Let $A$ be a torsion abelian group. Then $A$ has no $p$-torsion iff $A \otimes \Bbb Z_{(p)} = 0$.

Question

Let $A$ be a torsion abelian group, $p$ a prime. Then $A$ has no $p$-torsion iff $A \otimes \Bbb Z_{(p)} = 0$.

I could prove one directioin $\Rightarrow$. But how does one prove the converse?

Answer 1

Lets consider the exact sequence $0\rightarrow T\rightarrow A\xrightarrow{\times p} A\rightarrow Q\rightarrow 0$, where $T$ is our $p$ torsion, and $Q$ is the quotient. Tensoring with $\mathbb{Z}_{(p)}$ is exact, so we have the following exact sequence:

$0\rightarrow T\otimes \mathbb{Z}_{(p)}\rightarrow A\otimes \mathbb{Z}_{(p)}\xrightarrow{\times p} A\otimes \mathbb{Z}_{(p)}\rightarrow Q\otimes \mathbb{Z}_{(p)}\rightarrow 0$.

Since $T$ is $p$-torsion, every $n$ coprime to $p$ already acts invertibly on $T$, so this left term is just $T$. So we see that if $A\otimes \mathbb{Z}_{(p)}$ is zero, then $T$ must be zero by exactness, giving the other direction.