Let $E/F$ be a finite Galois extension and $\alpha\in E$ and $p(t)=irr(\alpha,F)$, the monic irreducible polynomial defined over $F$ which has a root $\alpha$.

Let $\beta\in E$ be another root of $p(t)$.

My question arises here.

Are the two Galois groups $Gal(E/F(\alpha))$ and $Gal(E/F(\beta))$ isomorphic?

If this is not the case, could you suggest me some counter-example?

If they are always isomorphic, how can we detect the differences between each roots of an irreducible polynomial? Because, I know the all the roots of some irreducible polynomial are not of equal status from the view of total field $E$, not from the base field $F$.


1 Answer 1


$\newcommand{\Gal}{\mathrm{Gal}}$Yes, the groups will be isomorphic. This can be seen by noting that $E$ is the splitting field over $F$ of some separable polynomial $f \in F[x]$. Now clearly $E$ is also a splitting for $f$ over both $F(\alpha)$ and $F(\beta)$, and there is an isomorphism $\theta$ from $F(\alpha)$ to $F(\beta)$, that takes $\alpha$ to $\beta$. By the uniqueness theorem for the splitting field, $\theta$ can be extended to an isomorphism $\eta$ of $E$.

Now the inner automorphism $g \mapsto \eta^{-1} g \eta$ (left-to-right composition) of $\Gal(E/F)$ will induce an isomorphism $\Gal(E/F(\alpha)) \to \Gal(E/F(\beta))$.

  • $\begingroup$ Thank you for your kind answer.But there is some point to correct. You said that by the uniqueness theorem of splitting field,$theta$ can be extended to an isomorphism of $E$. But it is not due to uniqueness of splitting field but to isomorphism extension theorem. $\endgroup$
    – user29422
    Feb 26, 2016 at 8:08

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