$K$ is a normal extension of$ F$ implies $ K $contains all the roots of minimal polynomial of $a$

Question

Let $K$ be a field of characteristic 0.

How can i show that:

If $K$ is a normal extension of$ F$(whereby $K=F(a)$ for some $a$ in $K$) then it implies that $ K $contains all the roots of the minimal polynomial of $a$?

I know that $K$ is a normal extension of $F $iff $K$ is the splitting field of some polynomial over $F$.

Also,the definition of normal extension that I know is:

$K $is a normal extension of $F$ if dimension of $K$(as a vector space over $F$) is finite and the fixed field of $G(K,F)$(group of automorphisms of $K$ relative to $F$) is $F$.

I encountered this problem while reading Theorem 5.6.6,pg 247,Herstein.

Answer 1

Let $f$ be the minimal polynomial of $a$ over $F$. Let $a_1, \dots , a_n$ be the roots of $f$ that are contained in $K$. Clearly, $G(K,F)$ fixes all the elementary symmetric functions of $a_1, \dots , a_n$. Therefore if $K/F$ is normal, the polynomial $\prod_{i=1}^n (X-a_i)$ has coefficients in $F$ hence it is equal to $f$ and we are done.