For two algebraically closed fields, one is isomorphic to a subfield of the other 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? 
 A: 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 $L_2$.
A: What follows is meant to illuminate the intuition, and is neither a proof nor entirely formally accurate.
Both fields contain the same copy of either the algebraic closure of $\mathbb{Q}$ or of $\mathbb{F}_p$ depending on what the characteristic is. This is a straightforward consequence of the fact that the smallest subfield of any field is one of the $\mathbb F_p$ or is $\mathbb Q$.
All of the other elements of $L_1$ and $L_2$ are transcendental, by definition. However, all transcendental elements “look the same.” From an algebraic point of view, transcendental numbers interact with algebraic numbers just like formal symbols. For an illustrative example, we have: $$\mathbb R(x)\cong {\mathbb R(\pi)}\cong{\mathbb R(e)}$$
We can take an algebraically closed field, add a transcendental element, close under field operations, close under polynomials, and what we get back is another algebraically closed field. However, it doesn’t matter which element we added, any field obtained this way is isomorphic to every other one by the above equation.
If you iterate this process, you can construct any algebraically closed field. The process is straightforward:

Let $A$ be an algebraically closed field whose smallest sub field is $F$. Let $F_0=\overline{F}$. To construct $F_{s(\alpha)}$, pick any element in $A\setminus F_{\alpha}$ and call it $a$. We define $F_{s(\alpha)}=\overline{F_\alpha(a)}$. For limit ordinals, we define $F_\beta=\cup_{\alpha<\beta} F_\alpha$. It can be proven that every element of $A$ will be eventually added by this process, and so for some $\gamma$ we have that $A=F_\gamma$. Therefore every algebraically closed field has this form.

So all of the algebraically closed fields look the same, it just matters how many times we do this procedure. That’s what determines which one is isomorphic to a subset of the other.
