Find the minimum polynomial of $\sqrt{i+\sqrt{2}}$ over $ \mathbb{R}$ So here's what I know about the minimum polynomial $p(x)$ of $c=\sqrt{i+\sqrt{2}}$ over $\mathbb{R}$...


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*$p(x)$ has to be degree 2

*$p(x)$ must be a multiple of $q(x)=x-\sqrt{i+\sqrt{2}}$ because $q(x)$ is the generator of the kernel of the substitution function $\sigma_c$ that maps polynomials over $\mathbb{C}$ to $\mathbb{C}$.

*$p(x)$ must be a divisor of $r(x) = (x^2-i-\sqrt{2})(x^2+i-\sqrt{2})$, because $r(c)=0$. 


So I tried multiplying $q(x)$ by each degree-1 factor of $x^2+i-\sqrt{2}=(x+\sqrt{\sqrt{2}-i})(x-\sqrt{\sqrt{2}-i})$, but (I think) neither product is over $\mathbb{R}$... I must be doing something wrong. Any tips?

Robert Israel's answer is excellent and complete, but I thought I might offer some supporting computations. Let $\alpha = i+\sqrt{2}$, and let $\beta = \sqrt{\alpha}$. Then $|\alpha| = \sqrt{3}$, so we can rewrite $\alpha$ as
$$\alpha = \sqrt{3}(\cos(\theta)+i\sin(\theta))$$
where $\cos(\theta) = \frac{\sqrt{2}}{\sqrt{3}}$ and $\sin(\theta) = \frac{1}{\sqrt{3}}$. Then using the principal branch of the square root, we see
$$\beta = \sqrt{\alpha} =  \sqrt[4]{3}\left(\cos\left(\frac{\theta}{2}\right)+i\sin\left(\frac{\theta}{2}\right)\right)$$
The half-angle formulas give us
$$\cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1+\cos(\theta)}{2}} = \sqrt{\frac{\sqrt{3}+\sqrt{2}}{2\sqrt{3}}}$$
and
$$\sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1-\cos(\theta)}{2}} = \sqrt{\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}}$$
whence
$$\beta = \sqrt{\frac{\sqrt{3}+\sqrt{2}}{2}} + \sqrt{\frac{\sqrt{3}-\sqrt{2}}{2}}i$$
Then it is easy to see that the trace of $\beta$ is 
$$\mathrm{Tr}(\beta) = 2\sqrt{\frac{\sqrt{3}+\sqrt{2}}{2}} = \sqrt{2\sqrt{3}+2\sqrt{2}}$$
and the norm of $\beta$ is
$$|\beta| = (\sqrt[4]{3})^{2} = \sqrt{3}$$
Then the minimal polynomial of $\beta$ over $\mathbb{R}$ is
$$(X-\beta)(X-\overline{\beta}) = X^{2} - \mathrm{Tr}(\beta) + |\beta|^{2} = X^{2} - \sqrt{2\sqrt{3}+2\sqrt{2}}X + \sqrt{3}$$
