Fermat's Last Theorem and Kummer's Objection

In 1847 Lamé had announced that he had proven Fermat's Last Theorem. This "proof" was based on the unique factorization in $\mathbb{Z}[e^{2\pi i/p}]$. However, Kummer, proved that when $p=23$ we do not have unique factorization, and in fact Kummer proved this 3 years earlier in 1844.

My question is: how can you prove that $\mathbb{Z}[\zeta_p]$ does not have unique factorization when $p = 23$, but does for $p < 23$?

Solution 1:

Two remarks.

1. Class field theory can be eliminated from Paul's answer: if the degree $(L:K)$ of an extension of number fields is coprime to the class number $h_K$ of $K$, the $h_K \mid h_L$. The proof follows by observing that the transfer of ideal classes $j: Cl(K) \longrightarrow Cl(L)$ composed with the relative norm is just raising to the $(L:K)$-th power in the class group of $K$. Since the relative degree in the example at hand is $11$, this remark applies here.

2. Showing that the class number is $1$ for $p < 23$ is difficult. Kummer proved that the ring of integers inside the 5th roots of unity is Euclidean, but this method becomes impossible to use for $p > 7$. Kummer showed that the class number is the product of two factors $h^-$ and $h^+$, and gave a simple formula for $h^-$. Using this result it is easy to show that $h^- = 1$ for $p < 23$. Showing that the factor $h^+$ is trivial is much more difficult. You can get an idea of the difficulty by consulting Schoof's calculations in the appendix of the 2nd edition of Washington's book.

Solution 2:

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 $3$.