Surely not reiterating history, but giving a semi-conceptual reason for the non-PID-ness of the 23rd cyclotomic field: the field $k=\mathbb Q(\sqrt{-23})$ has class number 3, as follows not-too-labor-intensely from the class-number formula for complex-quadratic fields. Now we grant ourselves something a bit serious, but 100 years old, about the "Hilbert classfield": the maximal unramified abelian extension of a number field has Galois group isomorphic to the ideal class group. Thus, there is an unramified cubic abelian extension $H$ of $k$. Since the 23rd cyclotomic field $F$ is of degree 11 over $k$ (e.g., by Gauss sum considerations), that cyclotomic field does not contain $H$. Thus, the compositum $HF$ is unramified (by multiplicativity of ramification indices) cubic over $F$. Again by the Hilbert-classfield result, this implies that the class number of $F$ is divisible by $3$.