I'm considering the correctness of the following assertion, which is related to linear disjointness (I'm trying to generalize it to subalgebras), What does "linearly disjoint" mean for abstract field extensions? and When is the tensor product of two fields a field?.
Let $A,B$ be flat $R$-algebras, both of them are transcendental over $R$ (which means that there are injective $R$-algebra homomorphisms $R[X]\hookrightarrow A$ and $R[X]\hookrightarrow B$). Is $A\otimes_RB$ not a field?
Here is my proof attempt. Suppose otherwise. Then $A,B,R$ are not trivial. The injections $R[X]\hookrightarrow A$ and $R[X]\hookrightarrow B$ making $A$ and $B$ to be $R[X]$-algebras. Write $a,b$ the image of $X$ in $A,B$ respectively.
Since $R\hookrightarrow A$, $R\hookrightarrow B$ and $A,B$ flat over $R$, we have $A\hookrightarrow A\otimes_RB$, $B\hookrightarrow A\otimes_RB$, and their images are linearly disjoint in $A\otimes_RB$, which means that, denote $A',B'$ be their images, then the map $A'\otimes_RB'\to A\otimes_RB$, $x\otimes y\mapsto xy$ is injective (in fact bijective). This implies that $\operatorname{rank}_R(A'\cap B')\leq 1$, which means that if there is an injective $R$-linear map $R^{\oplus n}\hookrightarrow A'\cap B'$, then $n\leq 1$.
On the other hand, consider the surjective ring homomorphism $A\otimes_RB\twoheadrightarrow A\otimes_{R[X]}B$. The $a\otimes 1$ and $1\otimes b$ map to the same element under this map.
If $A\otimes_{R[X]}B\neq 0$, then since $A\otimes_RB$ is a field, this map must be injective, hence bijective. So $a\otimes 1=1\otimes b$ in $A\otimes_RB$, and we have $R[X]\hookrightarrow A'\cap B'$, $X\mapsto a\otimes 1=1\otimes b$, which contradicts with $\operatorname{rank}_R(A'\cap B')\leq 1$.
If $A\otimes_{R[X]}B=0$, then I'm stuck. Any suggestions?
When $A,B,R$ are fields, it's the problem considered in the above MO links, and we can consider $A\otimes_{R(X)}B$ instead, which must be non-trivial, so the above argument works seamlessly.
