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I was a reading a proof in Lang, and I think he seems to be using the following:

Let $k$ be a field and $E$ and algebraic extension lying within a given algebraic closure $k^a$ of $k$. Suppose, there is an embedding $\sigma$ of $E$ into $k^a$ that fixes $k$ s.t. $\sigma(E)\subseteq E$ then, $\sigma(E)=E$.

I am generalizing this from a specific statement, so I could be wrong, but I am unable to prove this when the extension $E/k$ is not finite. In the finite case, I can just compare the degrees of the extensions.

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I'm assuming you want $\sigma$ to be the identity on $k$.

Here's one way to see this: Let $x\in E$, and let $P$ be the minimal polynomial of $x$ over $k$. Let's say that $E$ contains $d$ roots of $P$. Then $\sigma(E)$ also contains $d$ roots of $P$, so $\sigma(E)$ must contain $x$.

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  • $\begingroup$ Thanks for the answer. I was assuming the embedding fixed $k$. $\endgroup$ – BMI Sep 28 '11 at 23:56
  • $\begingroup$ Your proof shows that this has nothing to do with algebraic closures. Theorem Let $E$ be an algebraic extension of the field $k$. Then any $k$-endomorphism of fields $E\to E$ is an isomorphism. Mnemonic ENDO=ISO $\endgroup$ – Georges Elencwajg Sep 29 '11 at 6:14

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