Let $\mathcal L = \{0,1,+,\cdot,a,b,c\}$ be the first order language of ring theory, augmented with three individual constants $a$, $b$ and $c$, and let $T$ consist of the usual axioms expressing that any models are integral domains, along with the following axioms:

1) $a$, $b$ and $c$ are nonzero

2) $a$ generates a prime ideal not containing $b$

3) $a$ and $b$ generate a proper prime ideal

4) $c$ is in every power of the prime ideal mentioned in (3); this requires infinitely many first order statements, one for each $n\in\mathbb N$

5) if $x_{0}$ is nonzero and not a unit, it is divisible by an irreducible element

6) factorization into irreducibles, as far as it exists, is unique; this requires infinitely many statements of the form *all* $x_{i},y_{j}$ *irreducible* $\rightarrow x_{0}\cdot … \cdot x_{n} \neq y_{0}\cdot … \cdot y_{m}$ (one statement for each $n \gt m$), plus *all* $ x_{i},y_{i}$ *irreducible, and* $x_{0}\cdot … \cdot x_{n} = y_{0}\cdot … \cdot y_{n}$ $\rightarrow\bigvee_{\sigma\in S_{n+1}} (x_{i}$ *and* $y_{\sigma(i)}$ *are associated*$)$, where the disjunction runs over the symmetric group on $0, 1, …, n$ (one statement for each $n$).

Clearly, $T$ is consistent: if $S\subseteq T$ is a finite subset, the polynomial ring $k[X,Y]$ over a field $k$ is a model of $S$, when $a$, $b$ and $c$ are interpreted as $X$, $Y$ and $X^{n}$, with $n$ sufficiently large; so we can use the Compactness Theorem.

Now let $\Sigma = \{x_{0}\neq 0$ *and* $\exists_{y_{0}\ … y_{n}}(x_{0} = y_{0}\cdot … \cdot y_{n}$ *and* $y_{i}$ *is irreducible for* $i \lt n)\, |\,n\in\mathbb N\}$. It is a set of $\mathcal L$-formulas having only the variable $x_{0}$ free. Finite subsets of it are realized in every $T$-model $A$, as any such $A$ contains the irreducible element $a^{A}$ (the interpretation of the $\mathcal L$-constant $a$), and for $n$ large enough its $n$th power will satisfy all formulas of the finite subset.

No single $\mathcal L$-formula $\phi(x_{0})$ with only $x_{0}$ free and consistent with $T$ can imply all statements of $\Sigma$ under $T$, for it would have to express that $x_{0}$ is divisible by products of arbitrarily many irreducible elements. Therefore, $\Sigma$ is not an *isolated type*.

By the Omitting Types Theorem of Model Theory, $T$ has a model $A$ which omits $\Sigma$. Such an $A$ must be a UFD, since none of its nonzero elements can be divisible by products of arbitrarily many irreducible elements, whereas all nonzero non-units do have an irreducible factor, in view of axiom (5); and the axiom schema (6) ensures that factorizations are unique.

A fortiori, $A$ is a Krull domain, and by axioms (1) through (4), $\mathfrak p: = Aa^{A}+Ab^{A}$ provides a counterexample. Indeed, $0\neq c^{A}\in \bigcap_{n}\mathfrak p^{n}$. Here, $a^{A}$, $b^{A}$ and $c^{A}$ again denote the interpretations of the constants $a$, $b$ and $c$ in the structure $A$.

**Edit:** superficially, the *same* argument would seem to yield an UFD $A$ having an irreducible element $a^{A}$ and a nonzero element $c^{A}$ such that $c^{A}$ is divisible by every power of $a^{A}$, if one replaces (4) by the schema $\exists_{x}(c = a^{n}\cdot x)$, with $n$ ranging over $\mathbb N$. But then one would have $T\vdash\forall_{x_{0}}(\phi(x_{0})\rightarrow\psi(x_{0}))$ for all $\psi(x_{0})\in\Sigma$, where $\phi(x_{0})$ is the formula $x_{0} = c$. So $\Sigma$ would be an *isolated* type, and the rest of the argument would fall apart.