Find basis and degree of splitting field of degree 8 (another question)
Solution 1:
Yes!
Definitely $\mathbb{Q}(\sqrt[4]{2},i)=\mathbb{Q}[\sqrt[4]{2},i]$, so your set is generating $\mathbb{Q}(\sqrt[4]{2},i)$ as a $\mathbb{Q}$-vector space. Since the $\mathbb{Q}$-diemension of $\mathbb{Q}(\sqrt[4]{2},i)$ (i.e., the degree of the extension) is 8 and the cardinality of your set is also 8, then it must be a $\mathbb{Q}$-basis.