Let $A_0,\dots,A_n$ be algebras over some fixed commutative ring $k$ (you may assume $k=\mathbb{Z}$ for simplicity). Let $M_i$ be an $(A_{i-1},A_i)$-bimodule for $i=1,\dots,n$. A multilinear map from $M_0\times\dots\times M_n$ (identifying a bimodule with its underlying set) to an $(A_0,A_n)$-bimodule $T$ is a map which is $k$-linear in every argument and such that $\alpha f(m_0,\dots,m_n)=f(\alpha m_0,m_1,\dots,m_n)$, $f(m_0,\dots,m_{i-1}\alpha,m_i,\dots,m_n)=f(m_0,\dots,m_i,\alpha m_{i+1},\dots,m_n)$ and $f(m_0,\dots,m_n\alpha)=f(m_0,\dots,m_n)\alpha$ whenever it makes sense (with a slight variation in the $n=0$ case).

Define the tensor product $M_1\otimes\dots\otimes M_n$ as an $(A_0,A_n)$-bimodule with a multilinear map into it, such that every other multilinear map into a bimodule $T$ factors uniquely $A_0-A_n$-linear over it.

One sees that for $n=0$ the tensor product is $A_0$ equipped with the obvious bimodule structure and for $n=1$ the tensor product ist $M_1$.

Is there an obvious way to see that $(M_1\otimes M_2)\otimes M_3\cong M_1\otimes M_2\otimes M_3$?

Of course, one needs an isomorphism of functors $\operatorname{Mult}(M_1\otimes M_2,M_3;-)\cong\operatorname{Mult}(M_1,M_2,M_3;-)$. The morphism $\to$ is easy, but starting from a multilinear map $M_1\times M_2\times M_3\longrightarrow T$, how do i get a multilinear map $M_1\otimes M_2\times M_3\longrightarrow T$?

  • $\begingroup$ Associativity for three factors is standard and essentially obvious; the general case follows by induction. $\endgroup$ – egreg Apr 25 '16 at 15:58

Given a trilinear map $\mu:M_1\times M_2\times M_3\to T$, for each $m_3\in M_3$ the map $(m_1,m_2)\mapsto \mu(m_1,m_2,m_3)$ is an $(A_0,A_1,k)$-bilinear map $M_1\times M_2\to T$. There is thus a unique map $\nu:(M_1\otimes_{A_1} M_2)\times M_3\to T$ such that $\nu(m_1\otimes m_2,m_3)=\mu(m_1,m_2,m_3)$ and $\nu$ is $A_0$-linear in the first variable. It is straightforward to check that $\nu$ is also $A_2$-balanced and $A_3$-linear in the second variable (since $\mu$ is $A_2$-balanced and $A_3$-linear), and so $\nu$ is $(A_0,A_2,A_3)$-bilinear. It is now straightforward to see that this construction is inverse to the obvious construction taking a bilinear map $(M_1\otimes_{A_1} M_2)\times M_3\to T$ to a trilinear map $M_1\times M_2\times M_3\to T$ (namely, sending $\nu$ to the map $\mu(m_1,m_2,m_3)=\nu(m_1\otimes m_2,m_3)$).

  • $\begingroup$ Thank you! But how can there be a bilinear map $M_1\times M_2\to T$, if $T$ is a $(A_0,A_3)$-module and $M_2$ is a $(A_1,A_2)$-bimodule? $\endgroup$ – user114885 Apr 25 '16 at 18:07
  • $\begingroup$ I have edited to clarify exactly what sort of linearity I am talking about at each step. $\endgroup$ – Eric Wofsey Apr 25 '16 at 18:18

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