About Krull domain If $A$ is a Krull domain in its field of fractions $F$, and if $F'\subset F$ is a subfield, then $A\cap F'$ is a Krull domain. But is it necessarily a Krull domain of $F'$? That is, is $F'$ necessarily the field of fractions of $A\cap F'$?
 A: The the answer is affirmative if I assume that no DVR defining $A$ contains the field of fractions of $A \cap F'$.
The collection of intersection of DVR's defining $A$ with the field $F'$ is the collection of DVR's defining $A\cap F'$. So let $(D_i,v_i)$ be the family of DVR's in $F'$ defining the Krull domain $A\cap F'$. I have to show that the field of fraction of $A\cap F'$ is $F'$. Choose any element $y$ in $F'$. Now by definition there exist only finitely many valuations $v_i$ such that $v_i(y) < 0$, say $v_{i_1}, v_{i_2}, \ldots, v_{i_n}$. Note that for rest of the values of $i$ the element $y$ is in $D_i$. Now consider the intersection of maximal ideals corresponding to $D_{i_1}, D_{i_2}, \ldots, D_{i_n}$ with $A \cap F'$ which is not empty due to our assumption. Choose an element $z$ in the intersection and multiply $y$ by $z$ to push it inside $A \cap F'$.
In general the answer is NO. Here is a counterexample: $K[T]$ is a Dedekind domain and therefore is a Krull domain with its field of fractions $K(T)$. Now consider the subfield to be $K(T + \frac{1}{T})$. Then the intersection $K(T + \frac{1}{T}) \cap K[T]$ is $K$.
