MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $R$ be an integral domain. Suppose that for any two nonzero ideals $I$ and $J$, we have $I \oplus J$ is isomorphic to $R \oplus IJ$ as $R$-modules. Does this implies $R$ is a Dedekind domain?

share|cite|improve this question
Suppose that $R$ is a local Noetherian integral domain with maximal ideal $\mathfrak{m}$ and residue field $k$. Your identity implies that $\dim_k(\mathfrak{m}/\mathfrak{m}^2)+\dim_k(\mathfrak{m}^{n-1}/\mathfrak{m}^n)=1‌​+\dim_k(\mathfrak{m}^{n+1}/\mathfrak{m}^n),$ for all $n\in\mathbb{Z}_{>0}$. This means that the Hilbert polynomial of $(R,\mathfrak{m})$ is (on the nose) $H(n)=(d-1)\binom{n+1}{2}+n$, where $d=\dim_k(\mathfrak{m}/\mathfrak{m}^2)$. Since $\dim R=\deg H(n)$, there are only two possibilities: Either $d=1$ and $H(n)=n$, so that $R$ is a d.v.r. and we are done. Or, $d>1$ and... – Keerthi Madapusi Pera Mar 24 '11 at 17:02
$\dim R=2$ (and regular in codimension one) with Hilbert-Samuel multiplicity $d-1$. I don't immediately see how to discount this possibility, though I've so far only used a very weak version of your hypothesis. – Keerthi Madapusi Pera Mar 24 '11 at 17:03
Fields fit the condition (just one nonzero ideal) and they are not Dedekind domains. – KConrad Mar 24 '11 at 19:54
I can never decide whether fields should be considered Dedekind domains. (This somewhat idle question has been debated at MO before.) – Tom Goodwillie Mar 24 '11 at 22:07
I don't know if this works, but did you try to find a counterexample by taking $R$ to be a direct limit of Dedekind domains ? – François Brunault Mar 25 '11 at 8:31

As Sampath has pointed out, we may assume that $R$ is local. Your hypothesis implies for any two non-zero ideals $I,J$, you have a surjection onto $R$. By Nakayama, this implies either $I\to R$ or $J\to R$ is surjective, since if neither is, then the both have images contained in the maximal ideal and so does their sum. But this means one of them is principal.

Having recognized the confusion I caused by my terseness, let me be more explicit. First, my definition of DD is: R a domain (not a field) and for any non-zero ideal $I$, there exists another ideal $J$ such that $IJ$ is isomorphic to $R$. Easy to see that $\text{Hom} (I,R)$ for any non-zero ideal $I$ can be identified naturally with $J=\{x\in K|xI\subset R\}$ where $K$ is the fraction field of $R$. Then, the definition of DD means that for any non-zero ideal $I$, defining $J$ as above, $IJ=R$.

Now assume that for any non-zero ideal $I$ of $R$, there exists a surjection $I\oplus I$ to $R$. Then, I claim that $R$ is a DD. The hypothesis implies, there exists $a,b\in I$, $x,y\in K$ with $xI,yI\subset R$ and $xa+yb=1$. Easy to check then that $I$ is generated by $a,b$. Thus all ideals are generated by atmost two elements and in particular $R$ is Noetherian. Now, by the above localization argument, $I$ is locally principal and the rest is clear. I hope this is clearer.

Even without localizing, letting $J$ as above, we have $x,y\in J$ and thus $1\in IJ\subset R$ and hence $IJ=R$.

share|cite|improve this answer
But the hypothesis includes the case $J=I$, in which case your argument shows that $I$ is principal. That seems to complete the proof, provided that $R$ is Noetherian. – Neil Strickland Mar 24 '11 at 18:42
How is Nakayama involved here? If the image of I+J is not contained in the max ideal, then the image of either I or J is not contained in it. Nothing to do with finite generation. – Tom Goodwillie Mar 24 '11 at 22:11
What's the reduction to the local case? – Tom Goodwillie Mar 24 '11 at 22:11
I still can't understand why we can assume $R$ is local. – yeshengkui Mar 25 '11 at 12:48
One can localize because: it suffices to show any local ring of $R$ is a d.v.r and the hypothesis localizes, as any ideal in a local ring of $R$ extends from its contraction to $R$. Once we are in the local case, the above argument shows that any ideal is principal. – Hailong Dao Mar 25 '11 at 19:25

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.