# Galois group of an irreducible , separable polynomial be abelian , then each of the roots of the polynomial generates the splitting field?

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 .

• @GalPorat : what definition ? do you mean we don't need the automorphism group to be abelian etc. ? – user228169 Apr 25 '17 at 8:08
• Related 1, 2 – Jyrki Lahtonen Apr 25 '17 at 10:25

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 $E=k(a)$.

• Ah yes ! Thanks a lot – user228169 Apr 25 '17 at 8:27
• Short and to the point. Nice. +1 – DonAntonio Apr 25 '17 at 8:41

Since $f$ is irreducible, the Galois group is transitive on the roots.
In a transitive commutative permutation group, for any $x$, $Stab_G(x) = \{id\}$ :

Let $x$ be a root. if some element $g$ fixes $x$ and $y$ is another root, then there is an $h$ such that $h(x)=y$ and so $g(y) = gh(x) = hg(x) = h(x) = y$. Therefore if $g$ fixes $x$, it fixes every root, and so $g$ must be the identity.

This shows that $k(x) = E^{Stab_G(x)} = E^{\{id\}} = E$