Three fields $F\leq F_1,~F_2$ and $F_1\neq F_2$ but $F_1\cong F_2$ I'm thinking about something in field theory. I extract the essence problem to ask here (so the problem may be weird). Given three fields $F,~F_1,~F_2$ with $F\leq F_1$, $F\leq F_2$, $F_1\neq F_2$ and $F_1\cong F_2$. Can it be shown that $F_1$ contains at least two distinct isomorphic copy of $F$? (i.e. there exists $F'\neq F$, $F'\leq F_1$ and $F'\cong F$).
 A: You can show, that for any ring $ R $ (commutative, unitary) there exists at most one morphism $ \Bbb{Q} \to R $. This implies that there can be at most one subring of R isomorphic to $ \Bbb{Q} $. Now it is easy to construct counterexamples. Just consider the following field extensions of $ \Bbb{Q} $ in $ \Bbb{C} $:
$$ \Bbb{Q}[\sqrt[3]{3}] \quad \text{and} \quad \Bbb{Q}[\zeta \sqrt[3]{3}], $$
where $ \zeta $ is a non-trivial third root of unity. Now those are field extensions of $ \Bbb{Q} $ and they are isomorphic because $ \sqrt[3]{3}, \sqrt[3]{3} \zeta $ have the same minimal polynomial $ X^{3} - 3 $. Furthermore they are not the same because $ \Bbb{Q}[\sqrt[3]{3}] \subset \Bbb{R} $, but $ \zeta \sqrt[3]{3} \in \Bbb{C}\backslash\Bbb{R} $.
Hint for the first statement: This is a fun little exercise. If you haven't seen it, I can recommend it. First you can show that there exists exactly one morphism $ \Bbb{Z} \to R $. For this you just have to remember what the symbols in  $ \Bbb{Z} $ actually stand for. To show, that there is at most one way to extend this to a morphism $ \Bbb{Q} \to R $ do the same: remember what the symbols in $ \Bbb{Q} $ stand for.
