MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A$ be a ${\bf valuation}$ ${\bf ring}$ in the classical sense: $A$ is a domain with quotient field $K$ and for every non-zero $x\in K$ one has $x\in A$ or $x^{-1}\in A$.

Now ${\bf Bourbaki}$ (Commutative Algebra, Chapter VI, exercise 1 for §4) suggests that if $\mathfrak{p}$ is any non-maximal prime ideal of $A$, then $A$ does not possess any $\mathfrak{p}$-primary ideals other than $\mathfrak{p}$ itself.

But $\mathfrak{p} = \mathfrak{p}A_\mathfrak{p}$ (cf. ${\bf Matsumura}$, Commutative ring theory, Theorem 10.1), which is the maximal ideal of $A_\mathfrak{p}$ (another valuation ring of $K$). Hence $\mathfrak{p}^2$ is $\mathfrak{p}$-primary in $A_\mathfrak{p}$, and it follows that $\mathfrak{p}^2\cap A = \mathfrak{p}^2$ is $\mathfrak{p}\cap A$ - primary in $A$ - that is, $\mathfrak{p}$-primary.

And it is easy to find examples where $\mathfrak{p}^2 \ne \mathfrak{p}$.

What am I missing?

share|cite|improve this question
Are Bourbaki and Matsumura working with the same definition of primary ideal? I know that, for non-Noetherian rings, there are at least 2 different definitions out there. – Dustin Cartwright Apr 12 '13 at 18:10
Yes, same definition: $\mathfrak{q}$ is $\mathfrak{p}$-primary in $A$ if $A/\mathfrak{q}$ is a non-zero ring in which every zero-divisor is nilpotent, and $\mathfrak{p}$ is the radical of $\mathfrak{q}$. – Matthé van der Lee Apr 12 '13 at 18:59
And $\mathfrak{p} = \mathfrak{p}A_\mathfrak{p}$ is easily seen. Take $x\in \mathfrak{p}$ and $s\in A-\mathfrak{p}$; then $s.x^{-1} \notin A$ (for else $s = (s.x^{-1}).x \in \mathfrak{p}$), so the inverse $x.s^{-1} \in A$ because $A$ is a valuation ring. As $(x.s^{-1}).s \in \mathfrak{p}$ and $s\notin \mathfrak{p}$, it follows that $x.s^{-1} \in \mathfrak{p}$. – Matthé van der Lee Apr 12 '13 at 19:15
Am I correct to think that in your example $A$ is not noetherian? For instance, I think $\mathfrak{p}$ is not a finitely generated ideal, is it? – Mahdi Majidi-Zolbanin Apr 13 '13 at 2:39
Also, as far as I can see, Bourbaki defines primary ideals only for noetherian rings. Since in the exercise that you refer to, they speak of primary ideals, perhaps a noetherian condition is missing there? Otherwise how can they speak of a $\mathfrak{p}$-primary ideal if the ring is not noetherian? On the other hand, will the exercise be trivial if a noetherian condition is added? – Mahdi Majidi-Zolbanin Apr 13 '13 at 3:28
up vote 0 down vote accepted

There are certainly valuation rings in which each non-maximal prime ideal has no primary ideal except itself. Any Noetherian valuation ring has this property for trivial reasons. Also, this is true for any valuation ring $R$ with value group of the form $G_1 \oplus \cdots \oplus G_n$ under lexicographic order, where each $G_i$ is a non-discrete (i.e. dense) subgroup of $\mathbb{R}$.

To see this, note that we can parametrize the entire prime spectrum of $R$ from looking at the value group. Indeed, the $\operatorname{Spec} R$ looks like $0 \subset \mathfrak p_1 \subset \mathfrak{p}_2 \subset \cdots \subset \mathfrak{p}_n$. To use the $G_i$, note that the nonzero elements of $\mathfrak p_1$ are the ones where the first coordinate of the value is positive; the elements of $\mathfrak p_2 \setminus \mathfrak p_1$ are those whose values' first coordinate is zero and second coordinate positive, those of $\mathfrak p_3 \setminus \mathfrak p_2$ have value with the first two coordinates $0$ and third coordinate positive, and so forth. The fact that $\mathfrak p_i^2= \mathfrak p_i$ comes directly from the fact that $G_i$ is dense in $\mathbb R$.

However, you are right, in that any non-Noetherian valuation domain with finite rank and discrete value group will fail this property. That is, every nonzero non-maximal prime has nontrivial primary ideals.

share|cite|improve this answer
Neil, thanks a lot! This is precisely what I was looking for. – Matthé van der Lee Apr 13 '13 at 22:12
You're very welcome! – Neil Epstein Apr 13 '13 at 23:43

Neil, I'm afraid to report that even in the example you present there exist non-trivial primary ideals to the non-zero non-maximal prime ideals:-)

Let us take value group $\mathbb{R}\oplus\mathbb{R}$. Then $\mathfrak{p}$ = {the elements of the associated valuation ring having value $(x,y)$ with $x\gt0$} is, as you state, the only non-zero non-maximal prime ideal. Therefore it is the only minimal overprime (and hence the radical) of any non-zero ideal contained in it. You are right that $\mathfrak{p}$ = $\mathfrak{p}^2$, but still there exist $\mathfrak{p}$-primary ideals other than $\mathfrak{p}$.

Indeed, consider $\mathfrak{q}$ = {elements having value $(u,v)$ with $u\geq$1}. Then $\mathfrak{q}$ is an ideal of $A$, and we have $\mathfrak{q}\subset\mathfrak{p}$. Hence $\surd\mathfrak{q}=\mathfrak{p}$ by the above. Now if $a$ and $b$ are in $A$ and $b\notin\mathfrak{p}$, the value of $b$ must be $(0,v)$ for some $v\in\mathbb{R}$ (with $v\geq0$). And if we also have $a\notin\mathfrak{q}$, and $(x,y)$ denotes $a$'s value, then necessarily $x \lt1$. And so the value of $ab$, being the sum $(x,y+v)$ of the values of $a$ and $b$, does not satisfy $x\geq1$, and therefore $ab\notin \mathfrak{q}$. This shows that $\mathfrak{q}$ is a primary ideal, and thus a $\mathfrak{p}$-primary ideal $\neq\mathfrak{p}$.

share|cite|improve this answer
Hm. I suppose you are right. I should have been more careful. Well, here's at least an example of a valuation ring where there exists a non-maximal nonzero prime $\mathfrak p$ with no proper non-primary subideals. Let's take a valuation ring with value group $\mathbb{Z} \oplus \mathbb{Z} \oplus \cdots \oplus \mathbb Z$, with lex order. That is, you have countably many copies of $\mathbb Z$ with order type $\mathbb N$, plus one at the end. Then let $\mathfrak p$ be the ideal of elements whose value has some nonzero entry before the last entry. Then $\mathfrak p$ is the union of ... – Neil Epstein Apr 14 '13 at 18:43
... the prime ideals that it properly contains. Indeed, these primes can be parametrized by $\mathbb N$, such that $\mathfrak{p}_{i-1} \subset \mathfrak{p}_i$ and $\mathfrak{p}_i \setminus \mathfrak{p}_{i-1}$ are the elements with zeros in the first $i-1$ places and positive value in the $i$th place. Clearly $\mathfrak{p} = \bigcup_i \mathfrak{p}_i$. Now, suppose $\mathfrak q$ is a $\mathfrak p$-primary ideal. Then since it is not contained in any prime properly contained in $\mathfrak p$, it must contain all such primes, and hence it contains their union $\mathfrak p$. That is, $\frak q=p$. – Neil Epstein Apr 14 '13 at 18:51
Neil, very clever. Thanks again! – Matthé van der Lee Apr 14 '13 at 19:22
You're welcome again. I should say that the property you are asking about is called "branched". A prime ideal $P$ in a commutative ring $R$ is called branched if there is some $P$-primary ideal properly contained in $P$; otherwise, it is unbranched. So, for instance, in a Noetherian integral domain, $0$ is always the only unbranched prime ideal – Neil Epstein Apr 14 '13 at 21:47

In the specific example set out in the above comments, $\mathfrak{p^2}$ is a $\mathfrak{p}$-primary ideal of $A$ - in contradiction with the statement quoted from Bourbaki (still, after all, of course, a very authoriative source).

share|cite|improve this answer
"Authoritative", I meant to say. – Matthé van der Lee Apr 13 '13 at 2:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.