# Évariste Galois and His Theory

• Insights
Mentor
2021 Award
Galois died in a duel at the age of twenty. Yet, he gave us what we now call Galois theory. It decides all three ancient classical problems, squaring the circle, doubling the cube, and partitioning angles into three equal parts, all with compass and ruler alone. Galois theory also tells us that there is no general formula to solve the integer equation
$$x^5+a_1x^4+a_2x^3+a_3x^2+a_4x+a_5=0$$

#### Attachments

• Galois_1.jpg
18.5 KB · Views: 9
Last edited:
Klystron, ohwilleke and Greg Bernhardt

Mentor
Great article @fresh_42!

Thanks for taking the time to write it and share with everyone on PF and beyond.

A brief note on origami and cubic equations:
While the classical trisecting an angle and doubling a cube were deemed impossible with only straight edge and compass. They were solvable via origami. Sadly the ancient mathematicians didn't know the art.

https://plus.maths.org/content/power-origami

Greg Bernhardt and fresh_42
swampwiz
Sorry, I still don't get this Galois-ian explanation.

BTW, I have figured out a different way that sloppily proves the unsolvability, and have posted it in one of the forums here. It uses the fact that a free-term polynomial has solutions that are the [ n ( n - 1 ) ]-th of the discriminant, and so any formulae for the solutions must have the discriminant tucked away somewhere - i.e., because if there is a continuous function solution, a small perturbation of the coefficients for a free-term polynomial must be of the form that uses a small perturbation of the discriminant. Then add in the fact that a general polynomial can be made monic, and then be substituted with an linear offset term that zeroes out the ( n - 1 )-th coefficient, leaving ( n - 1 ) # of coefficients, and so any continuous function would need to be nested terms of the sum of some value and a root of the deeper-nested sum, with each nesting removing a coefficient degree-of-freedom. The discriminant comes into play because the net root of it must be [ n ( n - 1 ) ].

For n = 2, the discriminant must be within a 2-root (i.e., quadratic formula)

For n = 3, the discriminant must be within a 6-root made up of the previous 2-root and a new 3-root (which this article shows)

For n = 4, the discriminant must be within a 12-root made up of the previous 6-root and a new 2-root (which it is for the quartic formula)

for n = 5, the discriminant must be within a 20-root made up of the previous 12-root and a new ... er ... 5/3 root ... OOPS, not possible!

Mentor
2021 Award
This doesn't make any sense. Whatever you mean by a 5/3 root, ##a^{p/q}=\sqrt[q]{a^p}## which is a legit expression within any allowed solution. Nevertheless, general polynomials from degree 5 or higher cannot be reduced to such expressions.

swampwiz
This doesn't make any sense. Whatever you mean by a 5/3 root, ##a^{p/q}=\sqrt[q]{a^p}## which is a legit expression within any allowed solution. Nevertheless, general polynomials from degree 5 or higher cannot be reduced to such expressions.
I said it was sloppy. The root operations must be as per integers. The quadratic, cubic & quartic formulae all correspond to this.

Maybe I shall look over my notes and make this case better. I should say that I think that I completely understand the concept of Lagrangian resolvents.

Last edited:
Mentor
2021 Award
Maybe I shall look over my notes and make this case better.
I think it would be more promising to study the theory than to save what already sounds lost.

Mentor
2021 Award
I still don't get this Galois-ian explanation.
For any solution ##x_0## to ##x^5+a_1x^4 +a_2x^3+a_3x^2+a_4x+ a_0=0## there belongs an extension of the rational numbers ##\mathbb{Q}+x_0\cdot \mathbb{Q}+x_0^2\cdot \mathbb{Q}+x_0^3\cdot \mathbb{Q}+x_0^4\cdot \mathbb{Q}.## Every such extension has a group of functions, that do not affect ##\mathbb{Q},## but map ##x_0## to another solution ##x_1.##

Whenever we can solve such an equation with ##+\, , \,- \, , \, : \, , \, \cdot \, , \,\sqrt[5]{\, . \,}## or other roots, then we get certain groups of those functions. And from five on, they simply do not exist anymore.

Sorry, but it can get shorter and simpler than that.

swampwiz
I think it would be more promising to study the theory than to save what already sounds lost.
Uh, I thought the whole idea of the article was a quickie discussion of Galois Theory.

For someone that is not super-refined in mathematics, I think I can follow stuff as about as anyone else, especially on the topic of the impossibility of the quintic, which has been a bit of a bit of an obsession of mine once I had discovered that there were formulae for the cubic & quartic.

My understanding is that to properly understand the theory, one would need to master undergraduate abstract algebra first, then tackle Galois theory. The author of this article seemed to be trying to do a super quickie discussion on it, and I was just pointing out that it was a little TOO quick.

Last edited:
swampwiz
For any solution ##x_0## to ##x^5+a_1x^4 +a_2x^3+a_3x^2+a_4x+ a_0=0## there belongs an extension of the rational numbers ##\mathbb{Q}+x_0\cdot \mathbb{Q}+x_0^2\cdot \mathbb{Q}+x_0^3\cdot \mathbb{Q}+x_0^4\cdot \mathbb{Q}.## Every such extension has a group of functions, that do not affect ##\mathbb{Q},## but map ##x_0## to another solution ##x_1.##

Whenever we can solve such an equation with ##+\, , \,- \, , \, : \, , \, \cdot \, , \,\sqrt[5]{\, . \,}## or other roots, then we get certain groups of those functions. And from five on, they simply do not exist anymore.

Sorry, but it can get shorter and simpler than that.
Is this mapping you are talking about the same thing as Lagrange and his resolvent method? I understand how that method works, but that method doesn't prove that something is impossible.

Mentor
2021 Award
Is this mapping you are talking about the same thing as Lagrange and his resolvent method? I understand how that method works, but that method doesn't prove that something is impossible.
I don't think so.

Those mappings are permutations of solutions, just like conjugation maps ## i ## to ## -i ## with complex numbers. Now permutation groups with ##2, 3## or ##4## elements are all commutative. In order for higher order solutions, we would also need commutative groups of those mappings because there is no specific order among the solutions. But form ##5## on, permutation groups are no longer commutative. This makes it impossible. The formal proof, of why we need commutative mappings, and the proof that there are no such groups, however, is where mathematics starts. That needs more preliminaries and preparations to show.