1.3 The Tensor Product
Definition 1.3.1 (Tensor Product). Let $R$ be a commutative ring and $\seqf{E_j}$ be $R$ modules, then there exists a pair $(\bigotimes_{j = 1}^{n} E_{j}, \iota)$ such that:
$\bigotimes_{j = 1}^{n} E_{j}$ is an $R$-module.
$\iota: \prod_{j = 1}^{n} E_{j} \to \bigotimes_{j = 1}^{n} E_{j}$ is a $n$-linear map.
For any pair $(F, \lambda)$ satisfying (1) and (2), there exists a unique $\Lambda \in \hom(\bigotimes_{j = 1}^{n} E_{j}; F)$ such that the following diagram commutes:
\[\xymatrix{ \prod_{j = 1}^n E_j \ar@{->}[rd]_{\lambda} \ar@{->}[r]^{\iota} & \bigotimes_{j = 1}^n E_j \ar@{->}[d]^{\Lambda} \\ & F }\]$\bigotimes_{j = 1}^{n} E_{j}$ is the linear span of $\iota(\prod_{j = 1}^{n} E_{j})$.
The module $\bigotimes_{j = 1}^{n} E_{j}$ is the tensor product of $\seqf{E_j}$, and $\iota: \prod_{j = 1}^{n} E_{j} \to \bigotimes_{j = 1}^{n} E_{j}$ is the canonical embedding. For any $(x_{1}, \cdots, d_{n}) \in \prod_{j = 1}^{n} E_{j}$, the image
is its tensor product.
Proof. Let $M$ be the free module generated by $\prod_{j = 1}^{n}E_{j}$, and $N \subset M$ be the submodule generated by elements of the following form:
For any $1 \le j \le n$, $(x_{1}, \cdots, x_{n}) \in \prod_{k = 1}^{n} E_{k}$, and $x_{j} \in E_{j}$,
\[(x_{1}, \cdots, x_{j} + x_{j}', \cdots, x_{n}) - (x_{1}, \cdots, x_{j}, \cdots, x_{n}) - (x_{1}, \cdots, x_{j}', \cdots, x_{n})\]For any $(x_{1}, \cdots, x_{n}) \in \prod_{k = 1}^{n} E_{k}$ and $\alpha \in R$,
\[(x_{1}, \cdots, \alpha x_{j}, \cdots, x_{n}) - \alpha(x_{1}, \cdots, x_{n})\]
(1), (2): Let $\bigotimes_{j = 1}^{n} E_{j} = M/N$ and
then by definition of $N$, $\iota$ is $n$-linear.
(U): Let $(F, \lambda)$ be a pair satisfying (1) and (2), then $\lambda$ admits a unique extension to a linear map $\ol \lambda: M \to F$. Since $\lambda$ is $n$-linear, $\ker \ol \lambda \supset N$. By the first isomorphism theorem, there exists a unique $\Lambda \in \hom(\bigotimes_{j = 1}^{n} E_{j}; F)$ such that the given diagram commutes.
(4): Since $M$ is the free module generated by $\prod_{j = 1}^{n} E_{j}$, $M/N$ is generated by $\iota(\prod_{j = 1}^{n} E_{j})$.$\square$