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I'm stuck with the following field extensions exercise:

Find two examples of algebraic extensions $k \subseteq F$, $k \subseteq K$ and embeddings $F \subseteq \overline{k}$, $\sigma_1 : K \rightarrow \overline{k}$, $\sigma_2 : K \rightarrow \overline{k}$ extending $k \subseteq \overline{k}$ (where $\overline{k}$ is the algebraic closure of $k$) such that the composites $F\sigma_1(K)$ and $F\sigma_2(K)$ are not isomorphic.

The examples I'm looking for should not be Galois extensions since otherwise no such example would exist. Hence, I was thinking about taking $k = \mathbb{Q}$, $F = \mathbb{Q}(\sqrt[4]2)$, $K = \mathbb{Q}(\sqrt[3]2)$, the trivial embedding of $\mathbb{Q}(\sqrt[4]2)$ into $\overline{\mathbb{Q}}$ and the embeddings $\sigma_1$, $\sigma_2$ of $\mathbb{Q}(\sqrt[3]2)$ into $\overline{\mathbb{Q}}$ given by $\sigma_1(\sqrt[3]2) = \sqrt[3]2$ and $\sigma_2(\sqrt[3]2) = \sqrt[3]2 \omega$, where $\omega = e^{\frac{2 \pi i}{3}}$. Then, it's clear that the composites $F\sigma_1(K)$ and $F\sigma_2(K)$ are different since one of them is a subfield of $\mathbb{R}$ while the other one includes the non-real number $\sqrt[3]2 \omega$. However, I believe they are still isomorphic since $\mathbb{Q}(\sqrt[3]2) \cong \mathbb{Q}(\sqrt[3]2 \omega)$. Therefore, I'm kind of lost and having difficulty finding an example in which the composites are not only different but satisfy the stronger condition of being non-isomorphic. Maybe I need to play more with the embedding of $F$ into the closure of $k$ or maybe I need to take some other more complicated fields?

Any help, hints, or insights would be greatly appreciated.

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You were on the right track.

Starting in the same way as you did, let \begin{align*} k&=\mathbb{Q}\\[4pt] K&=\mathbb{Q}(\sqrt[3]2)\\[4pt] \end{align*} and let $\sigma_1,\sigma_2$ be embeddings of $K$ into $\overline{k}$ such that \begin{align*} \sigma_1(\sqrt[3]2)&=\sqrt[3]2\\[4pt] \sigma_2(\sqrt[3]2)&=\omega\sqrt[3]2\\[4pt] \end{align*} where $ \omega = e^{\large{\frac{2\pi i}{3}}} $.

Now let $F=\mathbb{Q}(\sqrt[3]2)$, and let $F_1,F_2$ denote $F\sigma_1(K),F\sigma_2(K)$, respectively.

Then $\omega\in F_2$, so $F_2$ has an element satisfying $x^2+x+1=0$, whereas $F_1$ has no such element.

It follows that $F_1,F_2$ are not isomorphic.

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  • $\begingroup$ Thank you so much! Yeah, I was also thinking about somehow adding $\sqrt 3$ to maybe make one of the composites equal to $\mathbb{Q}(i)$ or to involve $i$ in some way but couldn't manage to do it and yes, your argument works perfectly well and thank you so much again. $\endgroup$ Commented Feb 9 at 23:04

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