Maximal ideal of Dedekind domain 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?
 A: 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.
A: 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..)
A: 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 $\Lambda_m$.
A: 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.
A: 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.
