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I'm having difficulty with section 6 of Lagrange's Réfléxions sur la résolution algébrique des équations. That's page 11 of the paper, 215 of his Oeuvres. Annoyingly, this is one of the most critical points of the paper, since it's where he mentions the connection to permutations for the first time. This very sentence that I'm struggling with is essentially the birthplace of Galois theory.

Lagrange begins with a general equation of degree 3 having roots a, b, and c. He has already derived an auxiliary equation whose root y can be used to express a, b and c, and now he inverts these expressions to obtain:

y = (a + α b + β c)/3

Where α and β are the primitive cube roots of 1. Firstly I don't understand why Lagrange, right before section 6, says that we can "exchange α and β" (I'm frankly not even sure what that assertion means), but my bigger problem is section 6, where he now claims that (my translation)

As the auxiliary [réduite] equation does not depend immediately on the roots a, b, c of the original equation, but only on its coefficeints m, n, p into which the three roots enter equally [également - see below], it is clear that in the expression of y we muts be able to exchange at will the quantities a, b and c, thus the quantity y should have as many different values as can be formed by all of the permutations of a, b and c, which we know from the theory of combinations to be 3 x 2 x 1...

What Lagrange seems to be claiming here is that if we permute the values a, b and c in the expression of y, the resulting value will always be another root of the auxiliary equation. If this is what's being claimed, then I can't see how to prove it. Lagrange only mentions the fact that the auxiliary equation only depends on the coefficients of the original equation, into which the roots a, b, c "enter equally". I suspect by "equally", he means "symmetrically" , in which case he may be claiming that the coefficients of the auxiliary equation are symmetric functions of the values a, b and c.

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    $\begingroup$ I have tried asking questions about this paper on math.SE but got no answers, I think HSM is better because there's a higher chance of finding people who have studied the paper already or would be interested in reading it in order to answer the question. $\endgroup$ – Jack M Nov 15 '14 at 1:46
  • $\begingroup$ I think it's clear he means it is symmetric under exchange of $a,b,c$. This should be straightforward to prove. Perhaps you should give the auxiliary equation in terms of $m,n,p$, There is a chance I could tell you explicitly how to see it. $\endgroup$ – Danu Nov 15 '14 at 11:51
  • $\begingroup$ @Danu If $m, n, p$ are the coefficients of the orginal equation, the auxiliary equation is $y^3+p'y^3-n'^3/27=0$, where $n'$ and $p'$ are polynomials in $m, n, p$. $\endgroup$ – Jack M Nov 15 '14 at 12:00
  • $\begingroup$ What'd $p'$? Perhaps we should continue this discussion in chat. $\endgroup$ – Danu Nov 15 '14 at 12:11
  • $\begingroup$ This question appears to be off-topic because it is about a mathematical detail from a historically important treatise on mathematics, rather than the history of science or mathematics. $\endgroup$ – Danu Nov 15 '14 at 15:17
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I will try to answer this both in 18th century terms, and in modern dress.

First off, the réduite equation was introduced on p.208 and named on p.213; it is

$$y^6 + py^3 - n^3/27 = 0$$

The coefficients of this equation are $p$ and $-n^3/27$, thus depending only on the coefficients $n,p$ of the original equation $x^3+nx+p=0$. (Lagrange briefly allows an $mx^2$ term, then transforms it away, resulting in primed coefficients; let's ignore that complication.)

Now $n,p$ can be expressed in terms of the roots of the original equation using Vieta's formulas: just expand out

$$(x-a)(x-b)(x-c) = 0$$

expressing $n,p$ as so-called elementary symmetric functions of $a,b,c$.

Since $n,p$ are symmetric functions of $a,b,c$, it follows that the coefficients of the réduite are too. Thus, $-n^3/27$ will be some messy expression in $a,b,c$, but clearly it will be symmetric in those roots.

Lagrange has shown that $y=(a + \alpha b + \beta c)/3$, so everything in sight can be written as a rational function of the roots $a,b,c$. (We regard the cube roots of unity as known constants). Imagine what we get if we substitute these functions for $y,n,p$ in the réduite: a big messy equation in rational functions of $a,b,c$. If we now permute $a,b,c$, we will get another such equation. But the coefficients of the réduite are symmetric in $a,b,c$, so they don't change. Just $y$ changes, to some new value. Six possible permutations, six values of $y$, all roots of the same equation. Thus the réduite must have degree 6.

Now let's translate this into modern terms. Lagrange works with two sorts of expressions: rational expressions in the coefficients $n,p$ (or $m,n,p$ if we don't remove the $x^2$ term), and rational expressions in the roots $a,b,c$. Nowadays we like to think set-theoretically, so let's say $E$ is the field of all rational expressions in three letters $a,b,c$. Let $F$ be the subfield generated by the elementary symmetric polynomials $s_1=a+b+c$, $s_2=ab+ac+bc$, $s_3=abc$. Then the polynomial equation

$$f(X) = (X-a)(X-b)(X-c) = X^3 -s_1 X^2 + s_2 X - s_3 = 0$$

has coefficients in $F$ and roots in $E$. It's not hard, with a modicum of field theory, to show that $E$ is the splitting field of $f(X)$ over $F$. Furthermore, the automorphism group of $E$ over $F$ is just the permutation group on the letters $a,b,c$. The value $y$ has this property: applying the six automorphism to it yields six different values. A standard trick in Galois theory is to pick an element $y$ of the splitting field and form the equation

$$g(X) = \prod_{\sigma\in\text{Aut}(E/F)} (X-\sigma y) = 0$$

The coefficents of $g(X)$ then lie in the fixed field. That's exactly what Lagrange is doing here.

In a comment, the OP raises an issue touching on the differences between the modern and 18th century viewpoints. Let $g(Y)$ be the réduite, where $Y$ is a variable; let $y$ be a rational expression in $a,b,c$ for which $g(y)=0$. Now we permute the roots. The coefficients of $g$ don't change, but $y$ becomes $y'$. How do we know that $g(y')=0$?

The modern answer: permuting the roots induces an automorphism of the field $E$ over $F$, call it $\sigma$. So $\sigma g(y)=g(\sigma y)$ because $\sigma$ is an automorphism leaving all the coefficients of $g$ fixed. So $g(y')=g(\sigma y)=\sigma g(y)=\sigma 0=0$.

Lagrange had no explicit notion of the fields $E$ or $F$, still less of an automorphism of $E$ over $F$. But his remarks about the roots entering equally into the coefficients, and being able to exchange the roots at will, show he had an intuitive grasp of the modern concepts.

Lagrange probably pictured $g(y)=0$ as a big messy rational expression on the left, boiling down formally to 0. Since the equation is formal, not depending on the particular values of $a,b,c$, permuting the roots will not destroy the validity of the formal identity.

Finally, I recommend two articles for help with the pre-history of Galois theory (beside Cox's book): "The Development of Galois Theory from Lagrange to Artin", by B. M. Kiernan (Archive for History of Exact Sciences, v.8 no. 1/2 p. 40-154; p.45-55 discuss Lagrange's work), and "Niels Henrik Abel and the theory of equations" by H. K. Sørensen. (I've also heard good things about Edwards' book Galois Theory.)

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  • $\begingroup$ I don't follow - what exactly is the connection between what you describe and the passage from Lagrange? For example, what in your notation corresponds to Lagrange's $y$, $x$, and his équation réduite? $\endgroup$ – Jack M Nov 15 '14 at 22:47
  • $\begingroup$ @JackM I rewrote the answer, hopefully to address your question more directly $\endgroup$ – Michael Weiss Nov 18 '14 at 0:30
  • $\begingroup$ This is much clearer, however the crucial point still eludes me. In the paragraph beginning "Lagrange has shown that...", you imagine substituting the expressions for $y$, $n$ and $p$ in terms of $a, b, c$ into the réduite, $F(y)=0$. As you say, permuting $a, b, c$ only changes $y$, not the coefficients, so if we write $y'$ for a value obtained by interchanging $a, b, c$ in the expression for $y$, the left hand side of $F(y)=0$ becomes $F(y')$. But now the central question: how can you guarantee that the right hand side is still $0$? $\endgroup$ – Jack M Nov 18 '14 at 0:57
  • $\begingroup$ @JackM OK, see the new paragraphs. $\endgroup$ – Michael Weiss Nov 18 '14 at 14:38

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