3

As we know $\mathbb Z/n \mathbb Z \otimes_{\mathbb Z} \mathbb Z/m\mathbb Z \cong Hom(\mathbb Z/n \mathbb Z, \mathbb Z/m\mathbb Z)$. Can we generalize this result, for example, is it true that $A\otimes_{K}B \cong Hom(A,B)$, where A and B are finite modules over a ring K? Or perhaps you have any idea about other sufficient conditions for this isomorphism to be true?

anna abasheva
  • 891
  • 1
    "As we know": I didn't. That's one nice isomorphism to confuse undergrads with! I fear this cannot be generalized very far. I suspect it will hold for finitely-generated torsion modules over PIDs. – darij grinberg Jan 02 '15 at 00:29
  • It's definitely not true in general, even for abelian groups: $A=\mathbb{Z}/2\mathbb{Z}$ and $B=\mathbb{Z}$. – egreg Jan 02 '15 at 01:07
  • 1
    It would imply $\DeclareMathOperator\Hom{Hom} \Hom(A,B)\simeq\Hom(B,A)$ since $A\otimes B\simeq B\otimes A$, which is clearly false in general. – Bernard Jan 02 '15 at 01:29

0 Answers0