abstract algebra - Show that the algebraic closure of $F$ in $K$ is an algebraic closure of $F$.

(a) Let $K$ be an algebraically closed field extension of $F$. Show that the algebraic closure of $F$ in $K$ is an algebraic closure of $F$.

What is algebraic closure of $F$ in $K$? The definition of algebraic closure is:

If $K$ is an algebraic extension of $F$ and is algebraically closed, then $K$ is said to be an algebraic closure of $F$

In this case, $K$ is more than algebraic extension so, what algebraically closed extension? I'm a little confused by this.

(b) If $\mathbb{A} = \lbrace a \in \mathbb{C}\,|\,a\,\text{is algebraic over}\,\mathbb{Q}\rbrace$, then, assuming that $\mathbb{C}$ is algebraically closed, show that $\mathbb{A}$ is an algebraic closure of $\mathbb{Q}$.

I imagine that $\mathbb{C}$ is an algebraically closed field extension of $\mathbb{Q}$ and $\mathbb{A}$ is the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$.

So, I would like some help to understand these definitions for can answer the two items. Thanks for the advance!

It may be useful to put some definitions:

Lema 1. If $K$ is a field, then the following statements are equivalente:

1. There are no algebraic extensions of $K$ other than $K$ itself.


2. There are no finite extensions of $K$ other than $K$ itself.


3. If $L$ is a field extension of $K$, then $K = \lbrace a \in L\,|\,a\,\text{is algebraic over}\,K\rbrace$.


4. Every $f(x) \in K[x]$ splits over $K$.


5. Every $f(x) \in K[x]$ has a root in $K$.


6. Every irreducible polynomial over $K$ has degree $1$.

Definition 1. If $K$ satisfies the equivalent conditions of Lema 1, then $K$ is said to be algebraically closed.