Algebraic closedness in field of fractions

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 $\mathfrak{p}Q(A)[x_1,...,x_n]$.

• I was going to answer this at MSE but unfortunately the migration got rejected before I could. Any $x/y\in Q(C)$ is defined on a dense open subset of almost every fiber of $\operatorname{Spec} A\times \operatorname{Spec} B\to \operatorname{Spec} A$. If $x/y$ is algebraic over $A$, then its restriction to almost every fiber will be algebraic over $k$ and hence constant. – Eric Wofsey Apr 25 '15 at 13:18