Determine:$[\mathbb Q(\sqrt2,e^{\frac{2\pi i}{3}}):\mathbb Q]?$

Question

Determine:$[\mathbb Q(\sqrt2,e^{\frac{2\pi i}{3}}):\mathbb Q]?$

My try: Because $e^{2\pi i/3}=\dfrac{-1+i\sqrt3}2$ we know that $\Bbb{Q}(\sqrt2,e^{2\pi i/3})=\Bbb{Q}(\sqrt2,\sqrt{-3})$.Then $[\Bbb{Q}(\sqrt2,\sqrt{-3}):\mathbb Q]=4$.plz check or suggest any other method.Thank you

Answer 1

Note that $\mathbb{Q}(\sqrt{2}) \neq \mathbb{Q}$ since the former contains the irrational number $\sqrt{2}$. Therefore, $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}] \geq 2$. To show that it is exactly $2$, observe that $x^2 - 2$ serves as a degree $2$ minimal polynomial for $\sqrt{2}$ with coefficients in $\mathbb{Q}$.

Similarly, $\mathbb{Q}(\sqrt{2}, \sqrt{-3}) \neq \mathbb{Q}(\sqrt{2})$ since the former contains an imaginary number, whereas the latter does not. Therefore, $[\mathbb{Q}(\sqrt{2}, \sqrt{-3}):\mathbb{Q}(\sqrt{2})] \geq 2$. To show that it is exactly $2$, observe that $x^2 + 3$ serves as a degree $2$ minimal polynomial for $\sqrt{-3}$ with coefficients in $\mathbb{Q}(\sqrt{2})$.

By the Tower Theorem, we have that

$$[\mathbb{Q}(\sqrt{2}, \sqrt{-3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2}, \sqrt{-3}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2 \cdot 2 = 4$$