Question

Let K be an extension field of F and E the subfield of all element of K that are algebraic over F . If K is algebraically closed prove that E is an algebraic closure of F.

Short answer

E is the algebraically closure of F.

Step-by-step solution

Definition of algebraic extension 

Let K an extension field of F . Then K is said to be an algebraic extension of F if every element of K is algebraic over F.

Showing that  E is the algebraically closure of  F

Let K be an extension field of F and E be the subfield of all element of K that are algebraic over F . Now it is enough to show that $E/F$is algebraic extension and E is algebraically closed.

Since E consists of all elements of K that are algebraic over F . This implies that$E/F$is algebraic extension. Now to prove thatis algebraically closed it is sufficient that E has no proper algebraic extension.

Let E be an algebraic extension of E . Since E is algebraic extension of E. So,

$E\subseteq E^{\text{'}}\left(1\right)$

Let $a\in E^{\text{'}}$ then is algebraic over E and $E/F$ is algebraic extension. Therefore a is algebraic over F . But consists of all elements of K that are algebraic over F.

This implies.

$a\in E$

Since a is arbitrary so$E^{\text{'}}\subseteq E\left(2\right)$

From 1 and 2

$E^{\text{'}}=E$

Thus E has no proper extension. This implies E is algebraically closed.

So, E is algebraic closure of F .