Factorizing $2$ in $\mathbb{Q}(\zeta_{12})$

Question

I'm trying to factorize $2$ in $K=\mathbb{Q}(\zeta_{12})$. Since $\Delta_{\zeta_{12}}=2^4\cdot 3^2$, I know that $2$ and $3$ are the only primes that ramify in $K$. As $\mathcal{O}_{K}=\mathbb{Z}[\zeta_{12}]$, we only have to analyze the minimal polynomial, $\Phi_{12}(x)=x^4+x^2+1$, modulo $2$ and $3$ to understand how each of these ramify.

I see that $\Phi_{12}(x)=(x+1)^2(x+2)^2$ in $\mathbb{F}_3[x]$, so $(3)=(2,(\zeta_{12}+1))^2(2,(\zeta_{12}+2))^2$.

My problem is with $(2)$. $\Phi_{12}(x)$ is irreducible in $\mathbb{F}_2[x]$ (so $2$ is inert, and does not ramify(?)). What am I missing here?

Answer 1

What you are "missing" is that $\Phi_{12}(x)=x^4+x^2+1$ is not irreducible over $\mathbb{F}_2$, because we have $$ x^4+x^2+1=(x^2 + x + 1)^2. $$ So $2$ ramifies as it should do.

Answer 2

Dietrich has found the error, so I won't comment on that. If you actually want to do the factorization, note that $\displaystyle\left({2\over 3}\right)=-1\,$ so $(2)$ is inert in $F=K(\zeta_3)$, which is the maximal subfield where $(2)$ does not ramify, hence $F$ is the splitting and inertial field for $(2)$ and therefore in $K$ we have $(2)=\mathfrak{p}^2$ where $\mathfrak{p}$ has norm $4$. But clearly $(2)=(1+i)^2$, so this is the factorization in $K$.