The algebraic closure of an algebraic extension of a field is just the algebraic closure of the field.

Question

I have to show that if $K$ is an algebraic extension of a field $F$, then $\overline{K} = \overline{F}$.

My attempt: obviously, $\overline{F} \subset \overline{K}$.

OTOH, take an element $a$ from $\overline{K}$. But because an algebraic extension of an algebraic extension is an algebraic extension, $\overline{K}$ is an algebraic extension of $F$. Therefore $a \in \overline{F}. \square$

Is there anything I need to fix or take care of? In particular, is the first statement really "obvious"?

Answer 1

“The” algebraic closure of a field is not a well defined concept.

What you want to prove is that

if $K$ is an algebraic extension of a field $F$ and $\bar{K}$ is an algebraic closure of $K$, then $\bar{K}$ is also an algebraic closure of $F$.

An algebraic closure of $F$ is characterized by the fact that it is an algebraically closed algebraic extension of $F$.

1. $\bar{K}$ is an extension of $F$: true, because $\bar{K}$ is an extension of $K$;
2. $\bar{K}$ is algebraically closed: true by assumption;
3. $\bar{K}$ is an algebraic extension of $F$: true, by the argument below.

If $a\in\bar{K}$, then $a$ is algebraic over $K$. But every algebraic element over $K$ is also algebraic over $F$, by well-known results.