Is $A[b/a]$ a Krull domain?

Is $$A[X]/(aX+b)$$ a Krull domain when $$A$$ is and when $$a\in A-\{0\}, b\in A-Aa$$ are such that $$Aa$$ and $$Aa+Ab$$ are prime ideals of $$A$$?

This is stated as Proposition 8 in Pierre Samuel, "Sur les anneaux factoriels", Bulletin de la S. M. F., tome 89 (1961), p. 155-173. (Much the same argument appears in Bourbaki, Commutative Algebra, as exercises 13 and 15 for §1 of Ch.VII.)

Samuel notes that $$A[X]/(aX+b) \cong B := A[b/a]$$, $$a$$ is a also a prime element of $$B$$, and the ring $$B[a^{-1}] = A[a^{-1}]$$ is again a Krull domain - all true enough.

He then refers back to Remark 2 to Proposition 2 of the same paper, to conclude from the fact that $$B = B[a^{-1}]\cap B_{Ba}$$ that $$B$$ is Krull. However, in the Remark it is assumed that $$B$$ has the ACCP (the maximal condition for principal ideals). Indeed, this will guarantee that $$B_{Ba}$$ is a DVR (discrete valuation ring).

For $$B_{Ba}$$ to be a DVR, it is necessary and sufficient that $$\bigcap_{n\in\mathbb{N}} Ba^{n} = 0$$, and this is easily seen to be equivalent to $$\bigcap_{n}\mathfrak p^{n} = 0$$ in $$A$$, where $$\mathfrak p$$ denotes the prime ideal $$Aa+Ab$$ of $$A$$.

Any counterexamples with $$\bigcap_{n}\mathfrak p^{n} \neq 0$$ for such a $$\mathfrak p$$?

If $$v$$ is any of the essential valuations of $$Q(A)$$ for the Krull domain $$A$$ (corresponding to the height 1 prime ideals of $$A$$), then $$v(a) = 1, v(b) = 0$$ if $$v$$ belongs to $$Aa$$, and $$v(a) = 0$$ for any other $$v$$. So not much can be said about the various $$v(x)$$ for a general element $$x$$ of $$\mathfrak p$$, it would seem...

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.