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$\newcommand{\fa}{\mathfrak{a}}\newcommand{fb}{\mathfrak{b}}\newcommand{\fp}{\mathfrak{p}}$ Motivation: In elementary number theory, one proves that for $a,b\in\mathbb{N}$, $a b=\gcd(a,b)\operatorname{lcm}(a,b)$. This has an obvious "categorification": for any two ideals $\fa,\fb\subset\mathbb{Z}$, one has $\fa\fb=(\fa\cap\fb)(\fa+\fb)$. One can easily see that this identity holds in any Dedekind domain $A$. (Just note that $$ v_\fp(\fa\fb) = v_\fp(\fa)+v_\fp(\fb) = \min\{v_\fp(\fa),v_\fp(\fb)\} + \max\{v_\fp(\fa),v_\fp(\fb)\} = v_\fp(\fa+\fb)+v_\fp(\fa\cap\fb) = v_\fp((\fa+\fb)(\fa\cap\fb)) $$ for all prime ideals $\fp\subset A$.)

I wonder if this identity characterizes Dedekind domains. Note that this is certainly not the case if we allow $A$ to be non-noetherian. For example, if $A$ is any valuation ring, then $\fa\fb=(\fa+\fb)(\fa\cap\fb)$ for all ideals $\fa\subset A$ for essentially the same reason (just make the same computation using the valuation for $A$ instead of $v_\fp$). So, suppose we restrict to the following case:

Question 1: If $A$ is a noetherian local domain such that $\fa\fb=(\fa+\fb)(\fa\cap\fb)$ for all ideals $\fa,\fb\subset A$, is $A$ a discrete valuation ring?

Of course, a positive answer would imply that any noetherian domain whose ideals satisfy the identity is Dedekind (just localize at each prime ideal).

The next question is a long shot, but here goes:

Question 2: If $A$ is a local domain such that $\fa\fb=(\fa+\fb)(\fa\cap\fb)$ for all ideals $\fa,\fb\subset A$, is $A$ a valuation ring?

Any progress towards answering either question will be greatly appreciated!

Daniel Miller
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  • I think that the answer to your first question is yes. I remember reading it somewhere, and the author said "I've never found this characterization Dedekind domains to be too useful," or something like that :) It's possible I imagined this, and I haven't really thought about it, so take this with a grain of salt. – Keenan Kidwell Jul 09 '13 at 17:32
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    Okay, I found it. It's near the bottom of page 13 in Keith Conrad's notes: http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/idealfactor.pdf One needs to assume the domain is integrally closed, at least in the non-local case. I'm not sure about this for local domains. – Keenan Kidwell Jul 09 '13 at 17:36
  • Very nice! It would be interesting to see a proof though... – Daniel Miller Jul 09 '13 at 17:40

1 Answers1

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Proposition. Let $A$ be an integral domain. Then $A$ is Prüfer iff $\mathfrak{a}\mathfrak{b}=(\mathfrak{a}+\mathfrak{b})(\mathfrak{a}\cap\mathfrak{b})$ for all ideals $\mathfrak{a},\mathfrak{b}\subset A$.

Let me recall that one of the equivalent characterizations of Prüfer domains is the following: $A$ is Prüfer iff $A_P$ is a valuation ring for any prime ideal $P$. This shows immediately the "only if" part of the Proposition.

For the converse, we prove that every non-zero $2$-generated ideal of $A$ is invertible. Let $\mathfrak{c}=(x,y)$. If $x=0$ or $y=0$ there is nothing to prove. Suppose $x\neq 0$, $y\neq 0$. Then $\mathfrak{a}=(x)$ and $\mathfrak{b}=(y)$ are invertible and $$\mathfrak{c}(\mathfrak{a}\cap\mathfrak{b})\mathfrak{b}^{-1}\mathfrak{a}^{-1}=(\mathfrak{a}+\mathfrak{b})(\mathfrak{a}\cap\mathfrak{b})\mathfrak{b}^{-1}\mathfrak{a}^{-1}=\mathfrak{a}\mathfrak{b}\mathfrak{b}^{-1}\mathfrak{a}^{-1}=A,$$ so $\mathfrak{c}$ is invertible.

Now the answer to your questions is obviously positive.