# Primary ideals in valuation rings

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?

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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

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.

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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 $\neq\mathfrak{p}$.

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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).

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"Authoritative", I meant to say. –  Matthé van der Lee Apr 13 '13 at 2:14