Does $A\otimes \Bbb Q=0$ imply finiteness of $A$? Let $A$ be an abelian group. Does $A\otimes \Bbb Q=0$ imply finiteness of $A$?
If $A$ is finitely generated it's clear from the structure theorem but I am not sure about the general case.

No, the condition $A\otimes\mathbb{Q}=0$ is equivalent to $A$ being torsion.
It's easy to prove that $B\otimes\mathbb{Q}=0$ when $B$ is torsion.
Now, if $t(A)$ denotes the torsion part of $A$, from the exact sequence
$$
0\to t(A)\to A\to A/t(A)\to 0
$$
we deduce the exact sequence
$$
t(A)\otimes\mathbb{Q}\to A\otimes\mathbb{Q}
\to A/t(A)\otimes\mathbb{Q}\to 0
$$
and so $A\otimes\mathbb{Q}=0$ implies $A/t(A)\otimes\mathbb{Q}=0$. However, a nonzero torsion free group $C$ contains (a copy of) $\mathbb{Z}$, and so $C\otimes\mathbb{Q}\ne0$ (here use the fact that $\mathbb{Q}$ is flat).
Thus $A\otimes\mathbb{Q}=0$ if and only if $A/t(A)$ (which is torsion free) is zero, that is, $A$ is torsion. However, $\mathbb{Q}/\mathbb{Z}$ is torsion and infinite.
