If $i_1 : k \rightarrow K$ and $i_2 : K \rightarrow k$ are fields extension, do we have $k \cong K$?

Question

The question is quite simple, but yet I don't succeed to prove or disprove it.

If $i_1 : k \rightarrow K$ and $i_2 : K \rightarrow k$ are fields extension, do we have $k \cong K$ ?

Someone could help me ?

Answer 1

Extend $k:=\mathbb C$ to $K:=\mathbb C(X)$ and then to its algebraic completion $\cong k$.

$k \not\cong K$ since every element of $k$ has a square root.

Answer 2

Let $E_1$ and $E_2$ be elliptic curves over $\Bbb Q$ which are isogenous, but not isomorphic. Let $K_1$ and $K_2$ be their function fields. There are then isogenies $i_1:E_1\to E_2$ and $i_2:E_2\to E_1$. These induce maps of function fields $i_1^*:K_2\to K_1$ and $i_2^*:K_1\to K_2$. Then $K_2$ is isomorphic to a finite extension of $K_1$, and vice versa yet $K_1$ and $K_2$ are not isomorphic.