# Do we need the fundamental theorem of Galois theory in Galois' proof of the Abel-Ruffini theorem?

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)?

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

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