Finding the 8 Automorphisms of $\mathbb{Q}[\sqrt[4]{2}, i]$

Question

I think I will have i with root that identity ,then -the root,-i with root ,the root alone I cannot understand finding 8 ??

Answer 1

First, note that $K = \mathbb{Q}(\sqrt[4]{2}, i)$ is the splitting field for $f(x) = x^4 - 2$, and because finite extensions of $\mathbb{Q}$ are separable, $K/\mathbb{Q}$ is a Galois extension. Thus, $|\text{Gal}(K/\mathbb{Q})| = [K:\mathbb{Q}] = 8$. So indeed there must be $8$ automorphisms. Thinking of these automorphisms in terms of their action on the roots of $f$, we can define one "base" automorphism, $\phi$, as follows:

$$\phi(\sqrt[4]{2}) = i\sqrt[4]{2}$$ $$\phi(i) = i$$

We can define another "base" automorphism, $\tau$, as follows:

$$\tau(i) = -i$$ $$\tau(\sqrt[4]{2}) = \sqrt[4]{2}$$

And, of course, we have the identity automorphism. Note that $\phi$ generates $3$ non-identity automorphisms under composition with itself. Next, we can generate more by composing $\phi$ with $\tau$ in a variety of ways. So, all told, here are the $8$:

$$\{\text{id}, \phi, \phi^2, \phi^3, \tau, \phi \tau, \phi^2 \tau, \phi^3 \tau \}$$

If you plot the roots of $x^4-2$ in the complex plane, notice these can be thought of as vertices of a square; visually, we can see that $\operatorname{Gal}(K/\mathbb{Q}) \cong D_4$, wherein $\phi$ provides the rotation and $\tau$ the reflection.

Answer 2

A couple of complements to the previous answer.

By Eisenstein's criterion, the polynomial $x^4 - 2$ is irreducible over $\mathbb{Q}$, thus $\lvert \mathbb{Q}(\sqrt[4]{2}): \mathbb{Q} \rvert = 4$. Clearly $i \notin \mathbb{Q}(\sqrt[4]{2})$, so that $\lvert \mathbb{Q}(\sqrt[4]{2}, i): \mathbb{Q}(\sqrt[4]{2}) \rvert = 2$, and $$ \lvert \mathbb{Q}(\sqrt[4]{2}, i): \mathbb{Q}| = \lvert \mathbb{Q}(\sqrt[4]{2}, i): \mathbb{Q}(\sqrt[4]{2}) \rvert \cdot \lvert \mathbb{Q}(\sqrt[4]{2}): \mathbb{Q} \rvert = 2 \cdot 4 = 8. $$ Since $\mathbb{Q}(\sqrt[4]{2}, i)$ is the splitting field over $\mathbb{Q}$ of a separable polynomial, you know that $\mathbb{Q}(\sqrt[4]{2}, i) / \mathbb{Q}$ is a Galois extension, and that the Galois group has order $8$, the same as the degree.

Also, any root of a rational polynomial is sent to another root of the same polynomial by an automorphism (that automatically fixes the rationals).

Hence if $\sigma$ is an automorphism, the candidates for the images of $\sqrt[4]{2}$ under $\sigma$ are $\sqrt[4]{2}, - \sqrt[4]{2}, i \sqrt[4]{2}, -i \sqrt[4]{2}$, and the candidates for the images of $i$ are $i, -i$. As you see, there are $4 \cdot 2 = 8$ possibilities. Since the Galois group has order $8$, all these $8$ possibilities are realized, and you get the eight automorphisms given by $$ \sqrt[4]{2} \mapsto \begin{cases}\sqrt[4]{2}\\ - \sqrt[4]{2}\\ i \sqrt[4]{2}\\ -i \sqrt[4]{2}\end{cases} \qquad i \mapsto \begin{cases}i\\ -i\end{cases} $$