I am aware that there is a result saying that if $L_1, L_2$ are two algebraically closed fields of the same characteristic, then either $L_1$ is isomorphic to the subfield of $L_2$, or $L_2$ is isomorphic to a subfield of $L_1$.
I am unsure however of how to prove this result. I am quite surprised that this is true, because I feel like these algebraic subfields can be so large that there's no reason why their structures have to have similarities like this. Could anyone offer some insight please?
Let $k$ be the prime field of the common characteristic of $L_1$ and $L_2$, let $B_1$ be a transcendence basis for $L_1$ over $k$, and let $B_2$ be a transcendence basis for $L_2$ over $k$. Then either $|B_1|\leq |B_2|$ or $|B_2|\leq |B_1|$; suppose WLOG that $|B_1|\leq |B_2|$. Choose an injection $i:B_1\to B_2$; this then gives a homomorphism $j:k(B_1)\to k(B_2)\subset L_2$. Since $L_1$ is algebraic over $k(B_1)$ and $L_2$ is algebraically closed, we can extend $j$ to a homomorphism $L_1\to L_2$. Thus $L_1$ embeds in $L_2$.