Lemma 10.12.7 (00D2)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.12: Tensor products
Lemma 10.12.7 (cite)

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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. $\square$

Comments (2)

Comment #393 by Fan on December 15, 2013 at 22:44

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

Comment #398 by Johan on December 19, 2013 at 02:30

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

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