Background: A Krull domain is an integral domain $A$ with a family $(v_i)$ of valuations on the field of fractions $K$ for $A$ satisfying the following conditions:
- Each $v_i$ is discrete.
- The intersection of the valuation rings for the $v_i$ is $A$.
- For every $x\in K^\times$, there are finitely many indices $i$ such that $v_i(x)\neq0$.
This definition is paraphrased from Bourbaki. We require discrete valuations to be nontrivial.
Question: Is every field a Krull domain?
Concerns: If a field $k$ were a Krull domain, the only way the definition could be satisfied would be if we took an empty family of valuations because of property 2. In this case, properties 1 and 3 hold trivially and 2 holds if we take the intersection over the empty set to be $k$. But I don't see how that argument uses field properties of $k$: it seems to show that any integral domain is Krull if we take an empty family of valuations.
Motivation: I'm working through the proof that a finite intersection of Krull domains (living in some common field) is a Krull domain. The rings $\mathbb F_2[X]$ and $\mathbb F_2[Y]$ are clearly Krull domains living in the common field $\mathbb F_2(X,Y)$, and their intersection is $\mathbb F_2$. So $\mathbb F_2$ must be a Krull domain, but I can't see why.
I think it's better to exclude the fields from the class of Krull domains as Larsen and McCarthy did in their Multiplicative Theory of Ideals.
A reason could be the following: if a field $K$ would be a Krull domain, then by definition it is an intersection of DVRs contained in $K$, that is, equal to $K$, but a field is not a DVR. (In some sense we consider the DVRs as being the "smallest" examples of Krull domains.)