Is $\alpha$ algebraic or not? I am trying to find out if $\alpha = \sqrt[]{2} - i$ for $F = \mathbb{Q}$ is algebraic or not. 
I found that $\alpha^2 = 1 - 2i\sqrt[]{2}$ and have been trying to construct a suitable polynomial, but I have so far been unable to. Does anyone have a nice way to do this?

Hint: The sum of algebraic numbers is algebraic. One proof uses the resultant.
Wolfram alpha calculates the minimal polynomial of $\sqrt{2}-i$ to be $x^4-2x^2+9$. One way to come up with this on your own is to consider $1,\alpha,\alpha^2,\alpha^3,\alpha^4$ as vectors in $\mathbb{Q}[1,\sqrt{2},i]$ and find a linear combination summing to zero. We know that such a linear combination must exist since the minimal polynomials of $\sqrt{2},i$ have degree 2, and so the minimal polynomial of $\sqrt{2}-i$ has degree at most 4.
