----- > [!proposition] Proposition. ([[the tensor product of torsion-free abelian groups is torsion-free]]) > Let $R$ be a [[commutative ring|commutative]] [[ring]]. Let $M,N$ be $R$-[[module|modules]]. Suppose we have a zero element of the [[tensor product of modules|tensor product]] $M \otimes_{R} N$, $\sum_{i}m_{i} \otimes n_{i}=0$. Then there exist [[submodule generated by a subset|finitely generated]] [[submodule|submodules]] $M' \subset M$, $N' \subset N$, such that the same expression $\sum_{i}m_{i} \otimes n_{i}=0$ is true in $M' \otimes _R N'$. ^proposition > [!basicnonexample] Warning. > The statement is that *there exist* such submodules. It is not true that any pair of submodules will do. > - [ ] counterexample in written notes ^nonexample > [!proposition] Corollary. > Let $A$ and $B$ be [[torsion element of a module|torsion-free]] [[abelian group|abelian groups]]. Then $A \otimes_{\mathbb{Z}}B$ is a torsion-free abelian group. > > > Indeed, assume $n \sum_{i=1}^{\ell}a_{i} \otimes b_{i}=0$ in $A \otimes_{\mathbb{Z}} B$ for some $n >0$. The proposition supplies finitely generated subgroups $A' \subset A$, $B' \subset B$ such that this expression holds in $A' \otimes_{\mathbb{Z}} B'$. But $A'$ and $B'$ are torsion-free [[free abelian group|finitely generated]] abelian groups, and thus $A \cong \mathbb{Z}^{k}$ and $B' \cong \mathbb{Z}^{\ell}$ for some $k,\ell \geq 0$. Thus $A' \otimes B' \cong \mathbb{Z}^{k \ell}$ is torsion-free, and so $\sum_{i=1}^{\ell} a_{i} \otimes b_{i}=0$ > when interpreted in $M' \otimes N'$. But then this expression also holds in $M \otimes N$, thus $M \otimes N$ is torsion-free. > [!proof]- Proof. ([[the tensor product of torsion-free abelian groups is torsion-free]]) > [[tensor product of modules|Recalling]] the notation of the explicit tensor product definition, saying $\sum_{i}m_{i} \otimes n_{i}=0$ is saying $\sum_{i}j(m_{i}, n_{i}) \in K$, where $j(m_{i}, n_{i})$ gives the standard basis of $F^{R}(M \times N)=R^{\oplus(M \times N)}$. Writing each $j(m_{i},n_{i})$ as a [[linear combination]] of [[submodule generated by a subset|generators]] of $K$, we moreover obtain $\sum_{i}j(m_{i}, n_{i})=\sum_{i} k_{i}$ for $k_{i}$ a generator of $K$. The expressions for the $k_{i}$ involve finitely many elements of $M$ and $N$, generating submodules $M',N'$ and the above equation holds when interpreted in $R^{\oplus(M' \times N')}$. The result follows. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```