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Let $R$ be a commutative local Noetherian ring and $\mathfrak{m}$ its maximal ideal.

Prove that, if $\mathfrak{m}$ is principal, then $\mathrm{dim}(R)\leq 1$ (the Krull dimension of the ring).

Thank you.

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    $\begingroup$ Hint: Can you list all the ideals? Once you do that can you identify which ones are prime? $\endgroup$
    – Matt
    Commented Jun 3, 2012 at 17:03
  • $\begingroup$ No, I'm sorry, but commutative algebra has always been my Achiles' heel. Could you be so kind and detail a bit more? Thank you! $\endgroup$ Commented Jun 3, 2012 at 17:10
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    $\begingroup$ Let $\mathfrak{p}$ be a prime ideal. Then it is contained in $\mathfrak{m}$. But $\mathfrak{m}$ is principal, so ... etc. $\endgroup$
    – Zhen Lin
    Commented Jun 3, 2012 at 17:26
  • $\begingroup$ It follows that the generator of $\mathfrak{m}$ divides all of the generators of $\mathfrak{p}$ and then it follows that $\mathfrak{p}$ is either 0, or $\mathfrak{m}$, right? (We used that $R$ is Noetherian for $\mathfrak{p}$ to be f.g., I suppose, and local, for the uniqueness of $\mathfrak{m}$) $\endgroup$ Commented Jun 3, 2012 at 18:19
  • $\begingroup$ Try writing up your answer in the answer below, and we can comment to help improve it. You'll have more room! $\endgroup$
    – rschwieb
    Commented Jun 3, 2012 at 18:57

2 Answers 2

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First, let me explicitly assume $R$ is a noetherian local domain with a principal maximal ideal $\mathfrak{m}$.

Proposition. The Krull dimension of $R$ is at most $1$.

Proof. Let $\mathfrak{m} = (t)$, and let $\mathfrak{p}$ be prime. We know $\mathfrak{p} \subseteq \mathfrak{m}$, so it is enough to show that either $\mathfrak{p} = \mathfrak{m}$ or $\mathfrak{p} = (0)$. Suppose $\mathfrak{p} \ne \mathfrak{m}$: then $t \notin \mathfrak{p}$. Let $a_0 \in \mathfrak{p}$. For each $a_n$, because $\mathfrak{p}$ is prime, there exists $a_{n+1}$ in $\mathfrak{p}$ such that $a_n = a_{n+1} t$. By the axiom of dependent choice, this yields an infinite ascending sequence of principal ideals: $$(a_0) \subseteq (a_1) \subseteq (a_2) \subseteq \cdots$$ Since $R$ is noetherian, for $n \gg 0$, we must have $(a_n) = (a_{n+1}) = (a_{n+2}) = \cdots$. Suppose, for a contradiction, that $a_0 \ne 0$. Then, $a_n \ne 0$ and $a_{n+1} \ne 0$, and there is $u \ne 0$ such that $a_{n+1} = a_n u$. But then $a_n = a_{n+1} t = a_n u t$, so cancelling $a_n$ (which we can do because $R$ is an integral domain), we get $1 = u t$, i.e. $t$ is a unit. But then $\mathfrak{m} = R$ – a contradiction. So $a_n = 0$. $\qquad \blacksquare$


Here's an elementary proof which shows why we can reduce to the case where $R$ is an integral domain.

Proposition. Any non-trivial ring $A$ has a minimal prime.

Proof. By Krull's theorem, $A$ has a maximal ideal, which is prime. Let $\Sigma$ be the set of all prime ideals of $A$, partially ordered by inclusion. The intersection of a decereasing chain of prime ideals is a prime ideal, so by Zorn's lemma, $\Sigma$ has a minimal element. $\qquad \blacksquare$

Thus, we can always assume that a maximal chain of prime ideals starts at a minimal prime and ends at a maximal ideal. But if $R$ is a noetherian local ring with principal maximal ideal $\mathfrak{m}$ and $\mathfrak{p}$ is a minimal prime of $R$, then $R / \mathfrak{p}$ is a noetherian local domain with a principal maximal ideal $\mathfrak{m}$, and $\dim R = \sup_\mathfrak{p} \dim R / \mathfrak{p}$, as $\mathfrak{p}$ varies over the minimal primes.


Update. Georges Elencwajg pointed out in a comment that the first proof actually works without the assumption that $R$ is a domain, because $(1 - u t)$ is always invertible.

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So, following Zhen Lin:

Let $\mathfrak{p}$ be a prime ideal of $R$. Then it is contained in $\mathfrak{m}$, the maximal ideal.

But $\mathfrak{m}$ is principal, say $\mathfrak{m}=(m)$ and $\mathfrak{p}$ must be f.g., since $R$ is Noetherian, say $\mathfrak{p}=(p_1,\dots,p_n)$.

Now $\mathfrak{p} \subseteq \mathfrak{m}$ implies $m \mid p_i, \ \forall i=1,\dots,n$, so $\mathfrak{p}$ is either zero or $(m)=\mathfrak{m}$.

So the chain that gives the Krull dimension contains either nothing (if $\mathfrak{m}=0$) or $\mathfrak{m}$ only.

Hope I got it right...

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    $\begingroup$ The idea is essentially correct, but you should expand the part where you deduce $\mathfrak{p} = (0)$ or $\mathfrak{p} = \mathfrak{m}$. For instance, suppose $p_1 = a_1 m$ and $m \notin \mathfrak{p}$. Then we must have $a_1 \in \mathfrak{p}$. But then $a_1 = a_2 m$, and so on. How do we conclude that $a_1 = a_2 = \cdots = 0$? $\endgroup$
    – Zhen Lin
    Commented Jun 3, 2012 at 19:18
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    $\begingroup$ Well... yes, it's part of the argument. But it's more subtle than that. You have containment, but you don't know that it's strict. So you only know that $(a_n) = (a_{n+1}) = (a_{n+2}) = \cdots$ for $n \gg 0$. Now suppose $a_n \ne 0$, $a_{n+1} \ne 0$. Why do we have a contradiction? $\endgroup$
    – Zhen Lin
    Commented Jun 3, 2012 at 19:38
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    $\begingroup$ No, not quite. First, $(a_n) = (a_{n+1})$ implies there is a non-zero $u$ such that $u a_n = a_{n+1}$. But $a_n = a_{n+1} m$, so $a_n = a_n u m$. Now, since $(0)$ is prime, $R$ is an integral domain, so we can cancel $a_n$ to get $1 = u m$. But then $m$ is a unit – contradiction. $\endgroup$
    – Zhen Lin
    Commented Jun 3, 2012 at 19:51
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    $\begingroup$ Dear @Zhen, no, don't worry! We don't need that $R$ is a domain: we have $a_n=a_{n+1}m$ and $a_{n+1}=ua_{n}$ so that $a_n(1-um)=0$. But $(1-um)$ is invertible because $\mathfrak m=Rad(R)$ , the Jacobson radical of $R$. So $a_n=0$ and then $a_{n-1}=...=a_1=0$. From which $\mathfrak p=0$ follows, just as you said. $\endgroup$ Commented Jun 3, 2012 at 20:28
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    $\begingroup$ Dear Adrian: sweet dreams! And let me tell you , in order to completely reassure you, that we still have the nuclear option: since $\mathfrak m$ is minimal over $m$, Krull's principal ideal theorem tells you that $height (\mathfrak m)=dim(R)\leq 1$ $\endgroup$ Commented Jun 3, 2012 at 20:52

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