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I am thinking about $\mathbb{C}/\mathbb{Q}$. But other than the identity map and taking complex conjugates, I cannot think of anything. Any ideas? Thanks.

Edit I just managed to show that if $L/K$ is algebraic then $\theta$ must be injective. For if $\theta(a)=0$ then so must $\theta(p(a))=0$ where $p(X)$ is the minimal polynomial of $a$ over $K$. But $\theta(p(a))=p(\theta(0))\neq0$ because the minimal polynomial cannot have zero constant term. And if this extension is finite that proves it is an isomorphism. So now I am asking if we can be more ambitious than just requiring finite extensions--what if $L/K$ is only assumed algebraic? And what if we drop the assumption that $L/K$ is algebraic?

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1 Answer 1

No. A simple counterexample is to consider the inclusion $K(t^2)/K \subset K(t)/K$. Using the fact that any algebraically closed field of characteristic $0$ with the same cardinality as $\mathbb{C}$ is isomorphic to $\mathbb{C}$, a counterexample for $\mathbb{C}$ is to consider the inclusion $\mathbb{C} \subset \overline{\mathbb{C}(t)} \cong \mathbb{C}$.

It is true if $L/K$ is in addition assumed to be algebraic. The point is that to show that a $K$-homomorphism is surjective in this case it suffices to show that its image contains every finite subextension of $L/K$, and any $K$-homomorphism $L \to L$ permutes the roots of any polynomial over $K$.

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I was just going to point out that $\mathbb Q(\pi)$ has the map defined by $\pi\mapsto\pi^2$. I guess it's useless now. (+1) –  Asaf Karagila Oct 22 '12 at 23:37

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