What is the quotient field of the integral closure of a domain $A$ in an arbitrary field extension of $\operatorname{Frac}(A)$?

Question

Suppose $A$ is an integral domain, with quotient field $K$, and let $L/K$ be an arbitrary field extension. If $B$ is the integral closure of $A$ in $L$, what is the quotient field of $B$?

If $L/K$ is algebraic, the argument in this question shows the quotient field of $B$ is $L$.

Example (d) on page 4 of Curtis and Reiner's Methods of Representation Theory Vol I says the quotient field of $B$ is $L$ for any field extension $L/K$, but I don't see a way to get a handle on showing any non-algebraic element of $L$ is a fraction of elements in $B$ without any polynomial relation to exploit.

Answer 1

Let $L/E/K$ be the largest intermediate algebraic extension. More explicitly, $E$ is the set of all elements of $L$ which are algebraic over $K$. This is often called the algebraic closure of $K$ in $L$. Then if $B$ is the integral closure of $A$ in $L$, then its elements are in particular algebraic over $K$, so $B \subseteq E$. Then in fact, we have that $B$ is the integral closure of $K$ in $E$, which is an algebraic extension, so the quotient field of $B$ is $E$.