Tensor product of associative ring $A$ is flat over $A \otimes_k A^{op}$.

Question

It is claimed in this notes line 4 pg 5 that

If $A$ is flat over $k$, $A^{\otimes n}$ is flat over $A \otimes_k A^\text{op}$ for $n \ge 2$.

I am stuck even at $n=2$. I tried some base changes but got nowhere.

Any hint would be appreciated!

Answer 1

Let's try to understand the left $A\otimes_k \newcommand\op{{\text{op}}}A^\op$-module structure on $A^{\otimes_k n}$. It's defined in the notes by $$(f\otimes g)\cdot (a_1\otimes\cdots \otimes a_n) = fa_1\otimes \cdots \otimes a_n g.$$

If $M$ is a (right) $A\otimes_k A^{\op}$-module, where we write $m\cdot (f,g) = gmf$ then $M\otimes_{A\otimes_k A^\op} A^{\otimes_k n}$ satisfies $$m\otimes (a_1\otimes\cdots \otimes a_n) = m\otimes (a_1,a_n)(1\otimes a_2\otimes\cdots \otimes a_{n-1}\otimes 1)=a_nma_1\otimes(1\otimes a_2\otimes \cdots \otimes a_{n-1}\otimes 1).$$ So as $k$-vector spaces, we have the natural isomorphism $$M\otimes_{A\otimes_k A^{\op}} A^{\otimes_k n} \simeq M\otimes_k A^{\otimes_k (n-2)}.$$ (Explicitly the maps are given by $(m,a_1,\ldots,a_n)\mapsto a_nma_1\otimes a_2\otimes\cdots\otimes a_{n-1}$ and $(m,a_2,\ldots,a_{n-1})\mapsto m\otimes 1\otimes a_2\cdots \otimes a_{n-1}\otimes 1$).

The right hand functor is exact by assumption, so the left hand functor is exact as well.

Essentially this is the standard isomorphism $(M\otimes_A A)\simeq M$, but a little dressed up, with opposite rings thrown in as well.