If $a\otimes 1=1\otimes b$, must it be of the form $r\cdot (1\otimes 1)$?

Question

Let $R$ be a reduced commutative ring, and $A,B$ be two faithfully flat $R$-algebras, note that any faithfully flat ring map is injective. (we can assume all of them are local ring and local maps or even Noetherian if necessary)

Assume there exists $a\in A,b\in B$ s.t. $a\otimes 1=1\otimes b\in A\otimes_R B$. Does it implie that $a\in R\subset A$ where $R$ is viewed as a subring of $A$?

If not provide a counter example please, and can we add other conditions to make it true?

Answer 1

So far I have only found a class of situations such that the result is true. That is

Let $R$ be a commutative Hermite ring (any stably free modules are free), let $A$ and $B$ be $R$-algebra which are also finite free $R$-modules. Then $1\otimes _RB\cap A\otimes_R 1=R$ in $A\otimes_R B$.

Note that any Dedikind domain is Hermite, and any local ring is Hermite, see I.4.7 in the following reference.

Lam, T. Y., Serre’s conjecture, Lecture Notes in Mathematics. 635. Berlin-Heidelberg-New York: Springer-Verlag. XV, 227 p. (1978). ZBL0373.13004.

The starategy is to find a basis containing $1$ in both $A$ and $B$, then we just expand and compare their coefficients.

Note that in a Hermite ring, any vector $[r_1,...,r_n]\in R^n$ satisfying that $1=\sum r_i s_i$ for some $s_i \in R$ can be expand to a basis of $R^n$, i.e. an invertible $n\times n$ matrix over $R$. See this link for an elementary proof of this fact. Actually this is an equivalent definition of Hermite ring.

Let $\{a_1,...,a_n \}$ be a $R$-basis of $A$. So $\exists r_i\in R$ s.t. $1=\sum_{i=1}^n r_i\cdot a_i$.

Note that For any faithfully flat ring map $R\to A$ and any ideal $I\subset R$, we have $I\otimes_{R}A\cap R=I$. And finite free modules are clearly faithfully flat.

Thus $(r_{1},...,r_{n})=(r_{1},...,r_{n})\otimes_{R}A\cap R=A\cap R=R$. In particular, there exists $c_{i}\in R$ s.t. $1=\sum_{i=1}^{n}r_{i}c_{i}$.

By property of Hermite ring, $R^{n}$ has a basis containing $[r_{1},...,r_{n}]$ and they form an invertible matrix over $R$, say $\mathbf{M}$.

Apply $\mathbf{M}$ to the basis of $A$, $[a_{1},...,a_{n}]$, we get another basis starting from $1$. The result follows.