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I've been asked to find $[ \mathbb{Q}(\sqrt{2}, i ): \mathbb{Q} ]$, and to write down a basis.

Now I know that $[ \mathbb{Q}(\sqrt{2}, i ): \mathbb{Q} ] = 4$, and that a basis could be $ \left \{ 1, i, \sqrt{2}, i\sqrt{2}\right \} $ it is whether the way I am explaining how I arrived here is satisfactory explanation.

Utilising the tower law I want to find both $[ \mathbb{Q}(\sqrt{2}, i ): \mathbb{Q(\sqrt{2})} ]$ and $[ \mathbb{Q}(\sqrt{2}): \mathbb{Q} ]$ and multiply.

For $[ \mathbb{Q}(\sqrt{2}): \mathbb{Q} ] = 2$ as $\mathbb{Q}(\sqrt{2}) = [a+b\sqrt{2} | a,b \in \mathbb{Q}]$ so $\mathbb{Q}(\sqrt{2})$ is a 2d vector space over $\mathbb{Q}$ with basis $\left \{1, \sqrt{2}\right \}$

I then use a similar argument for $[ \mathbb{Q}(\sqrt{2}, i ): \mathbb{Q(\sqrt{2})} ] = 2$ in that $\mathbb{Q}(i)(\sqrt{2}) = [a+bi | a,b \in \mathbb{Q(\sqrt{2})}]$

and the rest follows. I'm not sure if this is enough to show it rigidly. Some of my friends have been using minimal polynomials to show the different degrees and I don't entirely follow the logic!

Any help appreciated!

xx