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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).

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    $\begingroup$ 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. $\endgroup$ – MooS Apr 21 '15 at 12:45
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    $\begingroup$ Noetherian condition is superfluous; see here. $\endgroup$ – user26857 Nov 6 '15 at 6:46
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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.

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

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
$(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

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

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