Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\Lambda$ be a Dedekind domain and $\mathcal{m}$ be a maximal ideal of $\Lambda$.

Is it possible that $\mathcal{m}=\mathcal{m}^2$?

If not, how can I prove it?

share|improve this question
4  
Wow, the site is really in good form tonight: three good answers all coming immediately. –  Pete L. Clark Mar 28 '11 at 10:12

5 Answers 5

A finitely generated idempotent ideal in an commutative ring is generated by an idempotent element (we dealt with this here a little while ago), and this applies to your $m$ since $A$ is noetherian. There is then $e\in A$ such that $m=Ae$ and $e^2=e$. But then $(1-e)e=0$ and —since your ring is a domain,— we have either $e=0$ or $e=1$, that is, $m=0$ or $m=A$.

(Notice this requires much less than Dedekindness..)

share|improve this answer

The ideals in Dedekind domains have unique factorization into products of maximal ideals. The equation $m = m^2$ represents two distinct factorizations of $m$ ($m = m$ and $m = m^2$), which is impossible.

share|improve this answer

No, this is not possible. The localisation $\Lambda_m$ is a principal ideal domain, hence $m\Lambda_m=t\Lambda$ for some prime element $t\in\Lambda$. The relation $m=m^2$ lifts to $\Lambda_m$ and yields $t\Lambda=t^2\Lambda$ in contradiction with the unique factorisation in $\Lambda_m$.

share|improve this answer

Apply NAK Lemma to conclude that $(1 + x)\mathcal{m} = 0$ for some $x \in \mathcal{m}$. Now $R$ is a domain and $1 + x \neq 0$ since $-1 \not \in \mathcal{m}$. So $\mathcal{m} = 0$. which is a contradiction.

share|improve this answer

Nonzero ideals in a Dedekind domain are invertible so cancellative, so if the domain is not a field, then $\rm\:m\:$ maximal implies $\rm\:m\ne 0\:,\:$ so $\rm\ m^2 = m\ \Rightarrow\ m = 1\:$ by cancelling $\rm\:m\:.$ In fact this generalizes to any ring. Namely, finitely generated idempotent ideals are principal, generated by an idempotent. See the three proofs in my prior answer here using the determinant trick or integral dependence, and Nakayama's lemma.

share|improve this answer
    
When you write "In fact this remains true...", my best guess as a reader as to what "this" refers to is that nonzero finitely generated ideals in any ring are invertible. As I know you know, this is not true, so I'm sure it's not what you meant... –  Pete L. Clark Mar 28 '11 at 16:07
    
@Pete: Obviously it refers to the generalization in the linked post. –  Bill Dubuque Mar 28 '11 at 16:10
    
It is not obvious to me, so it may perhaps not be obvious to someone else... –  Pete L. Clark Mar 28 '11 at 16:14
    
@Pete: I clarified the remark to eliminate possible misinterpretations. Thanks for mentioning it. –  Bill Dubuque Mar 28 '11 at 16:22

Your Answer

 
discard

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