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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.

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    $\begingroup$ I don't have time to write out a full answer here, but basically yes! I would even argue that it's better to think about this more general setting of field morphisms than the specific case of field extensions. It's worthwhile to think about how each theorem about field extensions can be generalized to a statement about field morphisms. $\endgroup$ Commented Apr 26, 2021 at 20:56
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    $\begingroup$ A possible source of confusion: the more general definition must include $i$ as part of the data of the extension: if you are only given that $F$ can be embedded in $E$, then that does not characterise the field extension. E.g., the identity embedding of $K(x_1, x_2, \ldots)$ in itself does not give the same field extension as the embedding induced by $x_i \mapsto x_{i+1}$. GIven that, any argument that only assumes abstract field properties on $E$ and $F$ will transfer to the more general setting. $\endgroup$ Commented Apr 26, 2021 at 21:04

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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

Often-True Fact You can convert a proof in the subset setting to a proof in the generalized setting by replacing $x$ (for various expressions $x$) by $i(x)$ in finitely many places.

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...)

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