Lemma 10.12.7. For $A$-module $M$, $B$-module $P$ and $(A, B)$-bimodule $N$, the modules $(M \otimes _ A N)\otimes _ B P$ and $M \otimes _ A(N \otimes _ B P)$ can both be given $(A, B)$-bimodule structure, and moreover

$(M \otimes _ A N)\otimes _ B P \cong M \otimes _ A(N \otimes _ B P).$

Proof. A priori $M \otimes _ A N$ is an $A$-module, but we can give it a $B$-module structure by letting

$(x \otimes y)b = x \otimes yb, \quad x\in M, y\in N, b\in B$

Thus $M \otimes _ A N$ becomes an $(A, B)$-bimodule. Similarly for $N \otimes _ B P$, and thus for $(M \otimes _ A N)\otimes _ B P$ and $M \otimes _ A(N \otimes _ B P)$. By Lemma 10.12.5, these two modules are isomorphic as both as $A$-module and $B$-module via the same mapping. $\square$

Comment #393 by Fan on

In the first line, $P$ and $N$ are not in mathmode.

Comment #398 by on

Good catch! I wonder if there is a way to find all of these using a python script?

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