Suppose $K/F$ is a field extension with Galois closure $L$, and let $G=\operatorname{Gal}(L/F)$. Why is $L$ the same as the composite of the Galois conjugates $\sigma(K)$ for $\sigma\in G$?

I know that $K$ is certainly contained in some Galois extension, and then the Galois closure of $K/F$ is just the intersection of all Galois extensions of $F$ containing $K$. That's the definition I've always seen. Why is the above characterization the same?

I'm curious because it shows up in the technical lemma preceding Abel's solvability by radicals theorem.


1 Answer 1


A Galois closure of an extension $K/F$ in a fixed algebraic closure $\overline{F}$ is a field which is minimal among all Galois extensions of $F$ containing $K$. One can prove there is a unique such field, which will be the intersection of all Galois extensions of $F$ containing $K$.

Suppose $L$ is the Galois closure of $K/F$. Then $L$ must contain all ${\rm Gal}(L/F)$-conjugates of $K$ and hence $L$ must contain the compositum $C=\prod_\sigma \sigma K$. Conversely, by the primitive element theorem we may write $K=F(\theta)$ where $\theta$ satisfies its minimal polynomial $m(x)\in F[x]$. Since $C$ is the splitting field of $m(x)$ it is normal, if we assume further that it is separable (which is the situation for number fields in particular) then $C/F$ is Galois containing $K$. But $L$ is minimal with these properties and $L$ contains $C$, so we must therefore have $L=C$.

  • 1
    $\begingroup$ Thanks. And to be sure, $C$ is the splitting field of $m(x)$ since $\{\sigma(\theta):\sigma\in G\}$ is the set of roots since $G$ acts transitively on them? $\endgroup$ Mar 13, 2014 at 3:23
  • $\begingroup$ @Hobbie yes. ${}$ $\endgroup$
    – anon
    Mar 13, 2014 at 3:25
  • $\begingroup$ Why can you use the primitive element theorem without justifying it is a separable extension? Or we should assume $K/F$ is separable in addition? $\endgroup$
    – Bach
    Feb 4, 2020 at 13:52
  • 1
    $\begingroup$ If $K/F$ isn't separable, then no $L \supset K$ can be Galois. $\endgroup$ May 5, 2023 at 21:01
  • $\begingroup$ Is this result still true when characteristic isn't zero? $\endgroup$
    – MLe
    Dec 9, 2023 at 22:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .