<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] Since $\sqrt{6+ 2\sqrt{5}}$ is a real number, if it were "over R" then the minimum polynomial would be $x- \sqrt{6+ 2\sqrt{5}}$!
The way I would do this problem would be to note that $(x- \sqrt{6+ 2\sqrt{5}})(x+ \sqrt{6+ 2\sqrt{5}})= x^2- (6+ 2\sqrt{5})= (x^2- 6)- 2\sqrt{5}$.
Then $((x^2- 6)- 2\sqrt{5})((x^2- 6)+ 2\sqrt{5})= (x^2- 6)^2- 20= x^4- 12x^2+ 36- 20= x^4- 12x^2+ 16$.
Notice the use of $(a- b)(a+ b)= a^2- b^2$ to get rid of the square roots.
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