A natural isomomorphism of $G$-modules

Question

Let $A$ be a $G$ module, $\Bbb Z$ regarded as a trivial $\Bbb ZG$ module . Then $$ \Bbb Z \otimes _G A \rightarrow A/\mathcal{G}A, \quad m\otimes a \mapsto ma+ \mathcal{G}A $$ is an isomorphism. $\mathcal{G}$ is the defined as the kernel of the augmentation map, $$ 0 \rightarrow \mathcal{G} \rightarrow \Bbb Z G \xrightarrow{\epsilon} \Bbb Z \rightarrow 0 $$ $\epsilon (\sum m_g g): = \sum m_g $

So my strategy to this problem is $-\otimes_G A$ the exact sequence. So that $\Bbb Z G \otimes A/\mathcal{G} \otimes A \cong \Bbb Z \otimes A$

So I guess the general question is: if $I$ is an ideal of ring $R$, $A$ is an $R$ module $I \otimes_R A \cong IA$? But this does not seem to be true in general.

Answer 1

It's not necessary to have $I\otimes A\simeq IA$ for it to work.

Just note that the image of $I\otimes A$ under $i\otimes a\mapsto ia$ is $IA$ - this is easy.

Then you have an exact sequence $\mathcal{G}\otimes A\to A\to \mathbb{Z}\otimes A\to 0$ and the image of the first map is $\mathcal{G}A$; therefore the cokernel of this first map is $A/\mathcal{G}A$

Note that in general for an ideal $I$, you have a short exact sequence $0\to I\to R\to R/I\to 0$, so that tensoring with an $R$-module $A$ yields an exact sequence $\mathrm{Tor}^R_1(R/I,A) \to I\otimes A\to A\to A/IA\to 0$; so the kernel of the map $I\otimes A\to IA$ is $\mathrm{Tor}^R_1(R/I,A)$ : it may not be trivial.

Answer 2

This is not a special property of $G$-modules.

Let $R$ be a ring (not necessarily commutative), $I$ a right ideal of $R$ and $M$ a left $R$-module. Then $$ R/I\otimes_R M\cong M/IM $$ as abelian groups.

Just show that $M/IM$ has the required universal property of the tensor product.