For which countable successor ordinals $\alpha$ is the reverse order isomorphic to the ideals of a PID ordered by inclusion?

Question

Let $\alpha$ be a countable successor ordinal and $\alpha^{\mathrm{op}}$ the reverse order. For which $\alpha$ is there a commutative principal ideal ring $R$ such that the ideals of $R$ form a chain isomorphic to $\alpha^{\mathrm{op}}?$ When can we find a PID $R?$

Answer 1

This is possible only when $\alpha\leq\omega+1$. Indeed, suppose $R$ is a principal ideal ring and the ideals of $R$ form a chain of order type $\alpha^{op}$, and let $x$ generate the unique maximal ideal of $R$. Note that if $(y)\subset R$ is any nonzero proper ideal then we can write $y=xz$ for some $z\in R$. If $y\mid z$, then $z=wy=wxz$ for some $w$, which implies $z=0$ and hence $y=0$ since $1-wx$ is a unit (because $wx$ is in the unique maximal ideal of $R$). Thus $(z)$ is strictly larger than $(y)$. Moreover, if $w\in (z)\setminus (y)$, we can write $w=zv$ and we must have $v\not\in (x)$ since otherwise $w$ would be in $(y)$. So $v$ is a unit, and $(w)=(z)$. It follows that there are no ideals strictly between $(w)$ and $(z)$.

We have thus shown that every nonzero proper ideal in $R$ has a successor in the inclusion order. This implies immediately that $\alpha\leq\omega+1$ (otherwise the $\omega$th largest ideal would be nonzero and have no successor). To see that any successor ordinal $\alpha\leq \omega+1$ is possible, note that if $k$ is a field, then $k[[x]]$ gives you $\alpha=\omega+1$ and $k[x]/(x^{n-1})$ gives you $\alpha=n$ for any finite $n$. If $R$ is a domain then we can use the fact that a PID is a UFD to conclude that the only possibilities are $\alpha=\omega+1$ (if $R$ has one prime element, i.e. if $R$ is a DVR) and $\alpha=2$ (if $R$ has no prime elements, i.e. $R$ is a field). Alternatively, you can use arguments similar to those in the previous paragraph to show that if $\alpha=n$ is finite, then $x^{n-1}=0$, so in the domain case we must have $n=2$.