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In our algebra lecture we discussed the following corollary that came after the Theorem of Galois Correspondence:

Corollary: Every increasing sequence of fields $K \subseteq M_1 \subseteq \dots \subseteq M_n \subseteq L$ corresponds to a decreasing sequence of subgroups of $G:=Gal(L:K)$: $$Gal(L:K) \geq Gal(L:M_1) \geq \dots \geq Gal(L:L)$$

Now here $Gal(L:K)$ denotes all isomorphisms of $L$ that are the identity on $K$, i.e. the galois group of $L$ over $K$. $Gal(L:K)$ is assumed to be finite, normal and separable.

But our proof only has these two steps

Proof:

  • $L:M_1$ is finite, seperable and normal

  • $M_{i+1}$ is a intermediate field

I do not know how one then has proven the corollary. Thanks a lot in advance!

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$L/K$ is a finite extension. For any subgroup $G \le Gal(L/K)$ let the fixed field $$L^G = \{ x \in L, \forall g \in G, g x = x\}$$

The Galois correspondence follows from :

Iff $L/K$ is Galois then $K = L^{Gal(L/K)}$. In that case the map $L/M/K \mapsto Gal(L/M) \le Gal(L/K)$ is a bijection, which means $M = L^{Gal(L/M)}$ (so that $L/M$ is Galois) and the decreasing sequences of subgroups of $Gal(L/K)$ are in bijection with the increasing sequences of intermediate fields of $L/K$

That to $K \subseteq M_1 \subseteq \dots \subseteq M_n \subseteq L$ we can associate the decreasing sequence of subgroups $Gal(L:K) \geq Gal(L:M_1) \geq \dots \geq Gal(L:L)$ is always true, no matter that $L/K$ is Galois or not. The point is that this is a bijection only when $L/K$ is Galois.

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It uses induction on $n$. If $n=1$ then you have the sequence $K\subseteq M_1\subseteq L$ which corresponds to $Gal(L:K)\geq Gal(L:M_1)\geq Gal(L:L)=1$. You should be able to prove that the intermediate group is a subgroup. Now assuming that it works for $<n$, you have $$K \subseteq M_1\dots \subseteq M_n\subseteq L$$ and as it says, $L:M_1$ is finite separable and normal i.e. Galois, so you can take $$M_1\subseteq M_2\subseteq\dots \subseteq M_n\subseteq L$$ and apply the induction hypothesis.

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