Module structure of base extension via tensor product

Question

Let $A,B$ be commutative rings.

Defining a product of $B\otimes_{A}B$ as $(b_1 \otimes b_2)\cdot (b_3 \otimes b_4)=(b_1b_3)\otimes(b_2b_4)$, this becomes a commutative ring.

Defining $b\cdot(b_1 \otimes b_2)=(bb_1)\otimes b_2$ for $b,b_1,b_2 \in B$, this becomes a $B$-module.

Then, my textbook says, if I define as $b\cdot(b_1 \otimes b_2)=b_1\otimes (bb_2)$ instead, it defines a different $B$-module structure. I don't understand this.

Isn't the homomorphism defined by $\phi(b_1\otimes b_2)=b_2\otimes b_1$ an isomorphism of them?

Answer 1

Let $B$ be an $A$-algebra and $M$, $N$ be $B$-modules. Here is an example in which $M \otimes_A N$ and $N \otimes_A M$ are isomorphic as $A$-modules, but not as $B$-modules. $\mathbb Z[x]$ is a $\mathbb Z$-module via the inclusion $\mathbb Z \hookrightarrow \mathbb Z[x]$. Let $M = \mathbb Z[x]$ be a $\mathbb Z[x]$-module via the identity. Let $N = \mathbb Z[x]$ be a $\mathbb Z[x]$-module via evaluation at $0$. Then $M \otimes_\mathbb Z N \cong N \otimes_\mathbb Z M$ as $\mathbb Z$-modules. However, they aren't isomorphic as $\mathbb Z[x]$-modules, since one is annihilated by $x$ while the other isn't.

Answer 2

The ring $B\otimes _A B$ has a canonical $A$-algebra structure and TWO different structures of $B$-algebra, which I'll call $(B\otimes _A B)_l$ and $(B\otimes _A B)_r$, according as multiplication by elements of $B$ happens on the left or on the right.
These $B$-algebra structures are in general different since if we denote by $\bullet $ and $\circ$ the actions of $B$ on $(B\otimes _A B)_l$ and $(B\otimes _A B)_r$, we have $$b\bullet (b_1\otimes b_2)=bb_1\otimes b_2\stackrel {\text{in general}}{\neq} b_1\otimes bb_2=b\circ (b_1\otimes b_2)$$ However, as you correctly noticed the swapping morphism of $A$-algebras $$\mu:B\otimes _A B\to B\otimes _A B:b_1\otimes b_2\mapsto b_2\otimes b_1$$ is indeed an isomorphism of $B$-algebras $$\mu:(B\otimes _A B)_l\to (B\otimes _A B)_r$$ as shown by the following trivial calculation : $$\mu(b\bullet (b_1\otimes b_2))=\mu(bb_1\otimes b_2)=b_2\otimes bb_1=b\circ(b_2\otimes b_1)=b\circ\mu(b_1\otimes b_2)$$

SUMMARY
The $B$-algebras $(B\otimes _A B)_l$ and $(B\otimes _A B)_r$ are different but isomorphic.