Let $f(x)\in k[x]$ be an irreducible , separable polynomial , let $E$ be the splitting field of $f(x)$ , then $E/k$ is a Galois extension . If $Gal(E/k)$ is abelian then is it true that $E=k(a)$ for every root $a$ of $f(x)$ ?
My Attempt : As $G:=Gal(E/k)$ is normal so for any root $a$ of $f(x)$ , $k(a)$ is Galois over $k$ , then if $H:=Gal(E/k(a))$ then $|H|=[E:k(a)]$ and $[G:H]=Gal(k(a)/k)=[k(a):k]=n$ . Also , as $n:=\deg f =[k(a):k]|[E:k]=|G| $ , and $G$ is abelian , so $G$ has a subgroup $K$ of order $n$ , let $F$ be the fixed field of $H$ , then $E/F$ and $F/k$ are Galois and $[E:F]=|K|=n$ .
But this is not leading anywhere .
Please help
Best Answer
The proof goes as follows:
Since $\operatorname{Gal}(E/k)$ is abelian, any subgroup is normal, hence any intermediate field of $E/k$ is normal. In particular $k(a)/k$ is normal and by the definition of a normal extension, it follows that all roots of $f$ are contained in $k(a)$. Hence $E=k(a)$.