Or in the contrapositive form
Is every one-dimensional UFD noetherian?
I know how to construct a non-noetherian UFD (polynomials in infinite number of variables over a field) and I know that it is even possible to construct a finite dimensional non-notherian UFD from here, but this example is 3-dimensional.
Since a noetherian one-dimensional UFD is a PID and vice versa, the question could be also rephrased as
Is every one-dimensional UFD a PID?
I guess the answer is negative, since it is a nice and elegant statement that have never heard about.
Every one-dimensional UFD is a PID (hence noetherian).
Let $\mathfrak p$ be a non-zero prime ideal of $R$ and $a\in\mathfrak p$, $a\neq 0$. Then $a$ can be written as a product of prime elements, so $\mathfrak p$ contains a prime element, say $p$. Then $(p)\subseteq\mathfrak p$ and since the dimension of $R$ is $1$ we get $(p)=\mathfrak p$. This shows that all prime ideals of $R$ are principal, and therefore $R$ is a PID.