What is the norm function in $\mathcal{O}_{\mathbb{Q}(\zeta_8)}$? Given a number $z$ in the ring $\mathcal{O}$ of algebraic integers of $\mathbb{Q}(\zeta_8)$ expressed as $a + b \zeta_8 + ci + d (\zeta_8)^3$, with $a, b, c, d \in \mathbb{Z}$, how do you calculate the norm function $N(z) \in \mathbb{Z}$?
Since $$\zeta_8 = \frac{\sqrt{2} + \sqrt{-2}}{2}$$ and $$(\zeta_8)^3 = \frac{-\sqrt{2} + \sqrt{-2}}{2}$$ is it strictly necessary to include $d$? (I'm including that just to give some idea of what I have thought about and tried).
In his answer to Mr. King's question Is $(1+i)$ ramified in $\mathbb{Q}(i,\sqrt{2})$?, Mr. Soupe shows that $(1 - (\zeta_8)^3)(1 + (\zeta_8)^3) = 1 + i.$
It therefore makes sense that in this ring $2$ should have a norm of $16$, $N(1 + i) = 4$ and $N(1 + (\zeta_8)^3) = 2$. For a moment I thought maybe I could just take norms from the intermediate rings and multiply them.
Which I thought was a sensible plan on account of the fact that no real, rational prime is inert in $\mathcal{O}_{\mathbb{Q}(\zeta_8)}$. If a real, rational prime $p$ doesn't split in $\mathbb{Z}[\sqrt{-2}]$ or $\mathbb{Z}[\sqrt{2}]$, then it is congruent to $5 \bmod 8$, in which case it splits in $\mathbb{Z}[i]$. Obvious $2$ is not prime in any of these intermediate rings.
What is really confusing me here is figuring out when to add and when to subtract, e.g., $N(1 + 2 \sqrt{-2}) = 9$ and $N(1 + 2 \sqrt{2}) = 7$ in the respective intermediate rings. But then what about $(1 + 2 \sqrt{-2})(1 + 2 \sqrt{2}) = 1 + 2 \sqrt{2} + 8i + 
I don't know the norm function for "simpler" quartic rings like $\mathcal{O}_{\mathbb{Q}(\root 4 \of 2)}$ or $\mathcal{O}_{\mathbb{Q}(\root 4 \of 3)}$, so I would appreciate an explanation of that.
But my main question here is: what is the norm of a number $z \in \mathcal{O}_{\mathbb{Q}(\zeta_8)}$?
 A: You could just apply the definition of the Norm: multiply all conjugates.
Conjugates are obtained by applying automorphisms that map $\zeta_8$ to numbers with the same minimal polynomial over $\mathbb{Q}$, that is, $\zeta_8^k$ with $k$ odd. In the case of $\mathbb{Q}(\zeta_8)$, the automorphism group is isomorphic to Klein's Four group with the members:


*

*$\sigma_0\colon\ \zeta_8\mapsto\zeta_8$: Identity;

*$\sigma_1\colon\ \zeta_8\mapsto\zeta_8^3$: Swaps $\zeta_8$ with $\zeta_8^3$
(preserving $\sqrt{-2}$), changes sign of $\zeta_8^2=\mathrm{i}$;

*$\sigma_2\colon\ \zeta_8\mapsto\zeta_8^5$: Changes sign of $\zeta_8$ (preserving $\mathrm{i}$);

*$\sigma_3\colon\ \zeta_8\mapsto\zeta_8^7$: Complex conjugation (preserving $\sqrt{2}$).


For $z=a+b\zeta_8+c\mathrm{i}+d\zeta_8^3$ we obtain:
$$\begin{align}
    \operatorname{N}(z) &= \sigma_0(z)\,\sigma_3(z)\,\sigma_1(z)\,\sigma_2(z)
    = |z|^2 |\sigma_1(z)|^2 = |z|^2 |\sigma_2(z)|^2
\\ \operatorname{N}(a + b\zeta_8 + c\zeta_8^2 + d\zeta_8^3)
 &= |a + b\zeta_8 + c\zeta_8^2 + d\zeta_8^3|^2
 |a + d\zeta_8 - c\zeta_8^2 + b\zeta_8^3|^2
\\  &= |a + b\zeta_8 + c\zeta_8^2 + d\zeta_8^3|^2
 |a - b\zeta_8 + c\zeta_8^2 - d\zeta_8^3|^2
\\  &= |(a + \mathrm{i} c)^2 - \mathrm{i} (b + \mathrm{i} d)^2|^2
\\  &= (a^2 - c^2 + 2bd)^2 + (b^2 - d^2 - 2ac)^2
\\  &= (a^2 + b^2 + c^2 + d^2)^2 - 2(ab + bc + cd - da)^2
\\  &= (a^2 - b^2 + c^2 - d^2)^2 + 2(ab - bc + cd + da)^2
\\  &= (a^2 + c^2)^2 + (b^2 + d^2)^2 + 4 (ad - bc)(ab + cd)
\end{align}$$
As you can see, those expressions can be rearranged to emphasize some relationships with the norm expressions over the intermediate fields $\mathbb{Q}(\mathrm{i})$, $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{-2})$.
Other approaches that avoid non-rational numbers are given
in another thread.
Applied to $\mathbb{Q}(\zeta_8)$, we can write
$$\operatorname{N}(a + b\zeta_8 + c\zeta_8^2 + d\zeta_8^3)
    = \operatorname{Res}_X(X^4+1, a + bX + cX^2 + dX^3)$$
using polynomial resultants, or
$$\operatorname{N}(a + b\zeta_8 + c\zeta_8^2 + d\zeta_8^3)
    = \det(aI + bZ + cZ^2 + dZ^3)
    = \begin{vmatrix}a&-d&-c&-b\\b&a&-d&-c\\c&b&a&-d\\d&c&b&a\end{vmatrix}$$
using the companion matrix
$Z=\left(\begin{smallmatrix}0&0&0&-1\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{smallmatrix}\right)$.
