Suppose L/K is Galois, M an intermediate field between L and K. Suppose that no intermediate field between L and M is Galois over K except L itself. Prove that if N is a subfield of L which contains all fields $\sigma(M)$ for $\sigma\in Gal(L/M)$, then N=L. I am trying to prove this by contradiction. I know that I have to use Galois correspondence, how do I deal with N and M?


1 Answer 1


Translating the given pieces of information using Galois correspondence suggests the following plan of attack (you justify the steps/claims):

Let $G=Gal(L/K)$, and let $H$ be the subgroup corresponding to $M$, i.e. $H=Gal(L/M)$.

  1. An intermediate field $F$ is Galois over $K$, iff $Gal(L/F)\unlhd G$.
  2. If $T$ is a subgroup of $H$ such that $T\unlhd G$, then $T=\{1\}$.
  3. $N$ contains $\sigma(M)$ for all $\sigma\in G$, so $Gal(L/N)\le \sigma H\sigma^{-1}$ for all $\sigma\in G$.
  4. We have $$Gal(L/N)\le\bigcap_{\sigma \in G}\sigma H\sigma^{-1}.$$
  5. The subgroup $$\bigcap_{\sigma \in G}\sigma H\sigma^{-1}\unlhd G.$$
  6. $Gal(L/N)\le\bigcap_{\sigma \in G}\sigma H\sigma^{-1}=\{1\}.$
  7. $N=L$.

You will see that using Galois correspondence is actually quite straightforward. It does take a bit of practice getting used to (we have all been there). IIRC it took me a while to internalize the effects of order reversal (= bigger intermediate fields correspond to smaller subgroups).

  • $\begingroup$ thank you! Can you explain step 3? I understand the other 6 steps, but I can't see why Gal(L/N) is a subgroup of $\sigma H \sigma^{-1}$. $\endgroup$
    – user335468
    Apr 17, 2017 at 14:14
  • $\begingroup$ If $N$ contains $\sigma(M)$, then $Gal(L/N)$ is contained in $Gal(L/\sigma(M))$. $\endgroup$ Apr 17, 2017 at 14:53

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