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Can someone please tell me how to determine the minimal polynomial of $i+\sqrt{2}$ and $\sqrt{1+\sqrt{3}}$ over $\mathbb{Q}$ ?

My idea was to look at the iterates of these polynomials and find a linear combination of them to then get my minimal polynomial. I tried finding that relation using the recipe given to me in the previous question, but didn't quite manage it, since that contained too many points where I simply didn't knew what to do next and why to do it :(

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    An easy way is computing $1$, $i+\sqrt{2}$, $(i+\sqrt{2})^2$, $(i+\sqrt{2})^3$, $(i+\sqrt{2})^4$,... and try to impose that a linear combination of them with rational coefficients is zero. If you obtain that the coefficients must be zero, add another power of $i+\sqrt{2}$, and so on. This reduces your problem to linear algebra.2012-10-30

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Say $\alpha = \sqrt{1+\sqrt{3}}$. Then $\alpha^2=1+\sqrt{3}.$ So $\alpha^2-1=\sqrt{3}.$ Hence $(\alpha^2-1)^2=3$ which implies $\alpha^4+1-2\alpha=3.$ This means $\alpha$ is a root of $f(x)=x^4-2x-2.$ Note that $f(x)$ is Eisenstein at $p=2,$ hence irreducible. Thus the minimal polynomial of $\alpha$ over $\mathbb Q$ is $f(x).$ Similarly the other can be done.

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For example, take $x=i+\sqrt{2}$. Then $x^2=-1+2i\sqrt{2}+2$, so $x^2-1=2i\sqrt{2}$. Hence $x^4-2x^2+1=(x^2-1)^2=-8$. Hence the polynomial $x^4-2x^2+9$ vanishes at $i+\sqrt{2}$. It is left to check that it is irreducible over $\mathbb{Q}$.
This can be done by noticing that if $i+\sqrt{2}$ is a root of a polynomial over $\mathbb{Q}$ (or over $\mathbb{R}$) then so is $-i+\sqrt{2}$. So the minimal polynomial is divisible over $\mathbb{R}$ by $(x-i-\sqrt{2})(x+i-\sqrt{2})=x^2-2\sqrt2x+3$. From here you can easily check that the aforementioned polynomial is irreducible over $\mathbb{Q}$.