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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.

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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.

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  • $\begingroup$ Of course! thanks a lot (again). I was thinking in this direction, but somehow managed to confuse myself... $\endgroup$ – KotelKanim May 27 '13 at 11:49

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