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?
 A: $2=(\alpha+i)^2=\alpha^2+2i\alpha-1$.
So $2i\alpha=3-\alpha^2$
So $-4\alpha^2=9-6\alpha^2+\alpha^4$
A: Standard technique: you do $\alpha^2, \alpha^3, \dots$ until you can write something like $$\alpha^k = \sum_{i=0}^{k-1} b_i \alpha^i\qquad , \qquad b_i \in \mathbb{Q}$$
Then you have found the polynomial, which means $\alpha$ is algebraic over $\mathbb{Q}$
A: You can think about the conjugates of those roots that you'd expect to see; in particular, you'd expect roots at $\pm \sqrt{2}\pm i$, since $\sqrt{2}$ and $-\sqrt{2}$ are conjugate as are $i$ and $-i$. Then, if you just multiply out
$$(x-\sqrt{2}-i)(x+\sqrt{2}-i)(x-\sqrt{2}+i)(x+\sqrt{2}+i)$$
you get a polynomial with rational coefficients.
A: 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.
A: 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.
