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Let $\overline{\mathbb{Q}}$ be an algebraic closure of $\mathbb{Q}$. Then, does there exists finite extension $E_1,E_2$ of $\mathbb{Q}$ inside $\overline{\mathbb{Q}}$, such that $E_1\neq E_2$ but $E_2\cong E_2$?

Here, extensions $E_1,E_2$ of $\mathbb{Q}$ are isomorphic if there is a bijective map from $E_1$ to $E_2$ which preserves addition and multiplication in the fields.

Instead of $\mathbb{Q}$, we may take any field, but there we get no such examples (line $\mathbb{R}$, $\mathbb{C}$).

For example, for $\mathbb{R}$, we can take $\overline{\mathbb{R}}=\mathbb{C}$. Now between $\mathbb{R}$ and $\overline{\mathbb{R}}$, there are no other fields. Hence we do not get $E_1,E_2$ such that $\mathbb{R}\subseteq E_1 ,E_2 \subseteq \overline{\mathbb{R}}$ but $E_1\neq E_2$.

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  • $\begingroup$ Thanks for noticing important mistakes. The question now written is in correct form. $\endgroup$ – Groups Aug 13 '15 at 6:57
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What does unequal mean for isomorphic fields after all? Is $\mathbb Q[\sqrt 2]$ really so much different from $\mathbb Q[X]/(X^2-2)$? When considering field extensions of $\mathbb Q$ living inside $\mathbb C$, you might for example consider $\mathbb Q[\sqrt[3]2]$ and $\mathbb Q[\sqrt[3]2\,e^{\frac23\pi i}]$, which are isomorphic by the obvious map, but only the former is also a subset of $\mathbb R$.

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