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 $g$)?

  • $\begingroup$ One can rephrase the question as 'does $F\le K\le L$ and $F\cong L$ implies $K\cong L$?' $\endgroup$ – Berci Jun 10 '15 at 9:37
  • $\begingroup$ What if we choose $f\equiv g\equiv 0$? $\endgroup$ – gebruiker Jun 10 '15 at 9:40
  • $\begingroup$ $0$ is not a field homomorphism unless the codomain is the trivial field with $0=1$. $\endgroup$ – Berci Jun 10 '15 at 9:41
  • $\begingroup$ @Berci Which isn't a field anyways. $\endgroup$ – Hayden Jun 10 '15 at 10:05
  • $\begingroup$ @Hayden: Well, that's only a matter of taste / definition.. similar to the question whether $0$ is natural or not. $\endgroup$ – Berci Jun 10 '15 at 10:59

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 $L\not\cong K$.

  • $\begingroup$ This is a very nice example, thank you! $\endgroup$ – user148212 Jun 10 '15 at 13:48

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