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.
 A: 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 $\rm\:f(\alpha)=0\:.$
A: Since a comment suggested that I add my comment as an answer:
To show that $K=\mathbb{Q}(\sqrt{3},\sqrt{-1})$ is an algebraic extension of $\mathbb{Q}$, you need to show that every element in K is the root of some polynomial $f(x)$ with coefficients in $\mathbb{Q}$.
A generic element in K has the form $a+b\sqrt{3}+c\sqrt{-1}+d\sqrt{-3}$  (thanks to Henning Makholm for pointing out the last term). Call this expression $\alpha$ and try to get rid of the square roots by repeatedly squaring. Use this to find a polynomial with $\alpha$ as root.
Since $\alpha$ was arbitrary, you would have shown that K is an algebraic extension of $\mathbb{Q}$. (alternatively, try to prove Chaz' assertion that every finite extension is algebraic)
