Let $A$ and $B$ be abelian groups. For each $m>0$, show that $A\otimes Z_m \cong A/mA$.
And describe $A\otimes B$, when $A$ and $B$ are finitely generated.
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$(i)$ Since all elements of $ A \otimes Z_m$ is of the form $a\otimes 1$ for $a\in A$, there is a surjective map $\varphi : A\rightarrow A\otimes Z_m$ such that $\varphi (a)=a\otimes 1$. So, If $\ker (\varphi)=mA$, proof is done.
$(ii)$ I know that finitely generated abelian group is free or of the form
$$Z_{m_1}\oplus \cdots\oplus Z_{m_t}\oplus F $$ where $m_1 >1$ and $m_1 |m_2| \cdots | m_t$ and $F$ free abelian
Can you help me?
Best Answer
In general, if $A$ is a commutative ring, $I$ an ideal, and $M$ an $A$-module, then $A/I\otimes_A M\simeq M/IM$.
To see this, consider the bilinear mapping $A/I\times M\to M/IM$ given by $(\bar{x},m)\mapsto \overline{xm}$. By the universal property of the tensor product, this induces a well-defined map $A/I\otimes_A M\to M/IM$ given by $\bar{x}\otimes m\mapsto \overline{xm}$.
An explicit inverse is given by $\psi\colon M\to A/I\otimes M$ by $m\mapsto \bar{1}\otimes m$. This contains $IM$ in the kernel, so descends to a well defined inverse $M/IM\to A/I\otimes M$ .
So in this case, $$ A\otimes \mathbb{Z}_m=A\otimes \mathbb{Z}/(m)\simeq A/(m)A=A/mA $$
For when $A$ and $B$ are finitely generated, recall that $\otimes$ distributes over $\oplus$. Also, it is well known that $\mathbb{Z}_m\otimes\mathbb{Z}_n\simeq\mathbb{Z}_{(m,n)}$, and $\mathbb{Z}\otimes_\mathbb{Z} A\simeq A$ for any $\mathbb{Z}$-module $A$. After expanding out your tensor of direct sums, this will account for all terms.