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Sorry if this is a very trivial question but I can't find a proof or a counterexample to it.

If $K$ and $L$ are isomorphic subfields of $\mathbb C$ both containing $\mathbb Q$ then are they identical (as sets)?

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closed as off-topic by user26857, Claude Leibovici, Watson, Daniel W. Farlow, Silvia Ghinassi May 16 '16 at 14:58

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No, take $\mathbb{Q}(x)$ and $\mathbb{Q}(y)$ where $x$ is any transcendental number and $y$ is any transcendental number not in $\mathbb{Q}(x)$ (which exists since $\mathbb{Q}(x)$ is countable and transcendental numbers are uncountable) : they are both isomorphic to $\mathbb{Q}(X)$ but they are not equal.

PS : all subfields of $\mathbb{C}$ contain $\mathbb{Q}$.

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  • $\begingroup$ It's not obvious to me that $\mathbb Q(\pi)$ and $\mathbb Q(e)$ are not equal as sets -- for example, it is not known whether $\pi$ and $e$ are algebraically independent. It would be more robust to consider, for example, $\mathbb Q(\pi)$ and $\mathbb Q(\pi i)$, which are certainly not equal as sets. $\endgroup$ – Henning Makholm May 1 '16 at 16:10
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    $\begingroup$ You're right, I just took the first transcendental numbers that came to mind thinking "they have no reason to be related" but they could be. I edited to say something more rigourous. $\endgroup$ – Captain Lama May 1 '16 at 16:13
  • $\begingroup$ You may want to ensure that $y$ is in fact transcendental over $\mathbb{Q}(x)$, not just not contained in $\mathbb{Q}(x)$, because otherwise you could choose $y = \sqrt{x}$ for example $\endgroup$ – Anon May 1 '16 at 16:57
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    $\begingroup$ @McFry The fact that $y\not\in \mathbb{Q}(x)$ is enough to ensure that the two are not equal (since $y\in \mathbb{Q}(y)$ obviously). But you're right, we could go further and enforce that the field are linearly independent over $\mathbb{Q}$. $\endgroup$ – Captain Lama May 1 '16 at 17:51
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Here's another example. First, note that $x^3-2$ is irreducible over $\mathbb{Q}$, by Eisenstein's Criterion with $p=2$. This polynomial has the following roots: $$\sqrt[3]{2},\ \zeta_3\sqrt[3]{2},\ \zeta_3^2\sqrt[3]{2},$$ where $\zeta_3 = -\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$ is a primitive 3rd root of unity. Note that since each of the above numbers is a root of the same irreducible polynomial, we must have $$\mathbb{Q}(\sqrt[3]{2})\cong\mathbb{Q}(\zeta_3\sqrt[3]{2})\cong\mathbb{Q}(\zeta_3^2\sqrt[3]{2}).$$ Now, we consider the evaluation map $$\varepsilon_{\sqrt[3]{2}}:\mathbb{Q}[x]\to\mathbb{C},\ \ \varepsilon_{\sqrt[3]{2}}(p(x)) = p(\sqrt[3]{2}).$$ This factors uniquely through the (1 to 1) mapping: $$\overline{\varepsilon_{\sqrt[3]{2}}}:\frac{\mathbb{Q}[x]}{(x^3-2)}\to\mathbb{C},$$ whose image is $\mathbb{Q}(\sqrt[3]{2})\subseteq \mathbb{C}$, a subfield containing only real numbers. In addition to the above map, we have two other evaluation maps: $$\varepsilon_{\zeta_3\sqrt[3]{2}}:\mathbb{Q}[x]\to\mathbb{C},\ \ \varepsilon_{\zeta_3\sqrt[3]{2}}(p(x)) = p(\zeta_3\sqrt[3]{2}),$$ $$\varepsilon_{\zeta_3^2\sqrt[3]{2}}:\mathbb{Q}[x]\to\mathbb{C},\ \ \varepsilon_{\zeta_3^2\sqrt[3]{2}}(p(x)) = p(\zeta_3^2\sqrt[3]{2}).$$ These two maps induce the (injective!) field homomorphisms $$\overline{\varepsilon_{\zeta_3\sqrt[3]{2}}}\quad\text{and}\quad \overline{\varepsilon_{\zeta_3^2\sqrt[3]{2}}}$$ out of the quotient (by the universal property of quotients), with images: $$\mathbb{Q}(\zeta_3\sqrt[3]{2})\quad\text{and}\quad\mathbb{Q}(\zeta_3^2\sqrt[3]{2}),$$ respectively. These two fields are not equal, which is certainly not obvious. Furthermore, since each field above contains imaginary numbers, neither can be equal to $\mathbb{Q}(\sqrt[3]{2})$. In order to show that the above two fields are not equal, we apply the Tower Law.

Assume that $\mathbb{Q}(\zeta_3\sqrt[3]{2}) = \mathbb{Q}(\zeta_3^2\sqrt[3]{2}) = K$. Then, we must have $$\frac{\zeta_3^2\sqrt[3]{2}}{\zeta_3\sqrt[3]{2}} = \zeta_3\in K,$$ and it follows that $\sqrt[3]{2}\in K$. Hence, $\mathbb{Q}(\sqrt[3]{2})\subseteq K$, and this inclusion is proper since $\mathbb{Q}(\sqrt[3]{2})$ does not contain imaginary numbers. Hence, by the Tower Law, \begin{align} \notag 3 &= [K:\mathbb{Q}]\\ \notag &= [\mathbb{Q}(\zeta_2\sqrt[3]{2}):\mathbb{Q}(\sqrt[3]{2})][\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]\\ \notag &= [\mathbb{Q}(\zeta_2\sqrt[3]{2}):\mathbb{Q}(\sqrt[3]{2})]\cdot 3, \end{align} and it follows that $$[\mathbb{Q}(\zeta_3\sqrt[3]{2}):\mathbb{Q}(\sqrt[3]{2})]= 1.$$ Hence, we must have $$\mathbb{Q}(\zeta_3\sqrt[3]{2}) = \mathbb{Q}(\sqrt[3]{2}),$$ which is a contradiction. Thus, all 3 fields are distinct. An interesting property of these fields is that although all 3 are isomorphic to each other, $\mathbb{Q}(\sqrt[3]{2})$ is not dense in $\mathbb{C}$, while the two other fields are.

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