Friday, October 02, 2009

Galois' Memoir: Lemma 3

The following is taken from the translation of Galois' Memoir by Harold M. Edwards found in his book Galois Theory. The proof itself is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

Lemma 3:

When a function V is chosen as satisfies Lemma 2 (see here), it will have the property that all the roots of the given equation can be expressed as rational functions of V.

That is:

Let P be a function of degree n with the roots: r1, ..., rn where each root is distinct.

Let V(r1, ..., rn) be the Galois resolvent where each distinct permutation has a distinct value.

Then for all roots ri ∈ F(V).


(1) Let V be a Galois Resolvent [see Definition 2, here]

(2) Let r1 be any root of an equation P

Since r1 is any root, the proof is done if we can show that ri ∈ F(V).

(3) Let:

g(x1, ..., xn) = ∏ (for each permutation σ) [V - f(x1, σ(x2), ..., σ(xn)) ] ∈ F(V)[x1, ..., xn]

where σ runs over all permutations of x2, ..., xn

(4) Since g is symmetric in x2, ..., xn, g can be written as a polynomial in x1 and the elementary polynomials s1, ..., sn-1. (see Lemma, here)

(5) Let:

g(x1, x2, ..., xn) = h(x1, ..., sn-1)

for some polynomial h with coefficients in F(V).

(6) Substituting in various ways the roots r1, ..., rn of P for the indeterminates x1, ..., xn which has the effect of substituting a1, ..., an-1 ∈ F for s1, ..., sn-1 [see Theorem 1, here, for details on the mapping between elementary symmetric polynomials and the coefficents of a polynomial], we obtain:

g(r1, r2, ..., rn) = h(r1, a1, ..., an-1)

and generalizing this, we get:

g(ri, r1, ...., ri-1, ri+1, ..., rn) = h(ri, a1, ..., an-1)

(7) Since V is a Galois Resolvent for a function f such that V = f(r1, ..., rn), we have:

V ≠ f(ri, σ(r1), σ(r2), ..., σ(ri-1), σ(ri+1), ..., σ(rn))

for i ≠ 1 and for any permutation σ of { r1, ..., ri-1, ri+1, ..., rn }. [see Lemma 2, here]

(8) Therefore:

g(ri, r1, r2, ..., ri-1, ri+1, ...., rn) ≠ 0 for i ≠ 1.

(9) On the other hand, the definition of g and V show that [see definition of g in step #3 above]:

g(r1, ..., rn) = 0

(10) From step #9 above and step #6 above, we know that:

h(X,a1, ..., an-1) ∈ F(V)[X]

vanishes for X = r1 but not for X = ri with i ≠ 1.

(11) Therefore, it is divisible by X - r1 but not by X - ri for i ≠ 1. [This follows from Girard's Theorem, see here]

(12) We then consider the monic greatest common divisor D(X) of P(X) and h(X,a1, ..., an-1) in F(V)[X]. [see Theorem 1, here for proof of the existence of a greatest common divisor for polynomials]

(13) Since P(X) = (X - r1)*...*(X - rn), we know that X - r1 divides P(X).

(14) From step #11 above, we know that X - r1 divides h.

(15) Since x - r1 divides both P(X) and h(X,a1, ..., an-1), it divides D.

(15) On the other hand, h(X,a1, ..., an-1) is not divisible by X - ri for i ≠ 1. [see step #10 above]

(16) Hence, D has no other factor than X - r1.

(17) Thus, D = X - r1 whence r1 ∈ F(V) since D ∈ F(V)[X].



Monday, September 28, 2009

Galois' Memoir: Lemma 2 (Galois Resolvent)

The following is taken from the translation of Galois' Memoir by Harold M. Edwards found in his book Galois Theory. The proof itself is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

Definition 1: Galois Resolvent Function

For any equation f(x) with distinct roots, the Galois Resolvent Function is a function g(x1, ..., xn) of the roots that no matter how the roots are permuted on the function, no two of the values are equal.

Definition 2: Galois Resolvent

The Galois Resolvent is a value of the Galois Resolvent Function where the roots of the equation f(x) are passed in as parameters.

Lemma 2: Galois Resolvent Function Exists

Given any equation f(x) with distinct roots a,b,c,... one can always form a function V of the roots such that no two of the values one obtains by permuting the roots in this function are equal.

For example, one can take:

V = Aa + Bb + Cc + ...

A, B, C, ... being suitably chosen whole numbers.


(1) Let the n distinct roots of f(x) be denoted a,b,c, ...

(2) Since these roots are distinct, the discriminant (a - b)2(a - c)2(b - c)2*... = D is not zero [For review of the discriminant, see here].

(3) What needs to be shown is that n integers A,B,C, ... can be chosen so that the n! numbers AS(a) + BS(b) + CS(c) + ... + are distinct where S ranges over all n! permutations of the roots a,b,c,...

[for details on why count(n permutations) = n!, see Corollary 1.1, here]

(4) Let P be the product of the squares of the differences of these n! numbers that is:

P = ∏ (S,T) [ A(S(a) - T(a)) + B(S(b) - T(b)) + ... ]2

where the product is all over n!(n! - 1)/2 pairs (unordered) of permutations S and T in which S≠ T.

Note: The purpose of this equation is to verify that all n! numbers are distinct. P is the product of all possible differences between any two permutations.

We know that there are n! possible permutations (see step #3 above).

We pick one of these permutations (1 out of n!) and call it S. Then, we pick a second one (1 out of n! - 1) and call it T. Since ordering doesn't matter and there are two ways to pick the same combination, we only need to deal with n!*(n!-1)/2 comparisons between S and T.

(5) To complete the proof, we only need to show that we can pick A,B,C, ... etc. such that P is nonzero.

If any of the permutations are not distinct, then the difference between S and T will be 0. If any of the differences are 0, then P = 0. So if P ≠ 0, it follows that we have found values for A,B,C,... such that all permutations are distinct.

(6) Let A, B, C, ..., be regarded at first as variables.

(7) Then P is a polynomial in variables A, B, C,... whose coefficients are polynomials in the roots a,b,c,...

(8) P is symmetric in the roots (this follows directly from the definition of P and the definition of symmetric polynomials, see Definition 1, here).

(9) Since P is symmetric, the coefficients are symmetric in the roots so using Waring's Method (see Theorem 4, here), P is a polynomial in A, B, C, ... with the coefficients symmetric in roots.

(10) So, we can determine the coefficients of P based on the elementary symmetric polynomials using the roots [see Theorem 1, here].

(11) We can therefore assume that the coefficients are known since we are assuming that the roots are known.

(12) Since P is a product of nonzero polynomials, it is nonzero. [see Theorem, here]

(13) Therefore once can assign integer values to A,B,C,... in such a way as to make P ≠ 0 [see Theorem, here].



Galois' Memoir: Lemma 1

The following is taken from the translation of Galois' Memoir by Harold M. Edwards found in his book Galois Theory. The proof itself is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

Lemma 1: An irreducible equation g(x) cannot have a root in common with a rational equation h(x) without dividing it.


(1) Let g(x) be a polynomial with coefficients in a given field K that is irreducible over K.

(2) Let h(x) be a polynomial with coefficients in the field K.

(3) Let r be a root for both h(x),g(x) so that h(r)=0 and g(r)=0

(4) Assume that g(x) does not divide h(x).

(5) Then g(x),h(x) are relatively prime since g(x) is irreducible [see Lemma 1, here]

(6) Then (see Corollary 3.1, here), there exists polynomials A(x), B(x) such that A, B all have coefficients in K and:

1 = A(x)g(x) + B(x)h(x)

(7) Since g(r)=0 and h(r)=0, it follows that:

1 = A(r)*0 + B(r)*0

which is impossible.

(8) So we have a contradiction and we reject our assumption in step #4.



Sunday, September 27, 2009

Galois' Memoir on the Solvability of Equations

Evariste Galois died in 1832 when he was just 20. He had made many attempts to gain attention to his theory of equations but each time had failed.

In 1829 when he was 17, he presented his findings on the the solvability of equations to the Paris Academy. Augustin-Louis Cauchy was appointed referee. There is a popular legend that Cauchy did not appreciate the work or somehow lost it but this does not seem to be the case. It is believed that Cauchy presented detailed comments to Galois and suggested that he resubmit his work. For more information on this, see this article. It was at this time that tragedy struck and Galois' father committed suicide. Galois' submission of his work was delayed.

It is believed that being given encouragement from Cauchy, he rewrote his memoir. He resubmitted his memoir to the Paris Academy in 1830. Jean-Baptiste Joseph Fourier was appointed referee. Unfortunately, Fourier's health took a turn for the worse and he died a few weeks after being appointed referee. No assessment was made and the paper was lost among Fourier's other papers.

In 1831, Galois was asked by Simeon Poisson to resubmit his memoir. After reviewing, Poisson rejected the paper as not fully developed. It is interested to note that Poisson was quite impressed by the quality of work but was not able to verify that they were correct.

Here is what Poisson wrote (see reference below for source):
We have made ever effort to comprehend M. Galois's proof. His arguments are neither sufficiently clear nor developed for us to judge their rigor, and we are not in a position to even give an idea of them in this report...

The author claims that the proposition which is the subject of his memoir is part of a general theory rich in application. Often, different parts of a theory are mutually clarifying, and it is easier to understand them together than in isolation. One should rather wait for the author to publish his work in entirety before forming a definite opinion.

Galois took all this very hard as can be imagined. There is another myth at this point that on the night before Galois' duel where he would die, he wrote up his theory in a single night. While he did write a letter to his friend about the nature of his work, it is clear that he had been working on the ideas since he was 17 and the fullest expression of them up to this point had been the paper that was rejected by Poisson.

Galois would die on May 30, 1832. His funeral was on June 2. His good friend Auguste Chevalier and his brother Alfred collected all his papers in the effort to get notice for his work. They were submitted to the most famous mathematicians of the day: Carl Gauss, Jacobi, and others. By 1843, they had made their way to Joseph Liouville.

Finally, they got the attention they deserved when Liouville published Galois's memoir with additional comments in 1846.

In the next set of blogs, I will review the famous memoir by Galois that had been rejected by Poisson and which was modified the night before his death. For an English translation of Galois' memoir, see Harold Edwards' Galois Theory.