I have worked out a solution to exercise 2.15 of Atiyah-Macdonald, which is needed in the solution of 2.3 (see Atiyah-Macdonald 2.3). However, the solution seems overly complicated, and I am not entirely sure about the argument in part 2. Any corrections and/or improvements would be appreciated.


Let $A, B$ be rings (commutative, with $1$). Let $M$ be a module over $A$ and $P$ be a module over $B.$ Let $N$ be a bimodule over $A$ and $B$; that is, $N$ is simultaneously a module over $A$ and over $B$ and the operations are compatible in the sense that $(ax)b=a(xb)$ for $x\in N, a \in A, b \in B.$

Then $(M \otimes_A N) \otimes_B P \sim M \otimes_A ( N \otimes_B P).$ Here “$\sim$” denotes isomorphism, where both sides are viewed bimodules over $A$ and $B$.

Attempt at solution:

Step 1: For fixed $p \in P,$ define the map $f_p: M \times N \to M \otimes_A (N \otimes_B P)$ which takes $(m, n) \mapsto m \otimes_A (n \otimes_B p).$ It’s easy to see that $f_p$ is $A-$multilinear. Hence it induces a map $f’_p: M \otimes_A N \to M \otimes_A (N \otimes_B P)$ which is a homomorphism of $A-$modules such that $f’_p(m \otimes n) = m \otimes_A (n \otimes_B p)$.

Step 2: Define now $g: (M\otimes_A N) \times P \to M \otimes_A ( N \otimes_B P)$ by $ g(x,y)= f_y’(x).$ We want to show that $g$ is $B-$multilinear.

We show first that $g$ is $B-$multilinear in the first component. Note that an element $x\in M\otimes_A B$ is of the form $x = \sum_1^{n} m_i \otimes_A n_i,$ with the $m_i \in M$ and $n_i \in N.$

Hence, for $\lambda \in B,$ we have $\lambda x_0+ x_1= \lambda \sum_1^{n_0} m^0_i \otimes_A n^0_i + \sum_1^{n_1} m^1_i \otimes_A n^1_i = \sum_1^{n_0} m^0_i \otimes_A \lambda n^0_i + \sum_1^{n_1} m^1_i \otimes_A n^1_i.$

Hence, we have

\begin{align} g(\lambda x_0+x_1, p)&= g( \sum_1^{n_0}(m^0_i \otimes_A \lambda n^0_i) + \sum_1^{n_1}(m^1_i \otimes_A n^1_i) , p) \\ &=\sum_1^{n_0} m^0_i \otimes_A (\lambda n^0_i \otimes_B p) + \sum_1^{n_1} m^1_i \otimes_A (n^1_i \otimes_B p) \\ &=\lambda \sum_1^{n_0} m^0_i \otimes_A (n^0_i \otimes_B p) + \sum_1^{n_1} m^1_i \otimes_A (n^1_i \otimes_B p) \\ &= \lambda g(x_0,p)+ g(x_1,p), \end{align}

where we have used the fact that $g(\cdot,y)=f’_y(\cdot)$ is $A-$linear.

We can show that $g$ is $B-$linear in the second component by a similar argument.

Step 3: To show this is an isomorphism, we construct an inverse mapping by the same procedure.

  • $\begingroup$ The part of the proof that you have shown is correct; its main weakness is that it doesn't directly generalize to the case of noncommutative rings and bimodules (though, of course, it's easy to tweak it so that it does -- basically, multilinearity over a noncommutative $B$ is no longer something you can split up in $B$-linearity in the first component and $B$-linearity in the second component). But keep in mind that you don't just have to construct the inverse but you also need to show that it's an inverse; I hope you know how this works? $\endgroup$ – darij grinberg Jul 27 '14 at 10:42
  • $\begingroup$ Oh, and you should also show that the maps in the two directions are $A$-$B$-bimodule homomorphisms (it is enough to prove this for one of them, but neither is easier than the other). Yes, the proof in all detail is fairly long, even though much of it is straightforward. But the only argument that is significantly shorter is the one Martin Brandenburg gives in his answer (using Yoneda's lemma). $\endgroup$ – darij grinberg Jul 27 '14 at 10:46
  • $\begingroup$ Thanks! I understand and appreciate the comments. $\endgroup$ – user142700 Jul 27 '14 at 12:28

Here is the fastest proof I can think of. It doesn't use any elements. It is purely functorial (and therefore also works in more general situations).

Let $A$ be a ring (not assumed to be commutative), $M$ a right $A$-module and $N$ a left $A$-module. Then $M \otimes_A N$ is an abelian group which satisfies the universal property $$\hom(M \otimes_A N,T) \cong \hom_A(M,\hom(N,T)).$$ Here, we use the obvious right $A$-module structure on $\hom(N,T)$.

More generally, if $N$ is even a $(A,B)$-bimodule, then $M \otimes_A N$ is a right $B$-module satisfying the universal property $$\hom_B(M \otimes_A N,T) \cong \hom_A(M,\hom_B(N,T)).$$

Now if $P$ is a left $B$-module, it follows

$$\hom((M \otimes_A N) \otimes_B P,-) \cong \hom_B(M \otimes_A N,\hom(P,-)) \cong \hom_A(M,\hom_B(N,\hom(P,-)))$$ $$\cong \hom_A(M,\hom(N \otimes_B P,-)) \cong \hom(M \otimes_A (N \otimes_B P),-).$$

By Yoneda, this means $(M \otimes_A N) \otimes_B P \cong M \otimes_A (N \otimes_B P)$.

  • $\begingroup$ The only thing I'd add to this is that the isomorphism is an $A$-$B$-bimodule homomorphism (a pecularity of the commutative case). It's nothing particularly different, you just have to carry around some more module structures in the proof. $\endgroup$ – darij grinberg Jul 27 '14 at 10:47
  • $\begingroup$ No, these are just abelian groups. Of course, if $P$ is a $(B,C)$-bimodule, then both sides are right $C$-modules. That the isomorphism then will be $C$-linear is just because the original isomorphism is natural in $P$. $\endgroup$ – Martin Brandenburg Jul 27 '14 at 10:48
  • $\begingroup$ In your setting they are just abelian groups, but the OP wants an isomorphism of bimodules in the commutative case. $\endgroup$ – darij grinberg Jul 27 '14 at 10:52

I apologize if this is old, but I recently worked out this problem, found this question accidently online, and as I had my work written down on paper next to me I decided to share this.

Let $M$ be a module over $A$, $N$ a module over $B$, and $P$ a bimodule over $A$ and $B$.

(i) Construct $M\otimes_A P$, this is a module over both $A$ and $B$, and furthermore, it is not hard to show that it is bimodule.

(ii) Construct $P\otimes_B N$, this is a module over both $A$ and $B$, and furthermore, a bimodule.

(iii) With the first two constructions, $(M\otimes_A P)\otimes_B N$ and $M\otimes_A (P\otimes_B N)$ will be bimodules over $A$ and $B$. We claim they are isomorphic as bimodules.

(iv) Fix $n_0\in \mathbb{N}$. Define the map $g_{n_0}: M\oplus P\to M\otimes_A (P\otimes_B N)$ as $g_{n_0}(m,p) = m\otimes (p\otimes n_0)$. It is an easy verification that $g_{n_0}$ bilinear over $A$. Therefore, we get an induced map, $\overline{g_{n_0}}: M\otimes_A P\to M\otimes_A (P\otimes_B N)$ such that $\overline{g_{n_0}}(m\otimes p) = m\otimes (p\otimes n_0)$. This map is linear over $A$, but a little computation will show that it is also linear over $B$ as well.

(v) Consider the map, $f:(M\otimes_A P)\oplus N \to M\otimes_A (P\otimes_B N)$ defined by $f(\alpha,n) = \overline{g_n}(\alpha)$ where $\alpha \in M\otimes_A P$. This map can be checked to be bilinear over $B$, and satisfies the property that $f(m\otimes p,n) = \overline{g_n}(m\otimes p)$. Thus, we get the induced map $\overline{f}:(M\otimes_A P)\otimes_B N \to M\otimes_A (P\otimes_B N)$. This map is linear over $B$, but some easy computation will show it is linear over $A$ as well. It satisfies the property that $\overline{f}((m\otimes p)\otimes n) = m\otimes(p\otimes n)$.

(iv) By repeating steps (iii),(iv), and (v), by interchanging the roles of these modules we get a map $\overline{h}: M\otimes_A (P\otimes_B N)\to (M\otimes_A P)\otimes_B N$ with the property that it is linear over $A$ and $B$, and $h(m\otimes (p\otimes n)) (m\otimes p)\otimes n$.

Now we have two maps $\overline{f}$ and $\overline{h}$ which are map over bimodules with $\overline{f}\circ \overline{h} = 1$ and $\overline{h}\circ \overline{f} = 1$. This is enough to conclude the isomorphism that we wanted.


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