\to F$ which sends $a\otimes 1$ to the fraction $\frac{a}{1}$.  So to show that $a\otimes 1\neq 0$, it suffices to show the fraction $\frac{a}{1}$ is nonzero.  The zero element of $F$ is the fraction $\frac{0}{1}$, and $\frac{a}{1}=\frac{0}{1}$ iff there exists $k\in\mathbb{Z}\setminus\{0\}$ such that $k(1\cdot a-1\cdot 0)=0$, or in other words such that $ka=0$.  So if $a$ is a non-torsion element, then $a\otimes 1\neq 0$.  (In fact, this map $g$ is actually an isomorphism, but that takes more work to prove.)
The case of $\mathbb{R}$ follows from the case of $\mathbb{Q}$ since $\mathbb{Q}$ is a direct summand of $\mathbb{R}$ and tensor products distribute over direct sums.

From a broader perspective, what's really going on here is that $\mathbb{Q}$ (or $\mathbb{R}$) is a flat $\mathbb{Z}$-module, meaning that if $A$ is a $\mathbb{Z}$-module and $B\subseteq A$ is a submodule, then the map $B\otimes\mathbb{Q}\to A\otimes\mathbb{Q}$ induced by the bilinear map sending $(b,q)\in\mathbb{Q}$ to $b\otimes q\in A\otimes\mathbb{Q}$ is injective.  So, in particular, if $a\in A$ is any element such that $na\neq 0$ for all $n$, the submodule $B\subseteq A$ generated by $a$ is isomorphic to $\mathbb{Z}$.  Since $\mathbb{Z}\otimes\mathbb{Q}\cong\mathbb{Q}$ and we have an injective map $\mathbb{Z}\otimes\mathbb{Q}\cong B\otimes\mathbb{Q}\to A\otimes\mathbb{Q}$, this implies $A\otimes\mathbb{Q}$ is nontrivial (and in fact $a\otimes1\neq 0$, since it is the image of the nonzero element $1\otimes 1\in \mathbb{Z}\otimes\mathbb{Q}$).
The argument using fractions above can be generalized to show that if $R$ is any commutative ring, then any localization of $R$ is flat as an $R$-module.  For $\mathbb{Z}$-modules, flatness can be characterized quite simply in general: a $\mathbb{Z}$-module is flat iff it is torsion-free.  See Show that a Z-module A is flat if and only if it is torsion free? for some sketches of the proof.
A: Both $\Bbb Q$ and $\Bbb R$ are divisible Abelian groups, that is, $\forall x, \, \forall n\in\Bbb N, \ n>0\ \exists y: ny=x$. 
Now let $G$ be a divisible Abelin group,  and $a\in A$ a torsion element with $na= 0$. Then for any $x\in G$, we have
$$ a\otimes x = a\otimes(ny) =(na) \otimes y= 0$$
(Here I moved the 'scalar' $n\in\Bbb Z$ over the tensor sign.) 
So, if $A$ is a torsion group, $A\otimes G=0$. 
