Basis for the field extension $\mathbb{Q}(3^{1/4},i)$ over $\mathbb{Q}$

Question

Consider the field extension $\mathbb{Q}(3^{1/4},i)$ over $\mathbb{Q}$. Find a basis for this feild extension over $\mathbb{Q}$.

I was thinking I would find a basis for $\mathbb{Q}(i)$ and a basis for $\mathbb{Q}(3^{1/4})$ over $\mathbb{Q}$. Then to take the pairwise product of each element in the basis to get a basis for the whole field extension. So far I've managed to prove that the set $B_{1}=\{1,i\}$ is a basis for $\mathbb{Q}(i)$.

What I'm having trouble with is showing that the set $B_{2} = \{1, 3^{1/4}, 9^{1/4}, 27^{1/4}\}$ is a basis for $\mathbb{Q}(3^{1/4})$.

Once that is show I could conclude that we have the following basis for the whole feild extension.

$$\{1, 3^{1/4}, 9^{1/4}, 27^{1/4}, i , 3^{1/4}i, 9^{1/4}i, 27^{1/4}i\}$$ Thanks in advance.

Answer 1

Great idea. Now, put $\;w:=\sqrt[4]3\;$ . If there are $\;a,b,c,d\in\Bbb Q\;$ with $\;a+bw+cw^2+dw^3=0\;$ , then it must be that $\;a=b=c=d=0\;$ , otherwise $\;w\;$ is the root of a polynomial of degree$\,<4\;$, which is impossible since $\;w\;$ is a root of the irreducible $\;x^4-3\in\Bbb Q[x]\;$ .

Since $\;[\Bbb Q(w):\Bbb Q]=4\;$, the above proves your set indeed is a basis for $\;\Bbb Q(w)/\Bbb Q\;$ .

Answer 2

By Eisenstein Criteria, the Polynomial $X^4-3$ is irreducible over $\mathbb Q$. Therefore, it is the minimal polynomial of $\sqrt[4]{3}$ over $\mathbb Q$.

From here it is easy to conclude that $B_2$ is a basis (this probably is a result from the textbook/class so may need no proof).