Minimal Polynomial of $i + \sqrt{2}$ in $\mathbb{Q}$ 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.
 A: First, note that $i+\sqrt{2}\in\mathbb{Q}(i,\sqrt{2})$, so the degree is either $2$ or $4$. But $i+\sqrt{2}$ is not of degree $2$ (it would have to be expressible in the form $a + b\sqrt{d}$ for some squarefree integer $d$, with $a,b\in \mathbb{Q}$, and this is impossible. So you are certainly right that it is degree $4$.
Now, here your method is not actually "guessing", but "working." Suppose $f(x)$ is a the minimal polynomial of $i+\sqrt{2}$, and consider the splitting field of $f(x)$, $K$. Complex conjugation gives an automorphism of $K$, and maps one root of $f(x)$ to another root of $f(x)$. That means that $-i+\sqrt{2}$ must also be a root of $f(x)$.
Likewise, the automorphism of $\mathbb{Q}(\sqrt{2})$ that fixes $\mathbb{Q}$ and maps $\sqrt{2}$ to $-\sqrt{2}$ extends to an automorphism of $K$ which fixes $f(x)$, and maps any root of $f(x)$ to a root of $f(x)$. So both $i-\sqrt{2}$ (the image of $i+\sqrt{2}$) and $-i-\sqrt{2}$ (the image of $-i+\sqrt{2}$) must be roots of $f(x)$. 
This gives you four roots of $f(x)$, which you know to be of degree $4$, so that gives the four roots and hence $f(x)$.
For more general comments, see this previous question. You can consider the powers of $i+\sqrt{2}$ until you get that $1, \alpha,\alpha^2,\ldots,\alpha^n$ is linearly dependent over $\mathbb{Q}$, and use the linear dependency to get the polynomial.
Added. In this case, we have:
$$\begin{align*}
i+\sqrt{2} &= i+\sqrt{2}\\
(i+\sqrt{2})^2 &= 1 + 2i\sqrt{2}\\
(i+\sqrt{2})^3 &= 5i - \sqrt{2}\\
(i+\sqrt{2})^4 &= -7 + 4i\sqrt{2}\\
               &= -9 + 2(1+2i\sqrt{2}) = -9 + 2(i+\sqrt{2})^2.
\end{align*}$$
This means that $\alpha=i+\sqrt{2}$ satisfies $\alpha^4 = -9+2\alpha^2$, or that $\alpha$ is a root of $f(x)=x^4 -2x^2 + 9$. 
A: There are various methods, e.g. using automorphisms, taking the characteristic polynomial of $\ z \mapsto (i+\sqrt{2})\ z,\,$ undetermined coefficients, etc. Here the simplest may be repeated squaring: $ \ x - i = \sqrt{2}\, $ $\,\overset{(\ \ )^2}\Rightarrow$ $ x^2 - 2\ i\ x -1 = 2\ $ or $ \ x^2 - 3 = 2\ i\ x,\,$ which, squared, yields the result.
Uupdate $\ $ Since the terseness of the "repeated squaring" method seems to have possibly confused at least one reader, perhaps it may help to elaborate a bit.
Theorem $\ $ Suppose that $ \,R\,$ is a ring which contains two elements $ \, w,\, i\, $ that satisfy $ \, w^2 = 2,\ \ i^2 = -1\,.\ $ Then $\ x = w + i\ $ satisfies $\  x^4 + 9 = 2x^2\,.$
Proof $\ $ Squaring $ \ x - i = w\ $ yields $ \ x^2 - 2\,i\ x - 1\, =\, 2,\,$ or $ \ x^2 - 3\ =\ 2\,i\, x.\,$ Squaring this yields $ \ x^4 -6\, x^2 + 9\ =\, -4\, x^2,\ $ so $ \ x^4 + 9\, =\, 2\, x^2\,.\quad$ QED
Remark $\ $ Notice that the above theorem may apply in rings other than  $ \, \mathbb Q(i,\sqrt{2}).\, $ For example, $\,$ in $ \, \mathbb Z/17 =\,$  integers mod $17\,$ we can choose $ \ w = 6,\ i = -4\ $ and conclude that $ \ x = w+i = 2\, \Rightarrow\, x^4 -2\ x^2 + 9\, =\, 0.\ $ Indeed $ \ 16 - 8 + 9\, =\, 17 \equiv\ 0\pmod{ 17}.$
A: 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.
