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 $\mathbb{R}$?
 A: $$\sqrt{6+2\sqrt{5}} = \sqrt{(\sqrt{5})^2+2\sqrt{5}+1} = \sqrt{(\sqrt{5}+1)^2} = \sqrt{5}+1$$
A: 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.
A: 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.
A: Here’s an advanced way of handling this problem, perhaps beyond your current background. The advantage is that it generalizes.
You’re asking for the square root of an element $\xi=6+2\sqrt5$  of $k=\Bbb Q(\sqrt5\,)$, a real quadratic field, and you want to know whether $\xi$ is a square already there. It happens that the ring of integers $R$ of $k$ is a principal ideal domain. Its integral basis is $\bigl\{1,\frac{1+\sqrt5}2\bigr\}$, yes, that’s the Golden Ratio you see there. And it’s known (and fairly easy to show) that every unit of $R$ is of the form $\pm\bigl(\frac{1+\sqrt5}2\bigr)^n$, that is, $\frac{1+\sqrt5}2$ is a fundamental unit of $R$. One other fact from algebraic number theory: the prime $2$ is inert in $k$, that is, it remains prime there.
Finally, the norm down to $\Bbb Q$ of your number is $(6+2\sqrt5\,)(6-\sqrt5\,)=16$. This implies that $6+\sqrt5$ is $4$ times a unit $u$, but of course we don’t know yet whether $u$ is a square in $k$. But $u=\frac{3+\sqrt5}2$, and you see (by a trial division by the fundamental unit) that $u=\bigl(\frac{1+\sqrt5}2\bigr)^2$. So $6+2\sqrt5$ is already a square in $\Bbb Q(\sqrt5\,)$, and thus its square root, namely $2\frac{1+\sqrt5}2=1+\sqrt5$ has the minimal polynomial $X^2-2X-4$.
