Proposition

Let $A, B$ be rings, let $M$ be an $A$-module, $P$ a $B$-module and $N$ an $(A, B)$-bimodule. Then $M \otimes_A N$ is naturally a $B$-module, $N \otimes_B P$ an $A$-module, and we have

\[\begin{align*} (M \otimes_A N) \otimes_B P \simeq M \oplus (N \oplus_B P). \end{align*}\]

Solution

Part 1: Discussion on bimodules

$M \otimes_A N$ is a tensor product of two $A$-modules, so it is an $A$-module. $M \otimes_A N$ also has a $B$-module structure by defining the “scalar” multiplication by $b(\sum (m_i \otimes n_i)) = \sum (m \otimes bn_i)$. This satisfies the module axiom. For instance,

\[\begin{align*} (b_1b_2)(m \otimes n) &= m \otimes ((b_1b_2)n) \\ &= m \otimes (b_1(b_2n)) \\ &= b_1(m \otimes b_2n) \\ &= b_1(b_2(m \otimes n)). \end{align*}\]

It can be verified easily that this satisfies the other properties in the axiom.

Finally,

\[\begin{align*} a((\sum m_i \otimes n_i)b) &= a(\sum m_i \otimes (n_ib)) \\ &= \sum m_i \otimes a(n_ib) \\ &= \sum m_i \otimes (an_i)b \\ &= (\sum m_i \otimes (an_i))b \\ &= (a\sum m_i \otimes n_i)b, \end{align*}\]

so $M \otimes_A N$ is an $(A, B)$-bimodule.

A similar argument shows that $(M \otimes_A N) \otimes_B P, N \otimes_B P, M \otimes_A (N \otimes_B P)$ are all $(A, B)$-bimodules.

Part 2: Isomorphisms

Let $z \in P$ be given. Let $f_z:M \times N \rightarrow M \otimes_A (N \otimes_B P)$ be defined such that $f_z(x, y) = x \otimes (y \otimes z)$. We claim that $f_z$ is $A$-bilinear. For any $a \in A$ and $x, x’ \in M$,

\[\begin{align*} f_z(ax + x', y) &= (ax + x') \otimes (y \otimes z) \\ &= ax \otimes (y \otimes z) + x' \otimes (y \otimes z) \\ &= a(x \otimes (y \otimes z)) + x' \otimes (y \otimes z) \\ &= af_z(x, y) + f_z(x', y). \end{align*}\]

Similarly, $f_z(x, ay + y’) = af_z(x, y) + f_z(x, y’)$ for any $a \in A$ and $y, y’ \in N$.

Therefore, $f_z$ is $A$-bilinear. By Proposition 2.12 [Atiyah], there must exist an $A$-linear mapping $f_z’:M \otimes_A N \rightarrow (M \otimes_A (N \otimes_B P))$ such that $f_z(x, y) = f_z’(x \otimes y)$. Moreover, $f_z’$ is $B$-linear because

  1. \[\begin{align*} f_z'(x \otimes y + x' \otimes y) = f_z'(x \otimes y) + f(x' \otimes y) \end{align*}\]
  2. \[\begin{align*} f_z'(b(x \otimes y)) &= f_z'(x \otimes by) \\ &= x \otimes (by \otimes z) \\ &= x \otimes (b(y \otimes z)) \\ &= b(x \otimes (y \otimes z)) \\ &= bf_z'(x \otimes y). \end{align*}\]

The argument above works for any $z \in P$.

Now, we will define $f: (M \otimes_A N) \times B \rightarrow M \otimes_A (N \otimes_B P)$ such that $f(x \otimes y, z) = f_z’(x \otimes y)$. We claim that $f$ is $B$-bilinear.

For any $b \in B$, $x \otimes y, x’ \otimes y’ \in M \otimes_A N$,

\[\begin{align*} f(b(x \otimes y) + (x' \otimes y'), z) &= f_z'(b(x \otimes y) + (x' \otimes y')) \\ &= bf_z'(x \otimes y) + f_z'(x' \otimes y') \\ &= bf(x \otimes y, z) + f(x' \otimes y', z). \end{align*}\]

because $f_z’$ is $B$-linear as shown above.

TODO