Show a field extension is normal, then find the Galois group

Question

Let $K=\mathbb{Q}(2^{\frac{1}{4}}, i)$. Show that $K$ is normal over $\mathbb{Q}(i)$, and $Gal(K/ \mathbb{Q}(i))$ is cyclic of order 4.

What I have done:

I showed that $K$ is the splitting field for $x^4-2$ over $\mathbb{Q}$, and that $[K: \mathbb{Q}(i)]=4$.

I am a bit confused on how to show it's normal. By definition, we want to show $K/ \mathbb{Q}(i)$ is algebraic and the minimal polynomial of every element $\alpha $ of $K$ splits in $K[X]$. I don't see how that is happening.

Also, when constructing the Galois group, I know that $i$ must map to $i$, since any automorphism in the Galois group will fix the elements in $Q(i)$. But I'm still not sure how to show the Galois group is cyclic.

Any help will be appreciated!

Answer 1

There is an important result in field theory, which is the following:

Let $K/F$ be a field extension. Then $K/F$ is a finite normal extension if and only if $K$ is the splitting field over $F$ for some polynomial in $F[x]$.

The proof of this can be found in every undergraduate algebra textbook.

Note that if $K$ is the splitting field for $f$ over $F$ and $\alpha$ is a root of $f$ in $K$, then $K$ is the splitting field for $f$ over $F(\alpha)$. In your case, $K:=\mathbb{Q}(2^{1/4},i)$ is the splitting field for $f:=(x^4-2)(x^2+1)$ over $F:=\mathbb{Q}$, so $K$ is the splitting field for $f$ over $\mathbb{Q}(i)$. By the result stated above, $K/\mathbb{Q}(i)$ is a normal extension.

To compute the Galois group, you need to make good use of conjugate elements, namely:

Let $K/F$ be an algebraic field extension. Let $\alpha,\beta\in K$. Then there is a unique isomorphism $F(\alpha)\to F(\beta)$ fixing $F$ and mapping $\alpha$ to $\beta$ if and only if $\alpha$ and $\beta$ are conjugate elements over $F$.

Of course, any automorphism of $K$ fixing $\mathbb{Q}(i)$ is determined entirely by the image of $2^{1/4}$, and such automorphism must map every element to one of its conjugates over $\mathbb{Q}(i)$. The above result then implies that $\text{Gal}(K/\mathbb{Q}(i))$ has exactly 4 elements, each mapping $2^{1/4}$ to one of its 4 conjugates over $\mathbb{Q}(i)$. (This is true because $K=\mathbb{Q}(i,\beta)$ for any conjugate $\beta$ of $2^{1/4}$.) To show that the Galois group is cyclic, simply find an element in the group that has order 4, such as the one that maps $2^{1/4}$ to $2^{1/4}i$.