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Let $E/K$ be a finite Galois field extension. Then by the fundamental theorem of Galois theory, there is canonical bijection between the subgroups of $\mathrm{Gal}(E/K)$ and the intermediate field extensions $E/L/K$, sending a subgroup to the intermediate field, whose elements are fixed by the automorphisms of the subgroup.

Question: What is known about the converse statement, i.e. if for a finite field extension, the described correspondence holds, is it Galois?

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Yes. For every subgroup $G$ of $\textrm{Aut}\,E$, one has $[E:E^G]=|G|$. Take $G=\textrm{Gal}(E/K)$. Then $|G|=[E:E^G]=[E:K]$, where $E^G=K$ follows from the correspondence that you assume. But $|G|=[E:K]$ means $E/K$ is Galois.

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  • $\begingroup$ Can you explain the last implication? $\endgroup$
    – J.R.
    Oct 14, 2012 at 18:45
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    $\begingroup$ This is a theorem of Artin: it states that if $E$ is any field and $G\subset \textrm{Aut}\,E$ is a finite subgroup, then $E/E^G$ is Galois and of degree $|G|$ (with Galois group $G$). You can find a statement and a proof in Lang's "Algebra" (p. 194) or just the statement here: en.wikipedia.org/wiki/Galois_extension. $\endgroup$
    – Brenin
    Oct 14, 2012 at 23:54
  • $\begingroup$ Artin's theorem is overkill. If the standard mappings in both directions are inverses then $E^{{\rm Aut}(E/K)} = K$. That this equality is equivalent to $E/K$ being Galois is part of the development of Galois theory. $\endgroup$
    – KCd
    Mar 24, 2015 at 20:54

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