# Height of principal ideal in Valuation ring with non-principal maximal ideal

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 $$r$$ ?

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, $$p = m$$.