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?