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Admittedly, my question is a bit provocative. Let me be precise.

The fundamental theorem of Galois theory states three things (for a fixed finite normal separable extension $M/K$):

(1) The maps $L\mapsto \mathrm{Gal}(M/L)$ and $H\mapsto \mathrm{Fix}(H)$ determine an anti-isomorphism between the poset of intermediate fields of $M/K$ and the poset of subgroups of $\mathrm{Gal}(M/K)$.

(2) For each intermediate field $L$, $|\mathrm{Gal}(M/L)| = [M:L]$ and for each subgroup $H$ of $\mathrm{Gal}(M/K)$, $[M:\mathrm{Fix}(H)]=|H|$.

(3) $L$ is a normal extension of $K$ if and only if $\mathrm{Gal}(M/L)$ is a normal subgroup of $\mathrm{Gal}(M/K)$.

The main ingredient of Galois' proof of the Abel-Ruffini theorem is the following theorem that I will denote by $(*)$: a polynomial $f$ over $\mathbb Q$ is solvable if and only if $\mathrm{Gal}(f)$ is solvable.

Certainly the proof of $(*)$ uses (3) of the fundamental theorem of Galois theory to translate between the solvability of $f$ and $\mathrm{Gal}(f)$.

Question: Does the proof of $(*)$ use (1) somewhere? Or can one prove $(*)$ without proving (1)?

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  • $\begingroup$ Any reason why we should not use the fundamental theorem of Galois theory ? $\endgroup$
    – Peter
    Commented Jul 3, 2022 at 10:43
  • $\begingroup$ @Peter I don't say we should not use it. I ask whether we have to use it. $\endgroup$ Commented Jul 3, 2022 at 13:01

1 Answer 1

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At least one of the previous lemmas needed on the proof that I remember uses it:

Lemma 3 Let $L/K$ be a finite Galois extension of prime degree $p$. Suppose $char(K) = 0$. Then $L/K$ is a prime radically solvable extension.

Proof: Let $\Omega$ be an algebraic closure of $L$. Let $\zeta$ be a primitive $p$-th root of unity in $\Omega$. $\color{blue}{\textrm {Then $L(\zeta)/K(\zeta)$ is a Galois extension and\\ $Gal(L(\zeta)/K(\zeta))$ is isomorphic to a subgroup of $Gal(L/K)$.}}$ Hence it is a cyclic group of order $p$ or $1$. By Lemma 2, $L(\zeta)/K(\zeta)$ is a prime radically solvable extension. On the other hand, by another previous lemma $K(\zeta)/K$ is a prime radically solvable extension. Hence, by Lemma 1, $L(\zeta)/K$ is a prime radically solvable extension. Hence $L/K$ is a prime radically solvable extension. QED

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