When $E/F$ is a field extension, I've always assumed that $E \supseteq F$. However, there is a more general definition that $E/F$ is a field extension iff $E, F$ are fields and there is an embedding $i : F \to E$. For example, with $\omega := e^{2 \pi i/3}$, $\mathbb{Q}(\sqrt[3]{2}) \cong \mathbb{Q}(\omega \sqrt[3]{2})$ since we are adjoining roots of the same minimal polynomial $X^3 – 2$, so $\mathbb{Q}(\sqrt{\sqrt[3]{2}}) \subseteq \mathbb{R}$ can be viewed as an extension not only of $\mathbb{Q}(\sqrt[3]{2})$, but also of $\mathbb{Q}(\omega \sqrt[3]{2})$.
Suppose that $E_1, E_2$ are fields with embeddings $i_1 : F \to E_1$, $i_2 : F \to E_2$, so that $E_1/F$, $E_2/F$ are extensions. Define an $F$-homomorphism as a ring homomorphism $\phi : E_1 \to E_2$ such that $\phi \circ i_1 = i_2$. This definition agrees with the usual notion of field extensions as supersets, where an $F$-homomorphism restricts to the identity on $F$.
This new definition troubles me because I don't know if theorems I've proved using the old definition will still hold true for this generalized definition.
If a proof assumes field extensions are supersets, how can I translate it to a proof where the extensions are embeddings?
For instance, Brian Conrad proves here that if $E/F$ is algebraic, then there is an $F$-homomorphism $\phi : E \to \overline{F}$ embedding $E$ in the algebraic closure $\overline{F}$. Throughout this proof, I assumed $\overline{F} \supseteq F$. Is there a change in perspective that makes the proof hold with $F$ merely embedding in $\overline{F}$?
As a corollary, if $E'/E$ is an algebraic extension (in the superset sense), then there is an $F$-homomorphism $E' \to \overline{F}$ extending $\phi$. Proof: $E$ embeds in $\overline{F}$, so $\overline{F}$ is an extension (under the general definition) of $E$ that is an algebraic closure $\overline{E}$, so there is an $E$-homomorphism $\psi : E' \to \overline{E}$, which by definition says $\psi|_E = \psi \circ \mathrm{Id}_E = \phi$ since $\phi$ is the embedding of $E$ in $\overline{E}$.
It feels like there are a lot of moving parts to keep track of; I'm looking for a way to simplify my reasoning.
Best Answer
In fact the subset case is of course a special case of the more general notion, with $i(x)=x$.
Hence it will often happen that
Of course there may remain details verifying statements about $i(x)$. It will often happen that the details follow trivially from the hypotheses.
You should work out the details in a few examples; you may see why people leave these details out: nothing but straightforward tedium. Everything works because everything's an isomorphism.
(If one knew some logic one might attempt to formulate and prove a statement that this does work if the original proof is of a suitable form...)