A simple algebraic ring extension of a UFD having no prime elements Let $D$ be a UFD over a field $k$ of characteristic zero.
Assume that $w$ is algebraic over $D$.
Denote $R=D[w]$.
Observe that $R$ is not necessarily a UFD.

Can one find an example in which $R$ has no prime elements at all?

My example: $D=k[x^2]$, $w=x^3$ (its minimal polynomial over $k[x^2]$ is of degree $2$. Notice that $x^3$ is integral over $D$), $R=k[x^2][x^3]$;
is it true that $k[x^2,x^3]$ has no prime elements? Of course, as a Noetherian ring (or more generally, as a ring which satisfies ACCP), $R$ has irreducible elements, for example, $x^2$ and $x^3$, but these are not prime elements since $(x^2)(x^2)(x^2)=(x^3)(x^3)$.
See also this question.
Remark: If $w$ is transcendental over $D$, then $R=D[w]$ is a UFD, as a polynomial ring (in one variable) over a UFD. 
 A: A prime element is nothing else but a non-zero principal prime ideal.
$\mathbb C[x^2,x^3] \cong \mathbb C[u,v]/(u^2-v^3)$ is one-dimensional and the non-zero prime ideals are given by $(u,v)$ and $(u-U,v-V)$ where $U,V \in k \setminus \{0\}$ with $U^2=V^3$.
$(u,v)$ is not principal, because the curve is not regular at that point. All other points are regular, i.e. the corresponding prime ideals are locally principal. We still have to check whether they are principal.
Of course by rescaling, we find an automorphism which lets us assume $U=V=1$, i.e. we only have to show that $(u-1,v-1) \subset \mathbb C[u,v]/(u^2-v^3)$ is not principal. In other words we have to show that $(x^2-1,x^3-1) \subset \mathbb C[x^2,x^3]$ is not principal. This is easy because both generators are irreducible by the virtue of the absence of linear polynomials. Thus the only principal ideal, that contains $x^2-1$ and $x^3-1$ is the ideal $(1)$, but we already know that $(x^2-1,x^3-1) \neq (1)$, because it is a prime ideal.
