Showing $\mathbb{Q}(\sqrt 2, \sqrt 3 ) = \mathbb{Q}(\sqrt 2 + \sqrt 3)$ using information about Galois group

Question

Let $E/F = \mathbb{Q}(\sqrt 2, \sqrt 3 )/\mathbb{Q}$. This extension is Galois of degree 4, and one can easily verify that $G:= \operatorname{Gal}(\mathbb{Q}(\sqrt 2, \sqrt 3 ) /\mathbb{Q})$ is $\{ \operatorname{id}, (\sqrt2\quad-\sqrt 2), (\sqrt3\quad-\sqrt 3),(\sqrt2\quad-\sqrt 2)(\sqrt3\quad-\sqrt 3)\}\cong C_2\times C_2$.

One can show that $\mathbb{Q}(\sqrt 2, \sqrt 3 ) = \mathbb{Q}(\sqrt 2+ \sqrt 3 )$ using for example that $(\sqrt2 + \sqrt 3)^{-1}\in \mathbb{Q}(\sqrt 2, \sqrt 3 )$. Now, I want to use only $G$ to show this equality.

If $\sigma\in G$, then $\sigma(\sqrt 2 +\sqrt 3 ) = \pm\sqrt 2 \pm\sqrt 3$, and all four option are different. Can I conclude that the minimal polynomial of $\sqrt 2+\sqrt 3 $ over $\mathbb{Q}$ is $$ (x-(\sqrt2+\sqrt3))(x-(\sqrt2-\sqrt3))(x-(-\sqrt2+\sqrt3))(x-(-\sqrt2-\sqrt3)) =x^4-10x^2+1.$$ In that case $[\mathbb{Q}(\sqrt 2 +\sqrt 3 ):\mathbb{Q}]=4$, so that the equality follows as well.

Is there a way which avoids introducing minimal polynomials (explicitly), but which also uses only the information from $E/F$ and $G$?

Answer 1

$$\mathbb{Q}(\sqrt 2, \sqrt 3) \ge \mathbb{Q}(\sqrt 2 + \sqrt 3)$$ is trivial.

$$\mathbb{Q}(\sqrt 2, \sqrt 3) \le \mathbb{Q}(\sqrt 2 + \sqrt 3)$$ will hold if we can write:

• $\sqrt{2} = P_1(\sqrt 2 + \sqrt 3)$

• $\sqrt{3} = P_2(\sqrt 2 + \sqrt 3)$

for some polynomials $P_1, P_2$.

Put $\alpha = \sqrt 2 + \sqrt 3$ and

• $\alpha^2 = 5 + 2 \sqrt{2} \sqrt{3}$
• $\alpha^3 = 5 (\sqrt{2} + \sqrt{3}) + 4 \sqrt{3} + 6 \sqrt{2}$

Clearly $\alpha^3 - 5 \cdot \alpha - 4 \cdot \alpha = 2 \sqrt{2}$ so we are done.