Determine the maximal ideals of $R=\mathbb{Z}[\sqrt[4]{2},\sqrt[4]{2}\,i]$ containing 5.

Question

I want to determine the maximal ideals of $R=\mathbb{Z}[\sqrt[4]{2},\sqrt[4]{2}\,i]$ containing 5, and I'm not familiar with the ring structure of $R$, thus have completely no idea. I am trying to represent $R$ as a quotient of polynomial ring (with no progress), am I in the right direction?

Answer 1

The maximal ideals of $R$ containing $5$ are the kernels of the ring homomorphisms from $R$ to $k=\overline{\Bbb F_5}$, the algebraic closure of the field of $5$ elements. Each such $\phi$ is determined by $\alpha=\phi(\sqrt[4]2)$ and $\beta=\phi(i\sqrt[4]2)$. Two such isomorphisms have the same kernel if one is the other composed with an automorphism of $k$.

Both $\alpha$ and $\beta$ are fourth roots of $2$, and $2$ has order $4$ in $k$. Therefore $\alpha$ and $\beta$ have order $16$ in $k$. The smallest solution of $5^r\equiv1\pmod{16}$ is $r=4$, so $k(\alpha)=k(\beta)=\Bbb F_{5^4}$. In $k$, $\beta^2=-\alpha^2$, so $\beta=\pm2\alpha$. Fix a solution $\alpha_0$ of $\alpha^4=2$ in $k$. Then the possible $(\alpha,\beta)$ are $(\alpha_0^{5^s},2\alpha_0^{5^s})$ and $(\alpha_0^{5^s},-2\alpha_0^{5^s})$. So there are eight homomorphisms $\phi$, but up to composition with automorphisms of $k$, only two essentially different ones.

The kernels of these homomorphisms are thus $I_1=((1-2i)\sqrt[4]2,5)$ and $I_2=((1+2i)\sqrt[4]2,5)$. These are the maximal ideals you seek.