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Let $K$ be algebraically closed field, over $F$ (base field). Assume the trancendental degree of $K$ over $F$ is infinite. Please give an example of a $F$-homomorphism $K\to K$ which is not surjective.

(I proved the theorem that if Trancendental Degree of $K$ over $F$ is finite, $K$ algebrically closed, then any $F$-homomorphism from $K\to K$ is surjective.)


$\mathbb{C}$ has infinite trancendence degree over $\mathbb{Q}$.

Consider $S=\{\text{all elements in }\mathbb{C}\text{ that are algebraically indep. over }\mathbb{Q}\}$. $S$ is infinite set.

So there exists an injective nonsurjective map from $S$ to $S$.

This gives an obvious map from $F(S)$ to $F(S)$. This map is a homomorphism because elements in S acts like variables, there is no relation between them. Let $q\in S$ which has no preimage in $S$. $q$ has also no preimage in $F(S)$, since if so then by clearing denominators, transfering sides, we get a non-zero polynomial if $F$ satisfying a relation between algebraically independent elements. Contradiction. But how to extend the map to $\mathbb{C}$ such that it is a homomorphism?

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    If you pick a transcendence basis for $K$ over $F$, then $K$ is algebraic over $F(S)$ and is in fact the algebraic closure of $F(S)$; hence, any homomorphism from $F(S)$ into an algebraically closed field extends to all of $K$; in particular, any homomorphism $F(S)\hookrightarrow K$ extends to a homomorphism $K\hookrightarrow K$.2012-04-17

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Hint. If $S$ is a transcendence basis for $K$ over $F$, then there exists a nonsurjective injection $S\to S$ (since $S$ is infinite). This in turn induces a map $F(S)\hookrightarrow F(S)$ that is not onto; extend to $K$.