Unique factorization fails in cyclotomic integers for $p=23$

Question

I want to proof that unique factorization fails in $\mathbb{Z}[\zeta_{23}]$. The product the two fallowing cyclotomic integers is divisible by $2$ but neither of the two factors is. $$ \left( 1 + \zeta^2 + \zeta^4 + \zeta^5 + \zeta^6 + \zeta^{10} + \zeta^{11} \right) \left( 1 + \zeta + \zeta^5 + \zeta^6 + \zeta^7 + \zeta^9 + \zeta^{11} \right) =\\ 2\zeta^{17} + 2\zeta^{16} + 2\zeta^{15} + 2\zeta^{13} + 2\zeta^{12} + 6\zeta^{11} + 2\zeta^{10}+ 2\zeta^{9} + 2\zeta^{7} + 2\zeta^{6} + 2\zeta^{5}. $$ The only thing left to check is the irreducibility of $2$ in $\mathbb{Z}[\zeta_{23}]$ and my literature points out that this is "a non-trivial fact in this situation". Is there an (easy) argument to proof the irreducibility?

Answer 1

I don't know how much algebraic number theory you know, but after you have found the identity $\alpha \beta = 2 \gamma$ given above the rest of the argument is (almost) computation free.

Let $L = \mathbf{Q}(\zeta_{23})$, and let $K \subset L$ denote the subfield $\mathbf{Q}(\sqrt{-23})$. There is a surjection

$$ \mathbf{F}^{\times}_{23} = G:=\mathrm{Gal}(L/\mathbf{Q}) \rightarrow \mathrm{Gal}(K/\mathbf{Q}) = \mathbf{F}^{\times}_{23}/\mathbf{F}^{\times 2}_{23} = \mathbf{Z}/2 \mathbf{Z}.$$

The Frobenius element at $2$ is $[2]$, and since $2 = 25 \bmod 23$ is a square, it's easy enough to see that it has order $11$ in $G$ and maps trivially to $\mathrm{Gal}(K/\mathbf{Q})$.

It follows from basic algebraic number theory that:

1. There is a factorization $(2) = \mathfrak{P} \mathfrak{Q}$ as prime ideals in $L$.

2. There is a factorization $(2) = \mathfrak{p} \mathfrak{q}$ as prime ideals in $K$, and the relative norm of $\mathfrak{P}$ and $\mathfrak{Q}$ to $K$ is $\mathfrak{p}^{11}$ and $\mathfrak{q}^{11}$ respectively.

3. The norm of $\mathfrak{p}$ and $\mathfrak{q}$ are both $2$.

To show that $2$ is irreducible, you need to show that $\mathfrak{P}$ is not a principal ideal. Assume that $\mathfrak{P} = (\alpha)$. Then

$$\mathfrak{p}^{11} = (N_{L/K}(\alpha)).$$

So now you just need to show that $\mathfrak{p}^{11}$ is not principal. Note that

$$\mathfrak{p}^3 \mathfrak{q}^3 = (8) = \left(\frac{3 + \sqrt{23}}{2} \right)\left(\frac{3 - \sqrt{23}}{2} \right).$$

From this it follows that $\mathfrak{p}^3$ and $\mathfrak{q}^3$ are principal. But if $\mathfrak{p}^3$ and $\mathfrak{p}^{11}$ are both principal then so is $\mathfrak{p}$. But this is impossible because there is no integral element of $K$ of norm $2$.

Answer 2

Surely the factorisation of $(8)$ given by user748541 should involve $\sqrt{-23}$ rather than $\sqrt{23}$.