The main theorem of the Galois theory proven with ideas from the first M\'emoire of Galois

TL;DR

This paper presents a proof of the main theorem of Galois theory utilizing symmetric polynomial concepts, inspired by Galois's original work, emphasizing historical and mathematical connections.

Contribution

It offers a novel proof of Galois's main theorem based on symmetric polynomials, linking historical ideas with modern algebraic techniques.

Findings

Proof connects Galois's original ideas with symmetric polynomial theory.

Highlights the historical influence of Galois's memoir on modern proofs.

Demonstrates the elegance of algebraic structures in solving polynomial equations.

Abstract

A proof of the main theorem of the Galois theory is presented using the main theorem of symmetric polynomials. The idea originated from studying the "M\'emoire sur les conditions de r\'esolubilit\'e des \'equations par radicaux" of Evariste Galois. The motto "Read the masters" pays off.

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Full text

Introduction.

The Galois theory today is based on automorphisms of a field extension that leave the basic field invariant. Central in this theory is the well-known correspondence theorem usually proven with an Artin-Dedekind lemma. This article will provide a proof of this theorem which is inspired by ideas you can find in the Mémoire sur les conditions de résolubilité des équations par radicaux. For translations see [1, 2, 4], for the original manuscript see [3]. The article also refers to the main theorem of the symmetric polynomials wich Galois also often uses. This article therefore shows that the motto ”Read the masters”111Edwards and Tignol support this motto. works.

The idea originates as follows: if you look at the n$\times$m array below in proposition I of the Mémoire you see that $\phi$V,$\phi_{1}$V,$\phi_{2}$V,….$\phi_{m-1}$V, the roots of the given equation, are listed in different orders in n rows. One entry of the m roots in a row is called an arrangement222For an excellent explanation of the concepts arrangement, arrangementgroup, substitution and substitutiongroup see [6]. of the roots a,b,c,d,…. The n arrangements form a so called arrangementgroup and the corresponding substitutiongroup is the Galois group of F $\subset$ F[a,b,c,d,..] working on a,b,c,d,… For explanation: an arrangementgroup is a set of arrangements such that if $\alpha$, $\beta$ and \gamma$$\inG and $\phi_{\alpha,\beta}$ is the substitution (a bijection on a,b,c,d,….) that transforms $\alpha$ to $\beta$ then $\phi_{\alpha,\beta}$ working on $\gamma$ is also $\in$ G. If G is an arrangementgroup you can make $\Pi$(G)=$\{\phi_{\alpha,\beta}|\alpha,\beta\in G\}$, the substitutiongroup of G. This is a group in the modern sense. On the left of the n$\times$m array are the corresponding values from Lemma II; important is that these are all different values. The values V, V${}^{{}^{\prime}}$, V${}^{{}^{\prime\prime}}$ … are all primitive elements for the field F[a,b,c,d,…], this is lemma III. You can make the arrangements in the rows of the n$\times$m array as follows; let every automorfism of the Galois group transform V from Lemma II. If H is a subgroup of the Galois group we will add a field to H using an arrangementgroup. The idea is this: let H be a subgroup of order k of the Galoisgroup with H=$\{h_{1},h_{2},..,h_{k}\}$, choose a random arrangement from the n rows and let $\alpha$ be the corresponding primitive element on the left. We add to H the field $\{P(h_{1}(\alpha),h_{2}(\alpha),...,h_{k}(\alpha))|$ P a symmetric polynomial $\in F[x_{1},x_{2},...,x_{k}]\}$. This field appears to be independent of the choice of the arrangement. If you let H operate on the chosen arrangement this will result in an arrangementgroup with the same corresponding primitive elements. If you have an arrangementgroup you can make the substitutiongroup; in this case you get H. But this mapping is not injective; if you read the example before proposition VI where Galois shows that the roots of the quartic are radical, the arrangementgroup of 12 arrangements is partitioned in 3 arrangementgroups of 4 elements. The substitutiongroups of these 3 arrangementgroups are all the four-group of Klein. In case of a fixed arrangement $\epsilon$ then you can prove that there is a one-to-one correspondence between the arrangementgroups containing $\epsilon$ and the subfields of the splitting field mentioned above. We will start from a subgroup of automorphisms in the Galois group and prove the main theorem of the Galois theory in a modern way. The story above is only to illustrate that is very useful to study the original writings of Galois. I am also convinced that Galois knows everything about symmetric polynomials333Read the very interesting article [5]; Galois gives in proposition I a characterization of F that refers strongly to the main theorem of Galois theory. We will prove this proposition at the end.

The proof.

 A Galois extension can be seen as a splitting field over a basic field F of a polynomial $\in$ F [x] with different roots. We assume that the integers form part of that basic field F in connection with lemma II of Galois. Let R=$\{r_{1},r_{2},...,r_{n}\}$ be the different roots of the aforementioned polynomial then F $\subset$ F[$r_{1},r_{2},...,r_{n}$] is a Galois extension. Lemma’s II and III of Galois guarantee that there are primitive elements. Choose a primitive element $\omega$ for the Galois extension of F $\subset$ F[R]. We define a mapping which to every subroup in the Galois group of F $\subset$ F[R] assigns a field in F[R]; we prove that this mapping is surjective and injective. We also prove that this field is exactly the field referred to in the main theorem proven by E. Artin. Proposition I of in the Mémoire is thereby proven. We prove the main theorem of the Galois theory using the main theorem of the symmetric polynomials as you will see.

**The definition af the field LH,ω.

** [0] Let H be a group and H $\subset$ Galois group of F $\subset$ F[R]; assume H = $\{\sigma_{1},\sigma_{2},...,\sigma_{m}\}$ and $\sigma_{i}$ an automorphism of F[R]. Assign to H the field LH,ω= $\{P(\sigma_{1}(\omega),\sigma_{2}(\omega),...,\sigma_{m}(\omega))|$ with P a symmetric polynomial $\in$ $F[x_{1},x_{2},...,x_{m}]\}$; F $\subset$ LH,ω because if x $\in$ F use P$(x_{1},x_{2},..,x_{m})$=x. If x $\notin$ F and x $in$ LH,ω use the minimum polynomial of x over F to prove that $\frac{1}{x}$ $\in$LH,ω.

**Independence of the choice of the primitive element.

** [1]: If $\theta$ is an other primitive element then LH,θ=LH,ω; $\theta$=g($\omega$) for some g $\in$ F[x]; Substituting $\theta$=g($\omega$) implies LH,θ$\subset$LH,ω. The other inclusion mutatis mutandis.

Surjectivity. [2]: If L is a subfield of F[R] and g is the minimal polynomial of $\omega$ over L we consider the field LG,ω with G the Galois group of L $\subset$ F[R]. Let G=$\{\tau_{1},\tau_{2},...,\tau_{n}\}$ with n equal to the degree of the extension L $\subset$ F[R]. The polynomial g(x)=$\prod_{i=1}^{n}(x-\tau_{i}(\omega))$ is in L[x] and in LG,ω[x]. Using the main theorem of symmetric polynomials and the coefficients of g$\in$L[x] we know that LG,ω$\subset$L. The minimal polynomial of $\omega$ over LG,ω divides g and so [F[R]:LG,ω] $\leq$ [F[R]:L]. Therefore also LG,ω=L and the surjectivity is proven.

Injectivity. [3]: [2] has been proven for every subfield of F[R], we apply [2] on L=LH,ω with H as in [0]. We are going to use the equality L=LH,ω=LG,ω with G as in [2]. Every automorphism $\in$ H is the identity on L=LH,ω, so H$\subset$G, the Galois group of L=LG,ω $\subset$ F[R]. Therefore m$\leq$n. Let f(x)=$\prod_{i=1}^{m}(x-\sigma_{i}(\omega))$ then f $\in$ L[x]=LH,ω[x]; the minimal polynomial g of $\omega$ over L=LG,ω in L[x] has degree n and divides f; consequently n$\leq$m and so n=m and H=G. The injectivity is proven; the assumption H1$\neq$H2 and L${}_{H_{1},\omega}$=L${}_{H_{2},\omega}$ leads to a contradiction because then H1=H2=the Galois group of (L${}_{H_{1},\omega}$=L${}_{H_{2},\omega}$) $\subset$ F[R].

The main theorem of the Galois theory follows: the mapping which to a subgroup H of the Galois group of F $\subset$F[R] assigns the field LH,ω is is a one-to-one correspondence between the subgroups in the Galois group of F $\subset$F[R] and the subfields in F[R]. [F[R]:LH,ω]=$|H|$ with H the Galois group of LH,ω $\subset$ F[R].

Yet to prove LH,ω=LH where LH=$\{x|\sigma(x)=x$ for all $\sigma\in H\}$. Let x $\in$ LH; there exists a polynomial h in F[x] with x=h($\omega$) and so x=$\sigma_{i}(x)=h(\sigma_{i}(\omega))$. Consider the symmetric polynomial P$(x_{1},x_{2},..,x_{m})$ = ($x_{1}+x_{2}...+x_{m}$)/m; x=P(x,x,..,x)= ($h(\sigma_{1}(\omega))+h(\sigma_{2}(\omega))+...+h(\sigma_{m}(\omega)))$/m and consequently x $\in$ LH,ω. That LH,ω $\subset$ LH is evident. This proves Proposition I in the Mémoire of Galois.

My email address is louis.dicker@ziggo.nl