$\mathbb{Z}[\zeta]$ is not a UFD

Question

I am writing a short presentation (approximately 10 minutes long) on the topic of Fermat's Last Theorem. Obviously, the details presented will have to be pretty sparse but I am compiling a handout as well. Part of this handout will include theorems and results that I will use in my presentation but do not have time to prove.

One of the results I will be using is that $\mathbb{Z}[\zeta]$ is not a unique factorisation domain, where $\zeta$ is is the $p$-th root of unity $e^{\frac{2\pi i}{p}}$.

I have searched the internet and found no explicit proof. The book I am reading, Number Fields by Daniel A. Marcus, says that the unique factorisation of $\mathbb{Z}[\zeta]$ fails for $p=23$. So clearly the proof can use this as a counter-example. Does anyone have a formal proof for this?

score 5 · Accepted Answer · answered Mar 12 '17 at 20:07

5

Since $\mathbb{Z}[\zeta_p]$ is a Dedekind ring, UFD is equivalent to PID. For $p=23$ we can give an ideal which is not principal, e.g., $\mathfrak p := (2,(1+\sqrt{-23})/2)$. Hence $\mathbb{Z}[\zeta_{23}]$ is not a UFD. This is due to Kummer.

answered Mar 12 '17 at 20:07

Dietrich Burde

140,055

Thank you. Is there a simple way to prove that $\mathbb{Z}[\zeta_{p}]$ is a Dedekind ring? Or would I simply be showing that it is Noetherian etc? – PercyF2519 Mar 12 '17 at 20:12
It is easy. All rings of integers are Dedekind rings. And $\mathbb{Z}[\zeta_p]$ is the ring of integers of the cyclotomic field $\mathbb{Q}[\zeta_p]$ – Dietrich Burde Mar 12 '17 at 20:13
Of course, thank you very much. – PercyF2519 Mar 12 '17 at 20:13
On the other hand, all $\mathbb{Z}[\zeta_p]$ are UFD's for primes less than $23$. – Dietrich Burde Mar 12 '17 at 20:16
So my book tells me. I find that quite interesting. Although the book skips on the details. – PercyF2519 Mar 12 '17 at 20:20
This leaves open the question of how to show $(2,(1+\sqrt{-23})/2)$ is not principal. – Gerry Myerson Jan 17 '24 at 22:15
1

@GerryMyerson This has been shown here already, I think. Right now I only found this post with a reference to Washington's book. – Dietrich Burde Jan 18 '24 at 12:07

$\mathbb{Z}[\zeta]$ is not a UFD

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