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 $p < 23$?

Eric Naslund
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    Some of this should be in the first chapter of Washington's _Cyclotomic Fields_. – Dylan Moreland Nov 25 '11 at 17:22
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    Actually, the objection was raised "on the floor" by Liouville, who noted the implicit assumption of unique factorization. Lamee acknowledged the gap, and worked to try to fill it (Cauchy also wrote several notes on the problem). However, completely independently, Kummer had in fact *already* shown that there was no unique factorization in many of these cyclotomic fields, and published it in 1844 (in an obscure journal); he communicated this to Liouville, who communicated it to the Academy and re-published it in his journal. – Arturo Magidin Nov 25 '11 at 20:40

4 Answers4


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.

franz lemmermeyer
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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 $3$.

paul garrett
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We know that $\Bbb{Z}[e^{2\pi i/23}]$ is the ring of integers of the 23rd cyclotomic field. Since the ring of integers is a PID iff it is a UFD, it suffices to understand why this ring is not a PID. For this, you can see my answer given here which proves that the ideal $Q = (2, \frac{1 + \sqrt{-23}}{2})$ is not principal in $\Bbb{Z}[e^{2\pi i/23}]$.


Failure at 23 is proved here. There is a lengthy discussion of factorization in cyclotomic fields in Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory, by Harold M. Edwards. This book appears to be freely available at Google books.

Gerry Myerson
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  • Is there any intuition behind the proof which you linked to? It pulls $1-\alpha +\alpha^21$ out of nowhere and then shows it factors in two different ways. – Eric Naslund Nov 25 '11 at 22:21
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    Perhaps the presentation at http://www.mathstat.concordia.ca/faculty/ford/Student/Veres/vthp.pdf is more helpful. Another approach is to prove that 2 is irreducible but divides $(1+z^2+z^4+z^5+z^6+z^{10}+z^{11})(1+z+z^5+z^6+z^7+z^9+z^{11})$ without dividing either factor, where $z$ is a primitive 23rd root of unity. – Gerry Myerson Nov 25 '11 at 23:40