Let $(R, \mathfrak m)$ be a Valuation ring , of finite Krull dimension say $d$, such that $\mathfrak m$ is not principal, hence $\mathfrak m$ is not finitely generated and $\mathfrak m^2=\mathfrak m$ .
Then can we find a principal ideal $J=(r)$ in $R$ such that height $(J)=d$ ? i.e. can we find $r \in R $ such that $\mathfrak m$ is the only prime ideal containing $r$ ?
Best Answer
In a valuation domain, ideals are totally ordered. That is if $I,J$ are $R$-ideals, then $I \subset J$ or $J \subset I$.
Since $\dim R = d < \infty$, there exists a sturated chain of prime ideals $p_0 \subset p_1 \subset \cdots \subset p_d = m$. Thus we can find an element $x$ such that $x \in m \setminus p_{d-1}$. Let $I = xR$. Since $I \not\subset p_{d-1}$, $p_{d-1} \subset I$. Suppose that $p$ is a prime containing $I$. Since $p_{d-1} \subset I \subset p$, the height of $p$ is at least $d$. Thus, $p = m$.