Find a $\mathbb{Q}$-basis for the splitting field of $x^4-2$ over $\mathbb{Q}$

Question

Let $\mathbb{K}$ be the splitting field of $x^{4}-2$ over $\mathbb{Q}$ with $\mathbb{K}\leq \mathbb{C}$ a field extension. Determine a $\mathbb{Q}$-basis for $\mathbb{K}$.

First, denote $f(x)=x^{4}-2$, I've already shown that the spliting field of $f(x)$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[4]{2},\sqrt[4]{2}i)$ and is indeed a degree 8 extension of $\mathbb{Q}$.

However, I'm having trouble determining a $\mathbb{Q}$-basis for $\mathbb{K}$.

I believe $(1,\sqrt[4]{2},\sqrt[4]{4})$ would be a basis for $\mathbb{Q}[\sqrt[4]{2}]$, but that's all I've gotten so far. Any advice is appreciated!

Answer 1

We want to make use of the following theorem (which proves the degree formula for field extensions):

Let $F\le E\le K$ be finite extensions. Let $\{\alpha_i\in E: 1\le i\le n\}$ be an $F$-basis for $E$ and $\{\beta_j\in K: 1\le j\le m\}$ be a $E$-basis for $K$. Then $\{\alpha_i\beta_j:1\le i\le n, 1\le j\le m\}$ is an $F$-basis for $K$.

First, let's fix some mistakes. The set $\{1,2^{1/4},4^{1/4}\}$ is not a $\mathbb{Q}$-basis for $\mathbb{Q}(2^{1/4})$. We know that $x^4-2$ is the minimal polynomial of $2^{1/4}$, so any basis must have $4$ elements. In fact, one such basis is $\{1,2^{1/4},2^{1/2},2^{3/4}\}$.

Let $\mathbb{E}:=\mathbb{Q}(2^{1/4})$. It remains to find a $\mathbb{E}$-basis for $\mathbb{K}$. Note that $\mathbb{K}=\mathbb{Q}(2^{1/4},i)=\mathbb{E}(i)$. We know that $x^2+1$ remains irreducible over $\mathbb{E}$ (because the roots of $x^2+1$ are non-real, but $\mathbb{E}\subset\mathbb{R}$). Then $x^2+1$ is the minimal polynomial of $i$ over $\mathbb{E}$, so $\{1,i\}$ is a $\mathbb{E}$-basis for $\mathbb{K}$.

By the quoted theorem, we then have that $\{1,2^{1/4},2^{1/2},2^{3/4}, i, 2^{1/4}i,2^{1/2}i,2^{3/4}i\}$ is a $\mathbb{Q}$-basis for $\mathbb{K}$.