If $A$ is a torsion abelian group and $D$ is a divisible abelian group, show that $A\otimes_{\mathbb{Z}}D= 0$.
My solution:
since $A\otimes_{_\mathbb{Z}}\mathbb{Z} \cong A, D \cong D\otimes_{_\mathbb{Z}}\mathbb{Z} $, and $\mathbb{Z} \otimes_{_\mathbb{Z}} A \cong A, D \cong \mathbb{Z} \otimes_{_\mathbb{Z}} D,$ I would have the requirment holds.. is that true?
Thanks for any help..
Best Answer
This is wrong. You do not prove that the tensor product is $0$. Proof: $$a\otimes d = a\otimes nd'= na\otimes d=0\otimes d'=0. $$ Here $na=0$ ($n$ exists since $A$ is torsion) and $nd'=d$ ($d'$ exists since $D$ is divisible).