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...
 A: 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.
