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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.

3x89g2
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    It is enough, if you are putting your faith in Wolfram. Always $k(\alpha+\beta) \subseteq k(\alpha,\beta)$ and you have equality of dimensions. – Hoot Jul 09 '16 at 21:32
  • @Hoot Ah OK. By the way, how can I find dimension of $\mathbb{Q}(\sqrt{2}+e^{2i\pi/3})$ without using computer? – 3x89g2 Jul 09 '16 at 21:34
  • Zarrax gives the right answer. Do you know any Galois theory? – Hoot Jul 09 '16 at 21:56
  • @Hoot : you don't define $[\mathbb{Q}(\alpha):\mathbb{Q}]$ outside of Galois theory ? – reuns Jul 09 '16 at 22:09
  • @Hoot I just started learning it. – 3x89g2 Jul 09 '16 at 22:10
  • @Misakov: I wrote a very elementary solution proving the double inclusion in a completely algorithmic way. – Jack D'Aurizio Jul 09 '16 at 22:12
  • @JackD'Aurizio : I don't see what is elementary in your solution, since I don't understand it – reuns Jul 09 '16 at 22:13
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    @user1952009 I'm not sure what you mean. Anyway, my point was that if you know something about the Galois correspondence then you know that there's only one other intermediate field and you know what automorphism has to fix it; and it's easy to check that this element is not fixed. But usually one learns the basics of fields first, so I didn't want to assume it. – Hoot Jul 09 '16 at 22:13
  • @user1952009: I simply prove that both $\sqrt{2}$ and $\sqrt{2}+\omega$ are rational functions with rational coefficients of $\alpha=\omega+\sqrt{2}$. You may skip $(2)$ and $(3)$ if not interested. – Jack D'Aurizio Jul 09 '16 at 22:16
  • @Jack This is very cool. – Hoot Jul 09 '16 at 22:17
  • @JackD'Aurizio : yes this I understand, you prove explicetely how $\sqrt{2}$ is in $\mathbb{Q}(\sqrt{2} + e^{2 i \pi / 3})$, but I don't get the rest : your method for finding the minimal polynomial (or a certain polynomial) having $\alpha = \sqrt{2} + e^{2 i \pi / 3}$ as root – reuns Jul 09 '16 at 22:17
  • @user1952009: that is not crucial for your purpose, you may simply neglect it. – Jack D'Aurizio Jul 09 '16 at 22:18
  • @JackD'Aurizio : wait I think you edited, now I got it, more or less – reuns Jul 09 '16 at 22:20
  • Refs: http://math.stackexchange.com/questions/1504169/prove-that-sqrt35-sqrt43-is-irrational , http://math.stackexchange.com/questions/1510146/find-the-irreducible-polynomial-of-this-element-over-mathbb-q , http://math.stackexchange.com/questions/359054/constructing-a-degree-4-rational-polynomial-satisfying-f-sqrt2-sqrt3-0 – Jack D'Aurizio Jul 09 '16 at 22:21
  • $\sqrt{2}+e^{2i\pi/3}$ is a primitive element of the extension. I refrain from publishing me an answer because there are enough already. – Piquito Jul 10 '16 at 13:54

5 Answers5

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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 ${\mathbb Q}(\sqrt{2} + e^{2i\pi \over 3}) = {\mathbb Q}(\sqrt{2},\,e^{2i\pi \over 3})$.

Zarrax
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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.

Jack D'Aurizio
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  • " the fifth row of the last matrix is a linear combination of the previous rows, giving that $\alpha$ is a root of $z^4+2z^3-z^2-2z+7$ " needs an explanation – reuns Jul 09 '16 at 22:27
  • @user1952009: it is one of the first claim proven in linear algebra, that five non-zero vectors in $\mathbb{R}^4$ have to be linearly dependent. – Jack D'Aurizio Jul 09 '16 at 22:35
  • lol.... no that's not what I'm asking !!! I understand all the other answers on this page, I just don't get what reasoning you are doing for making your polynomial, and saying it is irreducible (minimal) – reuns Jul 09 '16 at 22:36
  • do You understand what you wrote ? I'm not sure, because otherwise you'd write the justifications – reuns Jul 09 '16 at 22:37
  • @user1952009: it is probably a problem you will meet very soon. Given the minimal polynomials of $\gamma$ and $\delta$, find the minimal polynomial of $\gamma+\delta$. But as already told, $(2)$ and $(3)$ are not needed to answer your question. Anyway, just replace $x$ by $\sqrt{2}$ and $y$ by $\omega$: can you get it now? $1,\sqrt{2},\omega$ and $\omega\sqrt{2}$ are linearly independent over $\mathbb{Q}$. – Jack D'Aurizio Jul 09 '16 at 22:38
  • I already gave you references, I do not find me compelled to fill the trivial explanations you are missing. If you do no like/understand my answer, just ignore it. And please try to be polite. – Jack D'Aurizio Jul 09 '16 at 22:39
  • you are un-polite my friend, don't come on this forum if you don't like to explain. and the minimal polynomial of $a+b$ given we know the min.poly of $a$ and $b$ is a good hint https://en.wikipedia.org/wiki/Resultant#Applications – reuns Jul 09 '16 at 22:43
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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.

Jyrki Lahtonen
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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.

Sungjin Kim
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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., $\Bbb Q(\sqrt 2+e^{2\pi i/3})=\Bbb Q(\sqrt 2,e^{2\pi i/3})$.

Camilo Arosemena-Serrato
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