If $A\subseteq B$ are affine domains over an algebraically closed field $k$ of characteristic zero, such that $Q(A)$ is algebraically closed in $Q(B)$, how can one show that $Q(A)$ is also algebraically closed in the field of fractions of $Q(A)\otimes_kB$?

The history behind this problem:

Starting from the fact that $Q(A)$ is algebraically closed in $Q(B)$, I intend to conclude that a general fiber of the morphism Spec $B \rightarrow$ Spec $A$ is irreducible. To do so, by first Bertini Theorem, as in Shafarevich's *Basic Algebraic Geometry*, vol. 1, it suffices to show that $Q(A)\otimes_k B$ is geometrically irreducible over $Q(A)$, which, in turn, by Zariski-Samuel's *Commutative Algebra*, vol. 2 (see page 230, thm. 39), is equivalent to showing that $Q(A)$ is algebraically closed in the field of fractions of $Q(A)\otimes_kB$.

Now, for the proof it's easy to see that $Q(A)$ is alg. closed in $C:=Q(A)\otimes_kB$. But I cannot settle the case where some $x/y\in Q(C)$ might be algebraic over $Q(A)$, where both $x$ and $y$ go to 0 under the natural map $Q(A)\otimes_k B \rightarrow Q(B)$.

By the way, if $B=k[x_1,...,x_n]/\mathfrak{p}$, with $\mathfrak{p}\in$ Spec $k[x_1,...,x_n]$, then the field of fractions of $Q(A)\otimes_k B$ is equal to the residue field of $Q(A)[x_1,...,x_n]$ at the point $\mathfrak{p}Q(A)[x_1,...,x_n]$.