Unique Prime Ideal in $\mathbb{Z}[\zeta_6]$ of Residual Characteristic 3

Question

I’m trying to find the unique prime ideal in $\mathbb{Z}[\zeta_6]$ of residual characteristic 3, but I’m having difficulty finding it.

I computed that $\mathbb{Z}[\zeta_6]\cong \mathbb{Z}[x]/(x^2-x+1)$. Therefore, let $P\subset \mathbb{Z}[\zeta_6]$ be a prime ideal, and $(3)\subset P$, I will show $1+\zeta_6\in P$, but I couldn’t…

Please let me know if there is any way I can solve.

Answer 1

Your stumbling block seems to be that $3$ is not prime in this ring. It factors as

$3=(-1)(i\sqrt3)^2.$

If $\omega$ is the unique sixth root of unity in the first quadrant of the Argand plane, then

$i\sqrt3=\omega(1+\omega)$

and therefore

$1+\omega=(i\sqrt3)\omega^5.$

Thereby $1+\omega$ lies in the actual prime ideal $(i\sqrt3)$.

Answer 2

By the way, this is part of a general pattern: for $\omega$ a primitive $2p$th root of unity, with $p$ prime (your $2\cdot 3$ case generalizes to $2\cdot p$), $\omega+1$ satisfies ${(x-1)^p-1\over (x-1)+1}=0$, so the product of $\omega^k+1$ over $k=1,3,5,\ldots,2p-1$ excluding $p$ itself is $p$. Further, the ratios of the those elements (in $\mathbb Z[\omega]$) are demonstrably units, so these all generate the same (prime) ideal, whose $p-1$ power (square, instead, for $p=2$) is $p\in\mathbb Z$.