$A \otimes_k A$ isomorphic to $A \otimes_k A^\text{op}$ as $A$-bimodules?

Question

Let $A $ be a associative $k$-algebra, is $A \otimes_k A$ isomorphic to $A \otimes_k A^\text{op}$ as $A$-bimodules?

This seems counter intuitive but is true?

There is a morphism that is identity as $k$-modules. Further since the action on the right of $A \otimes_k A^{op}$ by $A$ is given on basis element $$(x \otimes_k y) (x') = x \otimes (x' \cdot_\text{op} y) = x \otimes yx'$$ The morphism is in fact a bimodule morphism.

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It is claimed in many text of Hochschild homology, that $A \otimes_k A$ is a free $A^e$ module. In which case, how?

Answer 1

It seems you've messed up your actions a bit. The right $A$-action on $A\otimes_kA$ is given by $(x\otimes y)a:=x\otimes(ya)$, and the right $A$-action on $A\otimes_kA^{\mathrm{op}}$ is given by $(x\otimes y)\cdot a:=x\otimes (y\cdot a)=x\otimes(ay)$.

In fact, there are two left $A$-module structures on $A\otimes_kA$, an 'outer' one $a(x\otimes y):=(ax)\otimes y$ and an 'inner' one $a\diamond(x\otimes y):=x\otimes(ay)$, and similarly there two right $A$-module structures. These were all used to good effect in

William Crawley-Boevey, Preprojective algebras, differential operators and a Conze embedding for deformations of Kleinian singularities

when defining the $A$-bimodule structure on the space of derivations $\mathrm{Der}_k(A,A\otimes_kA)$.