Algebraic Extensions

Question

I have the following question: there is this statement i can't understand:

Let $A$ be an integral domain which is integrally closed ( in its field of franctions ) and let $K$ be its fraction field. Let $L|K$ be a finite extension and let $B$ be the integral closure of $A$ in $L$. Then $B$ is integrally closed (in its field o fractions).

I know that composition of finite extension is still a finite extension, but how do I know that $B$ is integrally closed?

My idea is the following: $B$ is a field containing $A$ and so is $F$. Thus [L:A]=[L:B] [B:A] and so $B|A$ can't be "not finite".

Is it correct?

Answer 1

By definition, $B$ is integrally closed in $L$, and we want to show that it is integrally closed in its field of fractions. So we can restate the question: Show that $L$ is the field of fractions of $B$.

Is this any easier?

Answer 2

$B$ is the integral closure of $A$ in $L$. Then $B$ is integrally closed in $L$ -- the (integral closure of a ring inside a fixed ring) operator is an idempotent. Let's just check that: take $l\in L$ integral over $B$. Since $B$ is integral over $A$ it follows by transitivity of integral extensions that $l$ is integral over $A$. Therefore $l \in B$.

We should be done now since the field $L$ contains $B$ so it contains its ring of fractions. However, we can prove that $L$ is the field of fractions of $B$ if we use that $K$ is the field of fractions of $A$ and $K\subset L$ is algebraic ( so we do not need $A$ integrally closed in $K$ for this, only if we want $B \cap K = A$ ).

Let $l \in L$. $l$ is algebraic over $K$ so we have an algebraic equation with coefficients in $K$. Now every such coefficient is a fraction with numerator and denominator from $A$. Bring all the fractions to a common denominator $a \ne 0$. \begin{eqnarray*} l^n + \frac{a_1}{a} l^{n-1} + \ldots + \frac{a_n}{a} = 0 \end{eqnarray*} Multiply the above equality by $a^n$ and get \begin{eqnarray*} (al)^n + a_1 (al)^{n-1} + a a_2 (al)^{n-2}+\ldots + a^{n-1} a_n = 0 \end{eqnarray*} It follows that $al$ is integral over $A$ and therefore lies in $B$. We get $$l = \frac{al}{a}$$ a fraction with numerator in $B$ and denominator in $A$, hence also in $B$.