An example of a special $1$-dimensional non-Noetherian valuation domain

Question

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 $langle a_1rangle subsetneqqlangle a_2rangle subsetneqqlangle a_3rangle subsetneqqcdots$ and for each $nin mathbb{N}$, $langle a_irangle subsetneqqlangle a_{i+1}^nrangle$ for all $i$.

Badam Baplan · Answer

This isn't possible in any $1$-dimensional quasi-local domain $D$. If $a,bin D$ 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 $(a_{i+1}) = D$.

An example of a special $1$-dimensional non-Noetherian valuation domain

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