I am looking for a $1$-dimensional non-Noetherian valuation domain $R$ such that there exists a sequence $\{a_i\}_{i=1}^\infty$ of elements of $R$ such that $\langle a_1\rangle \subsetneqq\langle a_2\rangle \subsetneqq\langle a_3\rangle \subsetneqq\cdots$ and for each $n\in \mathbb{N}$, $\langle a_i\rangle \subsetneqq\langle a_{i+1}^n\rangle$ for all $i$.
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7$\begingroup$ This does not exist. The fact that a valuation ring has dimension $1$ is equivalent to the value group being "rank $1$", i.e. archimedean, which in turns translates to the property of the valuation ring: For any two nonzero elements $a, b$, there is $n$ such that $(b^n) \subseteq (a)$. The condition on ideals you impose would in fact mean that the valuation is of "infinite rank", which translates to $R$ having infinite dimension. $\endgroup$– Pavel ČoupekCommented Jul 21, 2020 at 5:12
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This isn't possible in any $1$-dimensional quasi-local domain $D$. If $a,b\in D$ and $b$ is not a unit then consider the multiplicative set $S$ generated by $b$. Clearly $S$ is not disjoint from any nonzero prime of $D$, so $S = D \setminus 0$. Thus $a \in S$ which means some power of $b$ divides $a$.
If you had an ascending chain $(a_i)$ which satisfied, for any $i$, the condition $(a_i) \subsetneq (a_{i+1}^n)$ for all $n \in \mathbb{N}$, then that chain would have to stabilize at $(a_{i+1}) = D$.
