Maximal ideal of Dedekind domain

Solution 1:

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.

Solution 2:

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 $\Lambda_m$.

Solution 3:

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..)

Solution 4:

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.