Ring Sandwiched between PIDs

Question

If I have three commutative rings $R \subset S \subset T$, such that $R$ and $T$ are principal ideal domains, will this imply that $S$ itself is a principal ideal domain?

Answer 1

For the original question, one counterexample would be $\Bbb Z \subseteq \Bbb Z[x]\subseteq \Bbb Q (x)$.

The ring in the middle is Noetherian but isn't Bezout, and thus certainly isn't a principal ideal ring.

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If, as you mentioned in the comment, we add that $Frac(R)=Frac(T)$, then the picture is different. By gathering up all the denominators of fractions of $R$ lying in $S$, you have a multiplicative set $M$ such that $M^{-1}R=S$. It's elementary to show that a localization of a principal ring is principal, and the localization of a Bezout ring is Bezout, so $S$ will have either of these properties if $R$ does. In this situation, $T$ doesn't play any role.

Answer 2

$k[t^3]\subset k[t^2,t^3]\subset k[t]$ .