Can an algebraic extension be isomorphic to a proper subfield?

Question

Let $F\subseteq K$ be a field extension such that $K$ is algebraic over $F$. Can $K$ be isomorphic to a proper subfield $L$ containing $F$ as a $F$-algebra?

My solution is written below. It would be helpful if you can verify if it is correct. I am a beginner in field theory.

If there was indeed a $F$-algebra isomorphism $\phi:K\to L$ (in the general case) then for any $\alpha\in K\cap L^c$, $\alpha$ must have a conjugate $\alpha_1=\phi(\alpha)\in L$. Now $\alpha_1\in K$ so it will be mapped to another conjugate $\alpha_2=\phi(\alpha_1)$in $L$. Now $\alpha_1\neq \alpha_2$ because otherwise there would be two distinct elements in $K$, viz. $\alpha$ and $\alpha_1$ both mapping to $\alpha_1$. Now by similar reasoning, $\alpha_3=\phi(\alpha_2)\notin \{\alpha,\alpha_1,\alpha_2\}$, and this process will go on indefinitely getting infinitely many conjugates of $\alpha$ in $L$. But for any $\alpha\in K$ there are only finitely many conjugates possible. Thus no such $F$-algebra isomorphism can exist.

Answer 1

Your argument works.

I would prefer to use essentially the same argument for a different statement and derive the statement in your question from that. The claim is that

If $K$ is algebraic extension over $F$, then any $F$-algebra endomorphism of $K$ is surjective

Proof Let $\phi: K \to K$ be an $F$-algebra endomorphism of $K$ and let $\alpha \in K$. Then the set of conjugates to $\alpha$ given by $X:=\{\alpha = \alpha_1, \alpha_2, \dots , \alpha_n\}$ is a finite set and because $\phi$ is an $F$-algebra endomorphism it must satisfy $\phi(X) \subset X$. Now because $K$ is a field, $\phi$ is injective, thus $\phi_{\mid X}: X \to X$ is an injective map from a finite set to itself, thus it must be bijective. Thus there exists an $\alpha_i \in X \subset K$ such that $\phi(\alpha_i)=\alpha$; so $\phi$ is surjective as claimed.

I think this proof looks a bit better, because it avoids contradictions and "continuing the same reasoning", but the key idea is the same.

Finally, to see how this claim implies the claim that you're trying to show: Suppose $L \subset K$ is a subfield containg $F$, then if we have an $F$-algebra isomorphism $\phi:K \to L$, then the composition with the inclusion $K \xrightarrow{\phi} L \hookrightarrow K$ is an endomorphism of $F$-algebras, so it must be surjective, thus $L=K$.