Question in finding a new $\mathbb{Q}$-basis for $F/\mathbb{Q}$.

Question

Let $F$ be the splitting field of $x^4 - 2$ over $\mathbb{Q}$. Let $G$ be its Galois group. When viewed as a $\mathbb{Q}$- vectorspace, $F$ has the following basis:

$$\mathcal{B}=\{1,2^{1/4},2^{1/2},2^{3/4},i,2^{1/4}i,2^{1/2}i,2^{3/4}i\}$$

According to the Normal Basis Theorem, there exists some $x\in F$ such that the set $\{g(x)|g\in G\}$ is a $\mathbb{Q}$-basis of $F$.

My questions is how we can find such $x\in F$? It seemed natural to me that we start doing something to the basis we already have. But, for example, $x$ definitely cannot be $i$ or $2^{1/4}$, since for any $g\in G$, we must have $g(i)=\pm i$ and $g(2^{1/4})=\pm 2^{1/4}$. For the same reason, if you take some linear combination, for example, $i+ 2^{1/4}$, then $g(i+2^{1/4}) = \pm i \pm 2^{1/4}$. So how should we find such $x$?

Answer 1

We try applying a method in K. Conrad's note: http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/linearchar.pdf

Since $\pm i \pm i^k2^{1/4}$ does not work because they sum to $0$, we modify the choice of $\alpha$ as $$\alpha= 1+ i + 2^{1/4} + 2^{1/2}+ 2^{3/4}+i 2^{1/4} +i 2^{1/2} + i 2^{3/4}.$$

The Galois group $G\simeq D_{2\times 4}$ is generated by $\sigma$ and $\tau$ satisfying $$ \sigma(2^{1/4})= i2^{1/4}, \ \ \sigma(i)=i,$$

$$ \tau(2^{1/4})=2^{1/4}, \ \ \tau(i)=-i,$$ $$ \sigma^4=\tau^2=\mathrm{Id}, \ \ \tau\sigma= \sigma^{-1}\tau.$$

Then we have $$\alpha= 1+ i + 2^{1/4} + 2^{1/2}+ 2^{3/4}+i 2^{1/4} +i 2^{1/2} + i 2^{3/4},$$ $$ \sigma(\alpha)= 1+i-2^{1/4}-2^{1/2}+2^{3/4}+i2^{1/4}-i2^{1/2}-i2^{3/4}, $$ $$\sigma^2(\alpha)=1+i-2^{1/4}+2^{1/2}-2^{3/4}-i2^{1/4}+i2^{1/2}-i2^{3/4},$$ $$\sigma^3(\alpha)=1+i+2^{1/4}-2^{1/2}-2^{3/4}-i2^{1/4}-i2^{1/2}+i2^{3/4},$$ $$\tau(\alpha)=1-i+2^{1/4}+2^{1/2}+2^{3/4}-i2^{1/4}-i2^{1/2}-i2^{3/4},$$ $$\sigma\tau(\alpha)= 1-i+2^{1/4}-2^{1/2}-2^{3/4}+i2^{1/4}+i2^{1/2}-i2^{3/4},$$ $$\sigma^2\tau(\alpha)=1-i-2^{1/4}+2^{1/2}-2^{3/4}+i2^{1/4}-i2^{1/2}+i2^{3/4},$$ $$\sigma^3\tau(\alpha)=1-i-2^{1/4}-2^{1/2}+2^{3/4}-i2^{1/4}+i2^{1/2}+i2^{3/4}.$$

Now, set up a $\mathbb{Q}$-linear relation $$ a_1 \alpha + a_2 \sigma(\alpha) + a_3 \sigma^2(\alpha)+ a_4 \sigma^3(\alpha) +a_5 \tau(\alpha) + a_6 \sigma\tau(\alpha) + a_7 \sigma^2\tau(\alpha) + a_8 \sigma^3\tau(\alpha)=0$$ must result in the following linear system over $\mathbb{Q}$. $$ \begin{pmatrix} 1 &1 &1 &1 &1& 1& 1& 1 \\ 1 &1 &1 &1 &-1 &-1 &-1 &-1\\ 1 &-1 &-1 &1 &1& 1& -1& -1\\ 1 &-1 &1 &-1 &1 &-1 &1 &-1\\ 1 &1 &-1 &-1 &1 &-1 &-1 &1\\ 1 &1 &-1 &-1 &-1 &1 &1 &-1\\ 1 &-1 &1 &-1 &-1 &1 &-1 &1\\ 1 &-1 &-1 &1 &-1 &-1 &1 &1\end{pmatrix}\begin{pmatrix}a_1\\a_2\\a_3\\a_4\\a_5\\a_6\\a_7\\a_8\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\\0\\0\\0\\0\end{pmatrix}.$$

An online matrix determinant calculator: http://comnuan.com/cmnn0100b/ gives the determinant of this $8\times 8$ matrix as $-4096$. Thus, the system is nonsingular, and all $a_i$ must thus be all $0$.

Therefore, we see that $\alpha$ generates the normal basis of $\mathbb{Q}(i,2^{1/4})$ over $\mathbb{Q}$.

Answer 2

Here’s a sketch of a construction for a normal basis in this case, but I’m afraid it’s too late at night for me to go through all the details.

Let’s use the chain of field inclusions $\Bbb Q\subset k=\Bbb Q(i)\subset K=\Bbb Q(i,\lambda)$, where $\lambda^4=2$. Now if we’re just looking for a basis of $K$ over $\Bbb Q$, we know that we can take a (two-element) basis of $k$ over $\Bbb Q$ and a (four-element) basis of $K$ over $k$, and multiply them together element by element to get eight things that are $\Bbb Q$-linearly independent.

I propose to do the same thing with a normal basis of $K$ over $k$ and the well-known normal basis $\{1+i,1-i\}$ of $k$ over $\Bbb Q$. Roughy speaking, almost any random linear combination of all the basis elements of $K$ over $k$ should do the trick for the top layer of the chain, and I calculated the $4$-by-$4$ determinant for the four $k$-conjugates of $\mu=1+\lambda+2\lambda^2+4\lambda^3$, and found that this $\mu$ gave a normal basis. Remember that the Galois group is $\{e,\sigma,\sigma^2,\sigma^3\}$, where $\sigma(\lambda)=\lambda^\sigma=i\lambda$. That is, $\lambda^{\sigma^m}=i^m\lambda$.

Now I say that since $1+i$ gives a normal $\Bbb Q$-basis of $k$, and $\mu$ gives a normal $k$-basis of $K$, it’s fairly easy to see that $(1+i)\mu$ and its seven other conjugates under the total Galois group form a $\Bbb Q$-basis of $K$. It’s the details of this that I’m now too groggy to fill you in on.