Assume that:
(1) $A \subsetneq B$ are integral domains and finitely generated algebras over a field $k$ ($k$ is algebraically closed of characteristic zero, if this helps).
(2) $A$ is algebraically closed in $B$.
(3) The field of fractions of $A$, $Q(A)$, has transcendence degree over $k$ one less than the transcendence degree of $Q(B)$ over $k$, that is, $\dim A=\dim B-1$.
My question: Is it true that $Q(A)$ is algebraically closed in $Q(B)$? If not, it would be nice to have a counterexample. (If I am not wrong, $k[x] \subsetneq k[x,y]$ is an example to my question; just use Exercise 1.3.)
I would appreciate any help in solving my question.
Edit: Perhaps Exercise 1.4 is a counterexample to my question; I am not sure if it satisfies my assumption (3) or not.
The example I gave above is $A = \mathbb{C}[x,y]$, $B = A[z,w]/(xz^2 - yw^2)$. $xz^2 - yw^2$ is irreducible so $B$ is an $\mathbb{N}$-graded domain with $A$ as the degree $0$ part. So $A$ is algebraically closed in $B$ by Exercise 1.3 in your link (basically anything in $B \setminus A$ can't satisfy a polynomial over $A$ since the highest degree term always survives as $B$ is a domain). Also your assumption (3) is just that $\dim A = \dim B - 1$ which holds here. But $Q(A)$ is not algebraically closed in $Q(B)$ e.g. $(z/w)^2 = y/x$ is in $Q(A)$ but $z/w$ is not in $Q(A)$.