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?

Note that
$\alpha + i = \sqrt{2}; \tag{1}$
thus
$\alpha^2 + 2i \alpha - 1 = (\alpha + i)^2 = 2, \tag{2}$
whence
$\alpha^2 + 2i\alpha = 3, \tag{3}$
so that
$2i \alpha = 3 - \alpha^2;  \tag{4}$
squaring:
$-4\alpha^2 = \alpha^4 - 6\alpha^2 + 9, \tag{5}$
and at last
$\alpha^4 -2\alpha^2 + 9 = 0. \tag{6}$
So, satisfying as it does the quartic polynomial $x^4 - 2x^2 + 9 \in \Bbb Q[x]$, $\alpha$ is algebraic over the rationals.
We also have, from $\alpha^2 = 1 - 2i\sqrt{2}$, that $(\alpha^2 - 1)^2 = -8$, from which the same result follows.
