If $\operatorname{Hom}(A,\mathbb{Q})=0$ and $\operatorname{Ext}(A,\mathbb{Z}_p)=\operatorname{Hom}(A,\mathbb{Z}_p)=0$ implies $A=0$

algebraic-topologyhomological-algebra

I want to prove that

If $A$ is an abelian group such that
$$\operatorname{Hom}(A,\mathbb{Q})=0$$
and,
$$\text{Ext}(A,\mathbb{Z}_p)=\text{Hom}(A,\mathbb{Z}_p)=0 \text{ for every prime } p$$
then $A=0$.

I want to follow the reasoning in the response of

Does trivial cohomology imply trivial homology? Does $\operatorname{Hom}(A,\mathbb Z) = \operatorname{Ext}^1(A, \mathbb Z) = 0$ imply $A = 0$?

and to do that I have to show that $\text{Hom}(A,\mathbb{Z})=0$ and $\text{Ext}(A,\mathbb{Z})=0$.

I have not been able to do such a thing. But from the exact sequences of $\text{Ext}$ it is easy to deduce that $\text{Hom}(A,\mathbb{Z})=0$ and that $\text{Ext}(A,\mathbb{Z})$ is a torsionfree divisibility group. From this two conclusions it is possible to deduce that $A$ is a torsionfree divisibility group but I could not reach $A=0$.

Any advice?

PD: This problem appears in Hatcher in a more topological way.

Best Answer

Since $\operatorname{Ext}(-,\mathbb{Z}_p)$ turns injections into surjections, $\operatorname{Ext}(B,\mathbb{Z}_p)=0$ for every subgroup $B\subseteq A$. In particular, this means $A$ has no subgroup isomorphic to $\mathbb{Z}_p$. Taken over all primes $p$, this means that $A$ is torsion-free.

Since $A$ is torsion-free, the canonical map $i:A\to A\otimes\mathbb{Q}$ is injective. If $A$ were nonzero, then for any nonzero element $a\in A$ we could get a homomorphism $A\otimes\mathbb{Q}\to \mathbb{Q}$ which is nonzero on $i(a)$, since $A\otimes\mathbb{Q}$ is a $\mathbb{Q}$-vector space. This is impossible since $\operatorname{Hom}(A,\mathbb{Q})=0$, so we must have $A=0$.

(Note that the assumption that $\operatorname{Hom}(A,\mathbb{Z}_p)=0$ is not needed.)

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