Show that the splitting field of $x^8-3$ has degree 32 over $\mathbb{Q}$

Question

I have already determined that a splitting field for $f(x) = x^8 - 3$ over $\mathbb{Q}$ is $K= \mathbb{Q}(i , \sqrt{2}, 3^{\frac{1}{8}})$. I have the following tower relationship:

$$[K: \mathbb{Q}] = [K: \mathbb{Q}(\sqrt{2}, 3^{\frac{1}{8}})][\mathbb{Q}(\sqrt{2}, 3^{\frac{1}{8}}):\mathbb{Q}(3^{\frac{1}{8}})][\mathbb{Q}( 3^{\frac{1}{8}}): \mathbb{Q}] = 2[\mathbb{Q}(\sqrt{2}, 3^{\frac{1}{8}}):\mathbb{Q}(3^{\frac{1}{8}})] 8.$$

We have $[\mathbb{Q}( 3^{\frac{1}{8}}): \mathbb{Q}] = 8$ since $f(x)$ is irreducible over $\mathbb{Q}$ by Eisenstein applied to $p = 3$. And $[K: \mathbb{Q}(\sqrt{2}, 3^{\frac{1}{8}})] = 2$ since $\mathbb{Q}(\sqrt{2}, 3^{\frac{1}{8}}) \subseteq \mathbb{R}$ and we can go from $\mathbb{Q}(\sqrt{2}, 3^{\frac{1}{8}})$ to $K$ by adjoining the complex number $i$. So all that remains to show is that $[\mathbb{Q}(\sqrt{2}, 3^{\frac{1}{8}}):\mathbb{Q}(3^{\frac{1}{8}})] =2$, which can be accomplished by showing that $\sqrt{2} \notin \mathbb{Q}(3^{\frac{1}{8}})$. But I am not sure how to show this, particularly because arbitrary elements of $\mathbb{Q}(3^{\frac{1}{8}})$ are quite unwieldy (i.e., of the form $a_0 + a_1 3^{\frac{1}{8}}+ \cdots + a_7 3^{\frac{7}{8}}$ for $a_i \in \mathbb{Q}$).

Hints or solutions are greatly appreciated.

Answer 1

I would change the order you are doing things in. For example, it is also true that $$ [K:\mathbb{Q}]=[K:\mathbb{Q}(\sqrt{2},3^{1/8})][\mathbb{Q}(\sqrt{2},3^{1/8}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}]. $$ Now you just have to show that $3^{1/8}$ is not in $\mathbb{Q}(\sqrt{2})$, which I'm assuming will be a little easier.

Answer 2

To show that $\Bbb Q(\sqrt{2}) \not\subset \Bbb Q(3^{1/8})$, you can prove that $Q(3^{1/8})$ is not a degree four extension of $\Bbb Q(\sqrt{2})$. If that were the case, the polynomial $x^8-3$ would have a quartic factor with coefficients in $\Bbb Q(\sqrt{2})$. Note that the constant term would be the product of four roots of $x^8-3$, which are all of the form $\zeta_8^i3^{1/8}$. We have now reduced the problem to showing $\zeta_8^i \sqrt{3} \not\in \Bbb Q(\sqrt{2})$, which should be more manageable.

Answer 3

Let $$\sqrt 2=a_0 + a_1 3^{\frac{1}{8}}+ \cdots + a_7 3^{\frac{7}{8}}$$ $$2=(a_0 + a_1 3^{\frac{1}{8}}+ \cdots + a_7 3^{\frac{7}{8}})^2$$

We do not need to find the product exactly;

Notice that $(a_7)^23^{\frac{14}{8}}$ is the largest passible constitute. Hence we must have $a_7=0$

Then $(a_6)^2 3^{\frac{12}{8}}$ is the largest passible one. Then $a_6=0$.

By contuniuing like this, we will get contradiction.

Edit: After 8 years later, I noticed that there is a need for edit due to the comments. I do not remember what I have thought at that time, so I hope following should be helpful:

$$2=(a_0 + a_1 3^{\frac{1}{8}}+ \cdots + a_7 3^{\frac{7}{8}})^2=\sum_{i=0}^{14} c_i 3^{i/8}=\sum_{i=0}^{7} c_i 3^{i/8}+\sum_{i=8}^{14} c_i 3^{i/8}$$

$$\sum_{i=0}^{7} c_i 3^{i/8}+\sum_{i=0}^{6} 3c_{i+8} 3^{i/8}=\sum_{i=0}^{6} (c_i+3c_{i+8}) 3^{i/8}+c_73^{7/8}:=\sum_{i=0}^{7} b_i 3^{i/8}$$

where $$c_i=\sum a_ka_{i-k}$$ and $$b_i=c_i+3c_{i+8} \ for \ i=0,\ldots, 6 \ and \ b_7=c_7.$$

Since $x^8-3$ is irreducable, $ \{3^{i/8} \mid for \ i=0,..., 7 \}$ is a base, and so the writing $$2=\sum_{i=0}^{7} b_i 3^{i/8}$$ must be unique. Hence, $$b_0=2, b_i=0 \ for \ i=1,...,7 .$$

Possibly, at that time I thought that we can easily obtain a contradiction by knowing that $b_i=0$ for $i\geq 1$. I will think about whether there is a practical way of doing this.