If F/K is an algebraic field extension, is every endomorphism of F/K an automorphism? If F/K is a field extension of finite degree, then every endomorphism of F/K is an automorphism. Does the result also hold for algebraic extensions?
 A: As Georges Elencwajg said, a $K$-endomorphism of an algebraic extension $F$ is indeed an isomorphism. Here is another argument.
Let $f:F\rightarrow F$ be your $K$-endomorphism. First, note that it is into, because its kernel is an ideal of $F$, and since it is not $F$, it is $\{0\}$. We want to prove it is onto, so let $a\in F$. Because $a$ is algebraic over $K$, it is a root of a polynomial $P$. Let $F_P$ be the set of roots of $P$ in $F$. Note that this is a finite set ($P$ has only a finite number of roots in $F$), and that $f(F_P)\subset F_P$ ($f$ maps a root of $P$ to a root of $P$
So $f$ induces a injective map $F_P\rightarrow F_P$ and $F_P$ is finite, so it is also onto. Hence, there is $b\in F$ such that $f(b)=a$.
A: Yes, the result holds also for endomorphisms $f:F\to F$ of infinite algebraic extensions $F/K$.
Proof:
Let $a\in F$ be an arbitrary element and let's show that $a$ is in the image of $f$.
Consider  the field $N=K[a_1,\dots, a_r]$ obtained by adjoining to $K$ the roots $a=a_1,\dots, a_r$ in $F$ of the minimal polynomial $Irr(a,K,X)$ of $a$ over $K$.
 Notice that $N$ is a finite-dimensional extension field of $K$ containing $a$ and that $f(N)\subset N$  because for every $i\in \{1,\dots ,r\}$ we must have $f(a_i)=a_j$ for some $j\in \{1,\dots ,r\}$  .
Hence we may restrict $f$ to $N$ and obtain an endomorphism $f_N:N\to N$, to which we may apply the finite dimensional result  to deduce that there exists $b\in N$ for which $f_N(b)=a$.
But then $f(b)=f_N(b)=a$ and we see that, as desired, $a$ is in the image of $f$.
