----- > [!proposition] Proposition. ([[when is an element of a tensor product nonzero?]]) > Let $R$ be a ([[commutative ring|commutative]]) [[ring]] and $M,N$ be $R$-[[module|modules]]. Consider the [[tensor product of modules|tensor product]] $M \otimes_{R}N$. > An element $\sum_{i=1}^{\ell}m_{i} \otimes n_{i}$ in $M \otimes_{R} N$ equals zero if and only if for every $R$-[[module]] $L$ and $R$-[[bilinear map|bilinear map]] $f:M \times N \to L$ we have $\sum_{i=1}^{\ell} f(m_{i}, n_{i})=0$. ^proposition > [!proposition] Corollary. > Let $R^{\oplus n}$ and $R^{\oplus m}$ be [[free module|free modules]] with [[basis|bases]] $\{ e_{1},\dots,e_{m} \}$, $\{ f_{1},\dots,f_{n} \}$ respectively. [[tensor product of modules#^note|Manifestly,]] $R^{\oplus n} \otimes_{R} R^{\oplus m}$ is [[submodule generated by a subset|generated by]] $\mathcal{B}=\{ e_{i} \otimes f_{j} \}_{i=1,\dots,_{m}}^{j=1,\dots,n}$. In fact, $\mathcal{B}$ is a [[basis]] for $R^{\oplus n} \otimes_{R} R^{\otimes m}$. Indeed, define maps $f_{ij}:R^{\oplus n} \times {R}^{\oplus m} \to R$ as $f_{ij}=\pi_{i}\pi_{j}$ for $\pi_{i}$, $\pi_{j}$ the obvious coordinate projections. If $\sum_{i \in [m], j \in [n]}c_{ij} (e_{i} \otimes f_{j})=0$, then by the proposition we have $\sum_{i \in [m], j \in [n]} c_{ij} f_{ij}(e_{i}, f_{j})=0$ for all $i,j$, that is, $c_{ij}=0$ for all $i,j$. [[Linearly independent|Linear independence]] follows. > Thus, $R^{\oplus n} \otimes_{R} R^{\otimes m}\cong R^{\oplus mn}$. ^proposition > [!proof]- Proof. ([[when is an element of a tensor product nonzero?]]) Per the [[universal property]], $f=h \circ i_{M \otimes N}$ for an $R$-[[linear map|linear map]] $h:M \otimes_{R} N \to L$; so: > >$\begin{align}h\left( \sum_{i=1}^{\ell} m_{i} \otimes n_{i} \right)&=\sum_{i=1}^{\ell} h(m_{i} \otimes n_{i}) \\ &= \sum_{i=1}^{\ell} h \circ i_{M \otimes N} (m_{i}, n_{i}) \\ &= \sum_{i=1}^{\ell} f(m_{i}, n_{i}). \end{align}$ >It is clear from this that if $\sum_{i=1}^{\ell}m_{i} \otimes n_{i}=0$, $\sum_{i=1}^{\ell} f(m_{i}, n_{i})=h(0)=0$ for every $f$. Now conversely and contrapositively assume $\sum_{i=1}^{\ell}m_{i} \otimes n_{i} \neq 0$. Then choosing $f=i_{M \otimes N}$ induces $h$ to be the identity, so $\sum_{i=1}^{\ell}f(m_{i}, n_{i}) \neq 0$. ----- #### ---- #### 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 > ```