TITLE: When is tensor product isomorphic to product?
QUESTION [0 upvotes]: Let $A$ and $B$ be algebras.
When do we have $A\otimes B\cong AB$, where $$AB=\{\sum a_ib_i\mid a_i\in A, b_i\in B\}$$
Is commutativity $ab=ba$ for $a\in A$, $b\in B$ a sufficient condition?
Thanks for any help.

REPLY [1 votes]: Tensor product is an "external" operation. It takes two completely unrelated algebras $A$ and $B$ and it spits out a new algebra $A \otimes B$ that only depends on the data of $A$ and $B$ taken separately. $A$ and $B$ could be the same algebra, subalgebras of a bigger algebra, it won't matter from the point of view of $\otimes$.
The product you define is an "internal" operation. For what you have written to make sense, it is necessary for $A$ and $B$ to both be subalgebras of the same algebra (say) $C$. Then the product $AB$ is the set of all linear combinations of $ab$ with $a \in A$ and $b \in B$; for $ab$ to even mean something, you must have some way of multiplying elements of $A$ with elements of $B$, and this depends on more than just the algebra structure on $A$ and the algebra structure on $B$; it also depends on how they live in $C$, and the product $AB$ is a subalgebra of $C$ too.
It's like for groups, if that helps: given two groups $H$ and $K$, you can always define their direct product $H \times K$; it's like the tensor product $A \otimes B$. But to define their product $HK$, $H$ and $K$ must be subgroups of a bigger group $G$, and then $HK \subset G$ too; it's more like the product $AB \subset C$.
Now, just like for groups, it is very rarely the case that $A \otimes B$ is isomorphic to $AB$. Take for example $A = B = C$. Then $AB = C$, while $A \otimes B = C \otimes C$. Unless $C$ has dimension $0$ or $1$, there will be no isomorphism $C \cong C \otimes C$, even if $C$ is commutative. I don't think there are interesting conditions (other than trivial ones) where $A \otimes B \cong AB$ – they're very different constructions, the word "product" being in both is more of a coincidence than anything.
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