Field $\mathbb{Q}(\sqrt{3}, \sqrt{-1})$ Am I right to say that the field $\mathbb{Q}(\sqrt{3}, \sqrt{-1})$ is an algebraic extension of $\mathbb{Q}$?
Because $\mathbb{Q}\subset\mathbb{Q}(\sqrt{3})\subset\mathbb{Q}(\sqrt{3})( \sqrt{-1})=\mathbb{Q}(\sqrt{3}, \sqrt{-1})$.
Thanks.

HINT $\rm\ K =\: \mathbb Q(\sqrt 3, \sqrt{-1})\:$ is a $4$-dimensional vector space over $\rm\mathbb Q\:,\:$ viz. $\rm\: K =\: \mathbb Q\langle 1,\:\sqrt 3,\:\sqrt{-1},\:\sqrt{-3}\rangle\:.$ 
Hence $\rm\:\alpha\in K\ \Rightarrow\ 1,\ \alpha,\ \alpha^2,\ \alpha^3,\ \alpha^4\:$ are linearly dependent over $\rm\:\mathbb Q\:.\:$ This dependence relation yields a nonzero polynomial $\rm\:f(x)\in \mathbb Q[x]\:$ of degree $\le 4\:$ such that $\rm\:f(\alpha)=0\:.$
