Is it true that $\mathbb{Q}(\sqrt{2},e^{2i\pi/3}) = \mathbb{Q}(\sqrt{2}+e^{2i\pi/3})$?

Question

Is it true that $\mathbb{Q}(\sqrt{2},e^{2i\pi/3}) = \mathbb{Q}(\sqrt{2}+e^{2i\pi/3})$? I know that $[\mathbb{Q}(\sqrt{2},e^{2i\pi/3}):\mathbb{Q}]=2\times2=4$. By using WolframAlpha (cheating), I know that $[\mathbb{Q}(\sqrt{2}+e^{2i\pi/3}):\mathbb{Q}]=4$ as well. But this is not enough.

Answer 1

The order $ [{\mathbb Q}(\sqrt{2} + e^{2i\pi \over 3}):{\mathbb Q}]$ has to divide $[{\mathbb Q}(\sqrt{2},\,e^{2i\pi \over 3}):{\mathbb Q}] = 4$ since ${\mathbb Q}(\sqrt{2} + e^{2i\pi \over 3}) \subset {\mathbb Q}(\sqrt{2},\,e^{2i\pi \over 3})$. So it's either $1, 2,$ or $4$. It can't be $1$ since $\sqrt{2} + e^{2i\pi \over 3}$ is not real.

If it were $2$, there would be a quadratic polynomial with rational coefficients with $\sqrt{2} + e^{2i\pi \over 3}$ as a root. Since the coefficients are real, the complex conjugate $\sqrt{2} + e^{-{2i\pi \over 3}}$ would have to be the other root, which would mean the linear term of the polynomial would have coefficient $-2\sqrt{2} - 2 \cos{2\pi \over 3} = -2\sqrt{2} - 1$, which is not rational. Thus $ [{\mathbb Q}(\sqrt{2} + e^{2i\pi \over 3}):{\mathbb Q}]$ can't be $2$ either.

The only remaining possibility is that $ [{\mathbb Q}(\sqrt{2} + e^{2i\pi \over 3}):{\mathbb Q}]= [{\mathbb Q}(\sqrt{2},\,e^{2i\pi \over 3}):{\mathbb Q}] = 4$. Since ${\mathbb Q}(\sqrt{2} + e^{2i\pi \over 3}) \subset {\mathbb Q}(\sqrt{2},\,e^{2i\pi \over 3})$, we therefore have ${\mathbb Q}(\sqrt{2} + e^{2i\pi \over 3}) = {\mathbb Q}(\sqrt{2},\,e^{2i\pi \over 3})$.

Answer 2

Well, obviously $$\mathbb{Q}(\sqrt{2}+\omega)\subseteq\mathbb{Q}(\sqrt{2},\omega)\tag{1}$$ so we just need to prove the opposite inclusion. By expanding the LHS of the following identity $$ (\alpha-\sqrt{2})^3 = \omega^3=1 \tag{2} $$ through the binomial theorem we get: $$ \sqrt{2} = \frac{\alpha^3+6\alpha-1}{3\alpha^2+2},\qquad \omega=\alpha-\sqrt{2}=\frac{2\alpha^3-4\alpha-1}{3\alpha^2+2}\tag{3}$$ and that proves $$ \mathbb{Q}(\sqrt{2},\omega)\subseteq\mathbb{Q}(\alpha)\tag{4} $$ as wanted.

Answer 3

A different approach (assuming a bit of Galois theory).

Because $e^{2\pi i/3}=\dfrac{-1+i\sqrt3}2$ we know that $\Bbb{Q}(\sqrt2,e^{2\pi i/3})=\Bbb{Q}(\sqrt2,\sqrt{-3})$. This is a Galois extension of $\Bbb{Q}$ with a Galois group isomorphic to Klein four. The standard exercise shows that the intermediate fields are $$ \begin{aligned} \Bbb{Q}(\sqrt2)&=\{a+b\sqrt2\mid a,b\in\Bbb{Q}\},\\ \Bbb{Q}(\sqrt{-3})&=\{a+b\sqrt{-3}\mid a,b\in\Bbb{Q}\},\\ \Bbb{Q}(\sqrt{-6})&=\{a+b\sqrt{-6}\mid a,b\in\Bbb{Q}\}. \end{aligned} $$

It is easy to check that the (obvious not rational) number $\sqrt2+e^{2\pi i/3}$ is not an element of any of the above fields. Therefore it must generate the whole thing.

Answer 4

It is clear that $\mathbb{Q}(\sqrt 2 + w) \subseteq \mathbb{Q}( \sqrt 2, w)$. We prove the other inclusion.

Since $w=-\frac12 + \frac{i\sqrt{3}}2$, we see that $\sqrt 2 + \frac{i\sqrt{3}}2\in\mathbb{Q}(\sqrt 2 + w)$.

Since $(\sqrt 2 + \frac{i\sqrt{3}}2)(\sqrt 2 - \frac{i\sqrt{3}}2) = 2 + \frac34 \in \mathbb{Q}$, we have $\sqrt 2 - \frac{i\sqrt{3}}2\in\mathbb{Q}(\sqrt 2 + w)$.

Adding two elements, we have $\sqrt 2 \in \mathbb{Q}(\sqrt 2 + w)$, and clearly $\frac{i\sqrt 3}2 \in \mathbb{Q}(\sqrt 2 + w)$.

Thus, we have both elements $\sqrt 2$ and $w$ are in $\mathbb{Q}(\sqrt 2 + w)$, thereby proving the other inclusion.

Answer 5

First you have that $e^{2\pi i/3}\notin\Bbb Q(\sqrt 2)$, as $e^{2\pi i/3}$ is not real. Neither $\sqrt 2\notin\Bbb Q(e^{2\pi i/3}),$ for otherwise we would have $\Bbb Q(\sqrt 2)=\Bbb Q(e^{2\pi i/3})$ as these two field extensions have degree $2$ over $\Bbb Q$.

Thus $x^2+x+1$ is irreducible over $\Bbb Q(\sqrt 2)$. Hence there is an embedding $f:\Bbb Q(\sqrt 2)(e^{2\pi i/3})\rightarrow\overline{\Bbb Q}$ fixing $\Bbb Q(\sqrt 2)$ with $f(e^{2\pi i/3})=e^{4\pi i/3}$. Then as $f(\sqrt 2+e^{2\pi i/3})=\sqrt 2+e^{4\pi i/3}$, we get that $\sqrt 2+e^{4\pi i/3}$ is a conjugate of $\sqrt 2+e^{2\pi i/3}$ over $\Bbb Q$.

Similarly as $\sqrt 2\notin \Bbb Q(e^{2\pi i/3})$, we get that $-\sqrt 2+e^{2\pi i/3}$ is a conjugate of $\sqrt 2+e^{2\pi i/3}$ over $\Bbb Q$.

Therefore $\sqrt 2+e^{2\pi i/3}$ has at least three conjugates over $\Bbb Q$, hence $[\Bbb Q(\sqrt 2+e^{2\pi i/3}):\Bbb Q]\geq 3$, however as $[\Bbb Q(\sqrt 2+e^{2\pi i/3}):\Bbb Q]\mid [\Bbb Q(\sqrt 2,e^{2\pi i/3}):\Bbb Q],$ we obtain $[\Bbb Q(\sqrt 2+e^{2\pi i/3}):\Bbb Q]=4$, i.e., $\Bbb Q(\sqrt 2+e^{2\pi i/3})=\Bbb Q(\sqrt 2,e^{2\pi i/3})$.