If we have fields $K \subseteq E \subseteq F$, and $F$ is a field extension of both $E$ and $K$, is $E$ a field extension of $K$?

Question

If we have fields $K \subseteq E \subseteq F$, and $F$ is a field extension of both $E$ and $K$ (i.e. both $E$ and $K$ are subfields of $F$, is $E$ a field extension of $K$? This appears obviously true to me but I'm wondering if there is an easy way to show this rigorously.

Answer 1

Yes. It is true that $K$ and $E$ are subfields of $F$. In order to prove any field $K$ is field extension of a field $F$. It is sufficient to show that there is a function $f$ such that $f:F\to E$ is monomorphism. For example :$\Bbb{Q}(\sqrt{2})$ is extension of $\Bbb{Q}$ (why!).

Added:Let $E$ and $F$ be fields. Suppose there exists an injective ring homomorphism $f:F\to E$. Then by isomorphism theorem, $f(F) \cong F/\ker(f) = F$. Thus, $f(F)$ is a subfield of $E$ that is isomorphic to $F.$

Answer 2

We need the operations of $K$ to be those of $E$ restricted to $K$. But we have that the operations of both $E$ and $K$ are the restrictions of those of $F$. Thus, the operations of $K$ restricted to $E$ are the restriction of the operations of $F$ to $E$, restricted to $K$.

But isn't that the same as the restriction of the operations of $F$ to $K$? Both $E$ and $K$ have their operations inherited from $F$.

It's a mouthful, but appears to be yes.

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In terms of morphisms, you have a (necessarily injective, because fields have no nontrivial ideals) morphism from $K$ to $F$ (inclusion). But, actually, the range of that morphism is contained in $E$. So we have the required morphism from $K$ to $E$. So $K$ is a subfield of $E$.