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

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?

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.