<s>[INST] min polynomial of $\sqrt{6+2\sqrt{5}}$ min polynomial of $\sqrt{6+2\sqrt{5}}$ over $\mathbb{Q}$
I found $\alpha^4 -12\alpha^2 +16 = 0$ but wolframalpha does not agree with me. furthermore, would anything change is it was over $\mathbb{R}$?
 [/INST] You can try to factor $\alpha^4-12\alpha^2+16$.  Assuming that $\sqrt{6+2\sqrt{5}}$ is not in $\mathbb{Q}$, then the best that you can hope for is that this factors into quadratics.  We write the following:
$$
\alpha^4-12\alpha^2+16=(\alpha^2+r\alpha+s)(\alpha^2+t\alpha+u).
$$
Multiplying this out, one has
$$
\alpha^4+(r+t)\alpha^3+(s+rt+u)\alpha^2+(ru+st)\alpha+su.
$$
Since the coefficient $\alpha^3$ is zero, $t=-r$.  This makes the coefficient of $\alpha$ $r(s-u)$.  Assuming that $r\not=0$ (which can be checked separately), this means that $s=u$.  Therefore, the factorization (if it exists) is
$$
\alpha^4-12\alpha^2+16=(\alpha^2+r\alpha+s)(\alpha^2-r\alpha+s).
$$
Since $s^2=16$, $s=\pm 4$ and since $2s-r^2=-12$, we know that the only solution in the integers is $s=-4$ and $r=\pm 2$.
Therefore,
$$
\alpha^4-12\alpha^2+16=(\alpha^2+2\alpha-4)(\alpha^2-2\alpha-4).
$$
We can check that this factorization is correct, and the roots of one of these polynomials is the given expression.
The roots of the first factor are $-1\pm\sqrt{5}$ and the roots of the second factor are $1\pm\sqrt{5}$.  Since the given expression is positive, we can eliminate all negative possibilities, so the given expression is either $-1+\sqrt{5}$ or $1+\sqrt{5}$.  Since $\sqrt{5}>2$, the expression in the radical is at least $10$, so its square root is at least $3$.  Since $-1+\sqrt{5}$ is at most $2$ (since $5<9$), this means that the given expression is $1+\sqrt{5}$.
Therefore, $\alpha^2-2\alpha+4$ is the minimal polynomial for this expression.
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