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.


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 $L=C$.

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    $\begingroup$ 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? $\endgroup$ – Dani Hobbes Mar 13 '14 at 3:23
  • $\begingroup$ @Hobbie yes. ${}$ $\endgroup$ – blue Mar 13 '14 at 3:25

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