Algebraic closure of a subfield of an algebraically closed field.

Question

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.

Answer 1

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.