Find the minimal polynomial of $i\sqrt{-1+2\sqrt3} \in \Bbb C$ over $\Bbb R$

Question

I need to find the minimal polynomial of $i\sqrt{-1+2\sqrt3} \in \Bbb C$ over $\Bbb R$ and prove the polynomial that I find is actually the minimal polynomial. I know that the $\Bbb R$-basis for $\Bbb C$ is that $1,i$. Is it possible to find the minimal polynomial by usuing the basis? I also tried to square $i\sqrt{-1+2\sqrt3}$. Then I will get $1-2\sqrt3.$ But then I did not make much progress. (I have not learnt Galois Theory yet) Can someone help me? Thanks so much.

Answer 1

Expanding on the comment: if $z_0 \in \mathbb{C} \setminus \mathbb{R}$ then its minimal polynomial over $\mathbb{R}$ is $(z-z_0)(z-\overline{z_0})\,$.

• $(z-z_0)(z-\overline{z_0})=z^2-(z_0+ \overline{z_0})z + z_0 \overline{z_0}$ is a real polynomial since $z_0+\overline{z_0}\,,z_0\,\overline{z_0}\,\in\mathbb{R}\,$.

• It is the minimal polynomial, else if it were reducible it would mean that its roots are real.

Answer 2

Note that $\alpha := \sqrt{-1 + 2\sqrt{3}}$ is real, so $(X/\alpha)^2 + 1$ has real coefficients (and has $\alpha i$ as a root). Making this monic, the minimal polynomial of $\alpha i$ over the reals is $X^2 + \alpha^2 = X^2 + 2 \sqrt{3} - 1$.