For an algebra automorphism $\omega$, what is $A_\omega \otimes_A A$?

Question

Let $A$ be a finite dim. algebra over a field $k$. Obviously, $A\otimes_A A \cong A$. Let $\omega: A \rightarrow A$ be an algebra automorphism. Let $M$ be a right $A$-module. Then $M_\omega$ is a right $A$-module such that $M= M_\omega$ as $k$-vector space and the right $A$-module structure is defined by $m \cdot a = m \omega(a)$. So what can we say about $A_\omega \otimes_A A$?

Answer 1

Popyaitte definitely has the right answer to the question you asked.

I feel like it's also worth noting that if $M$ is a right $A$-module, then $$M\otimes_A A_\omega \simeq M_\omega,$$ and if $N$ is a left $A$-module, then $$A_\omega \otimes_A N \simeq {}_{\omega^{-1}}N,$$ with ${}_{\omega^{-1}} N$ being the left module analog of $M_{\omega^{-1}}$.

The isomorphism for right modules is $m\otimes a \mapsto ma$. The fact that this is an isomorphism follows since $A_\omega \simeq A$ as left $A$-modules, and this is just the isomorphism $M\otimes_A A \simeq M$ as abelian groups. Right $A$-linearity is not hard to check.

For left modules, we note that $A_\omega \simeq {}_{\omega^{-1}}A$ as an $A$ bimodule. The map is $\omega^{-1}: A_{\omega}\to {}_{\omega^{-1}}A$ to check left $A$-linearity, note that $$\omega^{-1}(\alpha \cdot a) = \omega^{-1}(\alpha a) = \omega^{-1}(\alpha)\omega^{-1}(a)=\alpha\cdot \omega^{-1}(a).$$ For right linearity, we have $$\omega^{-1}(a\cdot \alpha) = \omega^{-1}(a\omega(\alpha)) = \omega^{-1}\alpha = \omega^{-1}\cdot \alpha .$$ Then the statement for left modules follows from the symmetric version of the statement for right modules, proved above.

Answer 2

For any $A$-module $M$, you have that $M\otimes_A A$ is isomorphic to $M$. So $A_\omega \otimes_A A$ is isomorphic to $A_\omega$.