TITLE: On the existence of field morphisms
QUESTION [8 upvotes]: Let $K$ and $L$ be two fields, does the existence of two field morphisms $f\colon K\rightarrow L,\ g\colon L\rightarrow K$ imply that, as abstract fields, $K\cong L$ (not necessarily via $f$ or $g$)?

REPLY [6 votes]: I think I found a counterexample:
Let $F:=\overline{\Bbb Q(x_1,x_2,\dots)}$, i.e. the algebraic closure of the extension of $\Bbb Q$ by infinitely many independent transcendent elements.
Let $K:=F(x_0)$ a simple transcendent extension. This field is not algebraically closed.
Finally, let $L:=\overline K$, its algebraic closure.
Unless I am mistaken, $L=\overline{F(x_0)}=\overline{\Bbb Q(x_0,x_1,x_2,\dots)}\cong F$.
Thus we gain field morphisms $K\to L$ and $L\cong F\to K$, though $L\not\cong K$.