# One-dimensional [Noetherian] UFD is a PID

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).

• This is not really a duplicate. In the linked topic, there is a proof only showing that any prime ideal is principal and then using prime ideal factorization in Dedekind domains. The OP asked for a proof, which does not need this machinery in Dedekind domains. The answer is to forget about ideal factorization but rather directly (and totally elementary) show that in such a domain the gcd admits Bezout coefficients. – MooS Apr 21 '15 at 12:45
• Noetherian condition is superfluous; see here. – user26857 Nov 6 '15 at 6:46

Usually you would say that a one-dimensional noetherian UFD is a Dedekind domain and for Dedekind domains UFD and PID is the same thing. Let us recap the proof on an elementary level:

First of all we show that every prime ideal is principal: Let $0 \neq \mathfrak p$ be a prime ideal and $0 \neq f \in \mathfrak p$. Since we have an UFD, we can factorize $$f = p_1^{r_1} \dotsb p_n^{r_n}$$ with the $p_i$ being irreducible, hence prime, since we have an UFD. Since $\mathfrak p$ is prime, we deduce $p_i \in \mathfrak p$ for some $i$. So we have a chain of prime ideals: $(0) \subsetneq (p_i) \subset \mathfrak p$. From dimension one we deduce $\mathfrak p = (p_i)$ is principal. So far we have shown that any non-trivial prime ideal (precisely the maximal ideals) of our ring is principal.

To show the result, we only have to show that $(f,g)$ is principal, since we can argue by induction due to the fact that any ideal is a priori finitely generated. Let

$$f = p_1^{r_1} \dotsb p_n^{r_n}, ~ g = p_1^{s_1} \dotsb p_n^{s_n}$$

The natural candidate is $d := p_1^{e_1} \dotsb p_n^{e_n}$ with $e_i = \min \{r_i,s_i\}$. We want to show $(f,g)=(d)$, which is equivalent to $(\frac{f}{d},\frac{g}{d}) = (1)$.

By our choice $d$ has the following property: For any $i$ we have that $\frac{f}{d}$ or $\frac{g}{d}$ does not admit the factor $p_i$. So there is no $(p_i)$ which contains the ideal $(\frac{f}{d},\frac{g}{d})$. But we have shown that any maximal ideal of our ring is of the form $(p_i)$. Hence $(\frac{f}{d},\frac{g}{d})$ is not contained in a maximal ideal, thus we deduce $(\frac{f}{d},\frac{g}{d}) = (1)$.

Edit: Maybe one should point out the critical argument: To show that a UFD is a PID, we only have to show that the greatest common divisor of two elements can be written with Bezout coefficients (This is what $(f,g)=(d)$ precisely states). After cancelling the gcd, we can assume that our two elements $f,g$ have no common irreducible factor. From this we deduce that there is no principal prime ideal containing both $f$ and $g$. So any prime ideal containing $(f,g)$ must be non-principal. And then we use that in a 1-dimensional UFD any prime ideal is principal.

• Great! Thank you for such a complete answer :) – user233193 Apr 21 '15 at 12:23
• Why do we need to show that $(f,g)$ is principal? I mean if we show primes are principal, are not we done? – Ninja Jul 30 '18 at 23:28
• @Ninja: I think so. A principal prime ideal cannot have height $\ge 2$. – pyon May 26 '20 at 19:32

[The proofs below do not require the Noetherian hypothesis].

We induct on "length" $$\ell\,$$ of an ideal $$\,I\ne 0\,$$ ($$\ell$$ = least number of prime factors of any $$\,0\neq i\in I).\,$$ If $$\ \ell = 0\$$ then some $$\,i\in I\,$$ has no prime factors,  thus $$\,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
$$(6)\ \$$ $$\rm D$$ is a $$\rm PID$$

Proof $$\$$ (sketch of $$\,1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 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

• In $(3\Rightarrow 4)$ you meant $(a,b)\not\subseteq (p)$? – Gabriel Soranzo May 8 '19 at 18:34
• @Macadam Yes - typo now fixed, thanks! – Bill Dubuque May 8 '19 at 19:04