# How to determine the minimal polynomial of $\sqrt{3 + 2\sqrt{2}}$ over $\mathbb{Q}$?

I first let $\alpha = \sqrt{3 + 2\sqrt{2}}$ and $\alpha^2 - 3 = 2\sqrt{2}$. This gives us $(\alpha^3 - 2)^2 = 8$. Expand the polynomial we obtain that $x^4 - 6x^2 +1$ has $\sqrt{3 + 2\sqrt{2}}$ as a root. But apparently this polynomial is not an irreducible polynomial over $\mathbb{Q}$. How should we determine the minimal polynomial in the first place? Many thanks!

• Why is it $\alpha^2-3=2\sqrt2$? Shouldn't it be $\alpha^2-3=\sqrt2$ ? And where does $(\alpha^3-2)^2$ come from? – user228113 Apr 9 '16 at 17:15
• @G.Sassatelli Good catch! I had a typo in my problem statement. Just fixed it. – nekodesu Apr 9 '16 at 17:18
• There is another typo. It should be $(\alpha^2-3)^2=8$, not $(\alpha^3-2)^2=8$. – user228113 Apr 9 '16 at 17:19

Hint A minimal polynomial must be irreducible, but as you say, the polynomial $p$ you produced is not: $$p(x) = (x^2 + 2 x - 1) (x^2 - 2 x - 1).$$
Since $p(\alpha) = 0$, however, $\alpha$ must divide one of these factors (and since $\alpha \not\in \Bbb Q$, that factor must itself be the irreducible polynomial of $\alpha$).
To see what's going on here, expand $(1 + \sqrt{2})^2$.