TITLE: Minimal Polynomial of $i + \sqrt{2}$ in $\mathbb{Q}$
QUESTION [5 upvotes]: I am trying to find Minimal Polynomial of $i + \sqrt{2}$ in $\mathbb{Q}$.   I was able to determine the minimal polynomial is fourth degree with roots at  $i-\sqrt{2}$, $i+\sqrt{2}$,$-i-\sqrt{2}$,$-i+\sqrt{2}$.   However I got this answer by guessing at what the roots should be.   Is there a general technique for this type of problem.

REPLY [2 votes]: The way I like to look at such things... suppose $P(z)$ is a polynomial with rational coefficients. Then $P(\bar{z}) = \bar{P(z)}$ for any complex number $z$. So if $i + \sqrt{2}$ is a root of $P(z)$, so is $-i + \sqrt{2}$. Similarly, suppose $z = a + b\sqrt{2}$, with 
$a$ and $b$ both of the form $q_1 + q_2i$ for rational $q_1$ and $q_2$. Then if $P(a + b\sqrt{2}) = c + d\sqrt{2}$, one has $P(a - b\sqrt{2}) = c - d\sqrt{2}$. So if $i + \sqrt{2}$ is a root of $P(z)$, so is $i - \sqrt{2}$, and if $-i + \sqrt{2}$ is a root of $P(z)$, so is $-i - \sqrt{2}$.
The upshot is that if $i + \sqrt{2}$ is a root of $P(z)$, so are $-i + \sqrt{2}$, $i - \sqrt{2}$, and $-i - \sqrt{2}$. Thus the minimal polynomial of $i + \sqrt{2}$ over $Q$ will have to have these as roots and will be of degree at least 4. Then you can verify that the polynomial with these roots has rational coefficients and therefore is this minimal polynomial.
In some sense Arturo Magidin's answer is a way of describing the above phenomenon in terms of field automorphisms.