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Sep 2, 2018 at 12:46 comment added Henno Brandsma @JyrkiLahtonen thanks for showing me it's more subtle than I thought.
Sep 2, 2018 at 12:44 comment added Jyrki Lahtonen And the outcome of the process may depend on the choice of subfields isomorphic to $E_1$ and $E_2$. Basically because that choice is not unique unless $E_i/F$ are Galois. Consider $F=\Bbb{Q}$, $E_1=F(\root3\of2)$, $E_2=F[x]/\langle x^3-2\rangle$. If you choose $E_1$ as the copy of $E_2$ sitting inside $\Bbb{C}$ you get $F(E_1\cup E_2)=E_1$. OTOH if you choose $F(e^{2\pi i/3}\root3\of2)$ as the copy of $E_2$, you get the degree six splitting field of $x^3-2$ inside $\Bbb{C}$.
Sep 2, 2018 at 11:34 comment added Torsten Schoeneberg However, if we are that careful, one should note that $\bar F$ is not unique either, only up to isomorphism fixing $F$.
Sep 2, 2018 at 10:39 history answered Henno Brandsma CC BY-SA 4.0