Sandwich theorem for UFD's Let $A \subseteq A[w] \subseteq C$ be three Noetherian integral domains, over a field $k$ of characteristic zero, with $A \subseteq C$ an algebraic ring extension (in particular, $w$ algebraic over $A$).
Further assume that: (1) $A$ and $C$ are unique factorization domains (UFD's).
(2) $A \subseteq C$ is root closed, namely, for all $n$, if $c \in C$ satisfies $c^n \in A$, then $c \in A$.
(3) $w$ is an irreducible element of $C$ (hence it is also an irreducible element of $A[w]$). Notice that, since $C$ is assumed to be a UFD, the assumption that $w$ is irreducible in $C$ implies that it is prime in $C$
(but it can be non-prime in $A[w]$).
(4) $w$ is a prime element of $A[w]$.   

Is it true that $A[w]$ must be a UFD? Or, is there a counter-example?

Remark: Without assuming (2),(3),(4) it is easy to find a counter-example:
$k[t^2] \subset k[t^2][t^3] \subset k[t]$ ($w=t^3$). Clearly, $k[t^2]$ and $k[t]$ are UFD's, but $k[t^2][t^3]$ is not a UFD, since $t^3$ is an irreducible element of $k[t^2][t^3]$ which is not prime: $t^3t^3=t^2t^2t^2$.
Somewhat relevant questions are: 1, 2.
Edit: After receiving a nice counter-example to my original question, it would be nice to know which additional conditions are required in order to guarantee that $A[w]$ is a UFD; for example:
(i) Finding a special form of the minimal polynomial of $w$ over $A$.
(ii) Additional properties of the ring extension $A \subseteq C$, such as flatness/separability (I am not sure I know whether the counter-example is flat/separable or not).
(iii) The only invertible elements of $C$ are $k^*$ (this is clearly not satisfied by the counter-example).
 A: That is not true.  Let $A$ be a polynomial ring in two variables, $k[s,t]$.  Let $C$ be the $A$-subalgebra of the fraction field generated by the fraction $s/t$, i.e., $C=k[(s/t),t]$.  Both $A$ and $C$ are unique factorization domains (roughly by Gauss's Lemma).  Since $A$ is integrally closed in its fraction field, the extension $A\to C$ is "root closed".
Let $w$ be the element $(s/t)^2-t$, just for one example.  Then the $A$-algebra $A[w]$ equals $k[s,t][w]/\langle t^2w + (t^3-s^2) \rangle$.  The quotient of $A[w]$ by the principal ideal generated by $w$ equals $k[s,t]/\langle t^3-s^2 \rangle$.  This is an integral domain.  Thus, the principal ideal generated by $w$ in $A[w]$ is a prime ideal.  Of course $w$ is prime, hence irreducible, in $C$ since $C/\langle w\rangle$ is the polynomial ring $k[s/t]$.   
Finally, $A[w]$ is not a unique factorization domain.  In fact, it is not even integrally closed in its fraction field.  The element $f=s/t$ of the fraction field satisfies the monic polynomial, $$f^2 -(w+t) = 0.$$  If $f$ were an element of $A[w]$, then $A[w]$ would equal all of $C$.  However, $A[w]/\langle w \rangle$ equals $k[s,t]/\langle t^3-s^2\rangle$.  This is not regular, it is not a unique factorization domain, it is not integrally closed in its fraction field, etc.  On the other hand, $C/\langle w \rangle$ is isomorphic to $k[s/t]$.  This is regular, it is a unique factorization domain, it is integrally closed in its fraction field, etc.
Edit.  The OP asks for a different example.  In the following example, both $A\to A[w]$ and $A\to C$ are finite and flat.  Both $A$ and $C$ are regular Noetherian integral domains.  Yet $A[w]$ is not even integrally closed in its fraction field, because it fails the condition $R1$ of Serre's criterion.  That is the common feature of both examples: a very singular scheme can be "sandwiched" between two smooth schemes.  Once you understand the geometric nature of these examples, you can produce many other examples for yourself.
Let $n\geq 5$ be an integer that is prime to the characteristic.  Let $A$ be $k[s_1,\dots,s_n]$, a polynomial ring in $n$ variables.  Define $C$ to be the following $A$-algebra, $$C = A[x]/\langle p_n(x) \rangle, \ \ p_n(x) = x^n - s_1x^{n-1} + \dots + (-1)^{n-m}s_mx^{n-m} + \dots + (-1)^ns_n.$$  The equation $p_n(x)$ allows us to "eliminate" $s_n$, so that $C$ is isomorphic to $k[s_1,\dots,s_{n-1},x]$, a polynomial ring in $n$ variables.  Thus, both $A$ and $C$ are Noetherian, regular, integral domains of dimension $n$.  By construction, $C$ is a finite flat extension of $A$ of rank $n$ as an $A$-module.
Now let $v$ be $s_nx$ in $C$.  Let $w$ be $v-s_1$.  Thus, the $A$-subalgebra $A[w]$ of $C$ equals the $A$-subalgebra $A[v]$.  The element $v$ satisfies the monic polynomial equation $q_n(v)=0$, $$q_n(v) = v^n-s_1s_n^1v^{n-1}  + \dots + (-1)^{n-1}s_{n-1}s_n^{n-1}v + (-1)^ns_n^n.$$  By direct computation, for a variable $y$, the ring $B=A[y]/\langle q_n(y) \rangle$ is a complete intersection ring that a finite, flat extension of $A$, and whose singular locus is an irreducible closed subset of the spectrum with associated prime ideal $\langle s_n,v\rangle$.  Since $A\to B$ is finite and flat, every irreducible component of $\text{Spec}(B)$ dominates $\text{Spec}(A)$.  However, after inverting $s_n$, the natural map $B[1/s_n]\to C[1/s_n]$ is an isomorphism.  Hence, $B$ is generically reduced and has a single irreducible component.  Since $B$ is a complete intersection ring, it is everywhere reduced.  Thus $B$ is an integral domain, hence embeds in its fraction field.  Therefore, since the map of fraction rings, $B[1/s_n]\to C[1/s_n]$, is injective, also $B\to C$ is injective.  In conclusion, the natural map $B\to A[v]$ is both injective and surjective, hence an isomorphism.   
Note that $B$ is not integrally closed in its fraction field.  Of course we can find a specific fraction that satisfies a monic equation yet is not in $B$.  However, it is faster to use Serre's Criterion.  Since $B$ is a complete intersection ring, it is Cohen-Macaulay, and thus it is $S2$.  Therefore $B$ is integrally closed in its fraction field if and only if $B$ is $R1$.  However, the singular locus of $B$ is the irreducible closed subset with coordinate ring $$B/\langle v,s_n \rangle = A/\langle s_n \rangle \cong k[s_1,\dots,s_{n-1}].$$  This has dimension $n-1$.  So $B$ is singular in codimension $1$, thus it is not $R1$.  Since $B$ is not integrally closed in its fraction field, by Gauss's Lemma, $B$ is also not a unique factorization domain.
The fraction field extension $\text{Frac}(A)\to \text{Frac}(C)$ is a degree $n$ extension whose associated Galois group is the full symmetric group $\mathfrak{S}_n$.  In particular, for $n\geq 5$, this cannot be a solvable extension, much less a radical extension.  Thus, $\text{Frac}(A)$ is "radically closed" in $\text{Frac}(C)$.  Since $A$ is integrally closed in its fraction field (by Gauss's Lemma again), also $A$ is "radically closed" in $\text{Frac}(C)$.  
The quotient ring $C/\langle w \rangle$ is $$k[s_1,\dots,s_{n-1},x]/\langle r_n \rangle, $$ $$r_n = (1-x^n)s_1 + x^{n-1}s_2 - x^{n-2}s_3 + \dots + (-1)^{n}x^2s_{n-1} + x^{n+1}. $$  By the Jacobian criterion, this ring is smooth over $k$.  Moreover, the ring extension $k[x]\to C/\langle w \rangle$ is an affine space bundle (over $D(x)$, resp. $D(1-x^n)$, it is trivialized).  Thus, it is irreducible of dimension $n-1$.  Since the quotient ring is regular of dimension $n-1$, the ideal $wC$ in $C$ is a prime ideal.  Therefore $w$ is a prime element of $C$.
The argument that $B/\langle w \rangle$ is integral of dimension $n-1$ is similar, but slightly more involved.  The quotient ring is $$B/\langle w \rangle = k[s_1,\dots,s_{n-1}]/\langle g_n \rangle,$$ $$g_n = (1-s_n)s_1^n + s_1^{n-2}s_ns_2 + \dots + (-1)^{n-1}s_1s_n^{n-1}s_{n-1} + (-1)^ns_n^n.$$ By the Jacobian criterion, the singular locus comes from the prime ideal $\langle s_1,s_n \rangle$.  So the ring is reduced.  Moreover, the ring extension $k[s_1,s_{n-1}] \to B/\langle w \rangle$ is flat of relative dimension $n-3$ away from $\langle s_1, s_{n-1}\rangle$, with the fiber over $\langle s_1,s_{n-1}\rangle$ of dimension $n-2$.  Since a closed subset of dimension $n-2$ cannot be an irreducible component of the complete intersection ring $B/\langle w \rangle$ of dimension $n-1$, it follows that every fiber dominates $\text{Spec}k[s_1,s_{n-1}]\setminus \langle s_1,s_{n-1} \rangle$.  The restriction over this open set is again an affine space bundle, hence it is irreducible.  Therefore, since $B/\langle w \rangle$ is reduced and irreducible, the element $w$ of $B$ is prime.  
I want to emphasize once more the geometric nature of the example.  You have a number of hypotheses -- "radically closed", prime element in $B$, etc.  These are "generically" satisfied.  So if you write down a family of complete intersection rings that fail $R1$ yet are sandwiched between factorial rings, "typically" all of your hypotheses should hold.  In the example above, both $A\to B$ and $A\to C$ are finite and flat.  So neither finiteness nor flatness are the issue.  The issue is that you can "pinch" $C$ to make a complete intersection ring that is not $R1$.
