TITLE: One-dimensional [Noetherian] UFD is a PID
QUESTION [10 upvotes]: I am looking for a reference which has a self-contained (elementary, that is, at the "undergraduate algebra level") proof of the the fact that any one-dimensional Noetherian UFD is a PID. Does anyone know such a reference? 
(Note: I have looked at the similar questions on this website but don't find them particularly useful).

REPLY [8 votes]: We give a few proofs proofs below (which do not use your Noetherian hypothesis). Our first proof essentially leverages a generalization of the Euclidean algorithm to PIDs, which shows ideals $\neq 0$ are generated by any element having a minimal number of prime factors [Dedekind-Hasse test].
We induct on the "prime length" $\ell\, $ of an ideal $I\ne 0\,$ ($\ell$ = least #prime factors of any $\,0\neq i\in I).\, $ If $\ \ell = 0\ $ then some $\,i\in I\,$ has no prime factors, $ $ so $\,i\,$ is a unit, $ $ therefore $\,I = (1)\,$ is principal. $ $ Else $\,\ell \ge 1\,$ hence a least length $\,i = jp\,$ for some prime $\,p.\,$ By hypothesis $\,(p)\,$ is maximal, thus $\,\color{#c00}{I\!+\!(p) = (1)}\,$ or $\,\color{#0a0}{(p)}.\,$ It cannot be  $\,\color{#c00}{(1)}\,$ else $\ I\supseteq jI,(jp)\,$ $\Rightarrow$ $\, I\supseteq jI\!+\!(jp)=j(\color{#c00}{I\!+\!(p)})= (j),\ $ contra $\,jp\,$ has least length. So $\,I\!+\!(p) = \color{#0a0}{(p)}\,$ so $\,(p)\supseteq I\Rightarrow p\mid I,\,$ i.e. $\, I = p\, (I\!:\!p).\,$ Notice that $\,j\in I\!:\!p\,$ so $I\!:\!p\,$ has shorter length so by induction $\,I\!:\!p = (k)\,$ so $\,I = p\,(I\!:\!p) = p(k) = (pk).$

Remark $\ $ Below is an alternative proof, which proves a bit more along the way.
Theorem $\rm\ \ \ TFAE\ $ for a $\rm UFD\ D$ 
$(1)\ \ $ prime ideals are maximal if nonzero, $ $ i.e. $\rm\ dim\,\ D \le  1$
$(2)\ \ $ prime ideals are principal
$(3)\ \ $ maximal ideals are principal
$(4)\ \ \rm\ gcd(a,b) = 1\ \Rightarrow\ (a,b) = 1, $ i.e. $ $ coprime $\Rightarrow$ comaximal 
$(5)\ \ $ $\rm D$ is Bezout  [i.e. all  $\rm(a,b)\,$ are principal ideals]
$(6)\ \ $ $\rm D$ is a $\rm PID$
Proof $\ $ [sketch of $\,\color{#c00}1 \!\Rightarrow\! 2\! \Rightarrow\! 3 \!\Rightarrow\! 4\! \Rightarrow\! 5\! \Rightarrow\! 6\! \Rightarrow\! \color{#c00}1]\ $ where $\rm\,p_i,\:\!P\,$ denote primes $\neq 0$
$(1\Rightarrow 2)$ $\rm\ \  p_1^{e_1}\cdots p_n^{e_n}\in P\,\Rightarrow\,$ some $\rm\,p_j\in P\,$ so $\rm\,P\supseteq (p_j)\, \Rightarrow\, P = (p_j)\:$ by dim $\le1$ 
$(2\Rightarrow 3)$ $ \ $ max ideals are prime, so principal by $(2)$
$(3\Rightarrow 4)$ $\  \rm \gcd(a,b)=1\,\Rightarrow\,(a,b) \not\subseteq (p) $ for all max $\rm\,(p),\,$ so $\rm\ (a,b) = 1$ 
$(4\Rightarrow 5)$ $\ \ \rm c = \gcd(a,b)\, \Rightarrow\, (a,b) = c\ (a/c,b/c) = (c)$ 
$(5\Rightarrow 6)$ $\  $ Ideals $\neq 0\,$ in Bezout UFDs are generated by an elt with least #prime factors
$(6\Rightarrow 1)$ $\ \ \rm (d) \supsetneq (p)$ properly $\rm\Rightarrow\,d\mid p\,$ properly $\rm\,\Rightarrow\,d\,$ unit $\,\rm\Rightarrow\,(d)=(1),\,$ so $\rm\,(p)\,$ is max