let $G$ be an abelian group. suppose that $G\otimes \mathbb Q=0$. Does this imply that $G$ is necessarily a torsion group?
$G\otimes \mathbb Q=0$ implies torsion group
3
$\begingroup$
group-theory
abelian-groups
tensor-products
-
1Your accounts have been merged. If you're having trouble logging into your old account, it's because you're unregistered; that $p$roblem will be solved by registering. – 2011-05-18
1 Answers
11
Yes. Note that $G\otimes_\mathbb{Z}\mathbb{Q}\cong (\mathbb{Z}\setminus 0 )^{-1}G$ as $\mathbb{Z}$-modules. Now $\frac{g}{n}=\frac{0}{1}$ by definition means $zg=0$ in $G$ for some $z\in\mathbb{Z}\setminus 0$.
-
2Just want to mention this is true in greater generality (and the proof is essentially the same): if $D$ is an integral domain with field of fractions $F$, and $M$ is any $D$-module, then the kernel of the natural map $M\rightarrow M\otimes F$ is the torsion submodule of $M$. – 2011-05-20