Specifically, Lorenzini writes (in "An invitation to arithmetic geometry):
Let $\{ A_i \}_{i \in I}$ any set of subrings of $K$ the field of fractions of a commutative domain $A$, such that all the $A_i$ are integrally closed in $K$, and such that $K$ is also the field of fractions of $\bigcap\limits_{i \in I} A_i$. Then the ring $\bigcap\limits_{i \in I} A_i$ is also integrally closed in $K$.
It seems like a very basic fact, but I really want to get it down since it is so relevant to the rest of the book.
I wrote down:
$\forall i \in I \forall a \in K \exists b_1, ..., b_{n-1} \in A_i [a^n + a^{n-1} b_{n-1} + ... + ab_1 + a = 0] $ Since $A_i$ integrally closed in $K$ $\forall i \in I$, $a \in A_i$. Since $K = Frac(\bigcap\limits_i A_i)$, $a \in Frac(\bigcap\limits_i A_i)$, but since it's in $A_i$ it's not really a fraction, so it's properly in $\bigcap\limits_i A_i$. So $\bigcap\limits_i A_i$ is integrally closed in $K$ and we are done.
Is this proof ok? It doesn't seem like it. What is wrong?
If $a\in K$ is integral over $\cap_{i\in I}A_i$, then surely $a$ is integral over each $A_i$, so that $a\in A_i$ for all $i$. It follows that $a\in\cap_{i\in I}A_i$, so that $\cap_{i\in I}A_i$ is integrally closed. (I think this is roughly what you said, more simply stated.)