By tensoring with $\Bbb Q$ a short exact sequence, why does $A/B$ have rank $0$? Let $A, B$ be free abelian groups of rank $n$. That is, $A$ and $B$ are both isomorphic to $\Bbb Z^n.$
Now, $0\to B\to A\to A/B\to 0$ is short exact sequence.
By tensoring with $\Bbb Q$, we get 
$$0\to B\otimes \Bbb Q\to A\otimes \Bbb Q \to A/B\otimes \Bbb Q\to 0.$$
Thus, $A/B$ has rank $0$. 
But why?
I cannot follow logic.
I guess $A\otimes \Bbb Q\to A/B\otimes \Bbb Q$ is the zero mapping, but I cannot prove this claim.
And, in the proof above, the properties of short exact sequence seem to be not used. What is the merit of using short exact sequence in this situation?
Thank you very much for your kind help.
 A: Since $A, B$ are free abelian groups of rank $n$, both $B\otimes_{\mathbb{Z}}\mathbb{Q}$ and $A\otimes_{\mathbb{Z}}\mathbb{Q}$ are $n$-dimensional vector spaces over $\mathbb{Q}$. Next you accepted the fact that the sequence
$$0\rightarrow B\otimes_{\mathbb{Z}}\mathbb{Q} \rightarrow A\otimes_{\mathbb{Z}}\mathbb{Q}\rightarrow A/B\otimes_{\mathbb{Z}}\mathbb{Q} \rightarrow 0$$ 
is exact, which is nontrivial. Tensor product $(-)\otimes_{\mathbb{Z}}C$ does not in general preserve left exactness (it always preserves right exactness). It does iff you are tensoring with torsion-free abelian group $C$ and $\mathbb{Q}$ is indeed torsion-free. Since you accepted this fact, note that part of your exact sequence is a monomorphism $B\otimes_{\mathbb{Z}}\mathbb{Q} \rightarrow A\otimes_{\mathbb{Z}}\mathbb{Q}$ between two $\mathbb{Q}$-vector spaces of the same dimension. This implies that this monomorphism is an isomorphism and hence $A/B\otimes_{\mathbb{Z}}\mathbb{Q} = 0$. 
Machinery of short exact sequences is just a convenient language to speak about this particular subject matter. Link you provided contains arguments formulated in other but equivalent languages (they are translatable).
