# Linear Algebra over a Ring/Multilinear algebra

Definition (multilinear function):

Let ${\displaystyle R}$ be a ring, and let ${\displaystyle M_{1},\ldots ,M_{n},K}$ be ${\displaystyle R}$-modules. Then the set of ${\displaystyle R}$-multilinear functions from ${\displaystyle M_{1}\times \cdots \times M_{n}}$ to ${\displaystyle K}$ is the set

$\displaystyle L(M_1, \ldots, M_n, K) := \left\{ f: M_1 \times M_n \to K \middle| \forall j \in [n]: \forall m_1 \in M_1, \ldots, m_n \in M_n, l_j \in M_j, r \in R: f(m_1, \ldots, m_j + r l_j, \ldots, \right\}$ .

Proposition (equivalent definition of tensor product of free modules using multilinear functions):

Let ${\displaystyle R}$ be a ring, and let ${\displaystyle M_{1},\ldots ,M_{n}}$ be free, finitely generated ${\displaystyle R}$-modules. Then if we alternatively define

${\displaystyle M_{1}^{*}\otimes \cdots \otimes M_{n}^{*}:=L(M_{1},\ldots ,M_{n},R)}$,

and let the elementary tensors be ${\displaystyle m_{1}^{*}\otimes \cdots \otimes m_{n}^{*}(l_{1},\ldots ,l_{n}):=m_{1}^{*}(l_{1})\cdots m_{n}^{*}(l_{n})}$, then the ${\displaystyle M_{1}\otimes M_{n}}$ from this definition satisfies the same universal property as the usual tensor product ${\displaystyle M_{1}\otimes \cdots \otimes M_{n}}$. In particular, the two tensor products are canonically isomorphic.

Proof: For ${\displaystyle j\in [n]}$, let ${\displaystyle (e_{\lambda }^{j})_{\lambda \in \Lambda _{j}}}$ be a basis of ${\displaystyle M_{j}}$, where ${\displaystyle \Lambda _{j}}$ is the respective finite index set. Given any ${\displaystyle R}$-module ${\displaystyle K}$ and any multilinear map ${\displaystyle f:M_{1}\times \cdots \times M_{n}\to K}$, we want a unique linear function ${\displaystyle g:M_{1}\otimes \cdots \otimes M_{n}\to K}$ such that ${\displaystyle g\circ h=f}$, where ${\displaystyle h:M_{1}\times \cdots \times M_{n}\to M_{1}^{*}\otimes \cdots \otimes M_{n}^{*}}$ is the map that sends a tuple to the respective elementary tensor. ${\displaystyle \Box }$