# Algebraic closure of a subfield of an algebraically closed field.

clarification of algebraic closure and algebraically closed field

I have the same question as the OP, and I was able to understand the proof given in the top answer by Lubin, which I reproduce below:

Notation: $$L$$ is algebraically closed, $$K$$ a subfield of $$L$$, $$K'$$ the algebraic closure of $$K$$ in $$L$$. I use the definitions as in the top answer in the linked question.

Suppose $$K''$$ is an algebraic extension of $$K'$$. Since $$K'$$ is algebraic over $$K$$ and $$K''$$ is algebraic of $$K'$$, $$K''$$ must be algebraic over $$K$$, which implies $$K' = K''$$.

My question is where are we using that L is algebraically closed? It seems to me the proposition is true for K a subfield of any field L.

EDIT: Ok, I think the example of $$L = \mathbb{R}$$, $$K = \mathbb{Q}$$ gives a counterexample for what I proposed, but I still don't see where the proof uses L is algebraically closed.

In your notation, we are using $$L$$ algebraically closed when we conclude from "$$K''$$ must be algebraic over $$K$$" via "$$K''$$ must be a subfield of $$L$$" to "$$K'=K''$$". If $$L$$ were not algebraically closed, there was no reason to expect $$K''$$ is a subfield of $$L$$.
Indeed, if $$L=\mathbb{R}$$ and $$K=\mathbb{Q}$$, then $$K'=\bar{\mathbb{Q}}\cap\mathbb{R}$$ does not contain $$\sqrt{-1}$$, whereas $$K''=\overline{K'}$$ would.