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

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.

• You already have the answer $Frac(A)=L\cap \overline{K}$ May 15, 2021 at 20:15

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 $$E$$.