3 ms·

The Abel–Ruffini theorem, that general quintics and higher have no solutions via radicals, feels to me like such a fundamental result, that there must be some i

The Abel–Ruffini theorem, that general quintics and higher have no solutions via radicals, feels to me like such a fundamental result, that there must be some intuitive explanation for it. I have tried to understand the topic at a level that shows me that intuition, but I have never quite reached it. Any suggestions?

go_elmo 5y ago
The proof reduces it to galois-groups and shows some easy properties, which are intuitive in the galois-group domain (only divisor of S_5 is A_5 which is not of prime-order). Maybe studying the relationship / reduction to / from polynomial to galois will help
jstx1 5y ago
It's apparently the topic of 3blue1brown's next video - https://twitter.com/3blue1brown/status/1420461376896520193 https://twitter.com/3blue1brown/status/1420461376896520193
gmadsen 5y ago
I think the intuition only comes from understanding the arguments intuitively. so basically an entire algebra class, you need to understand field extensions and PIDs.
kortex 5y ago
Gonna butcher this, but it's good practice, so I'm gonna take a stab at it. The roots of polynomials are intimately related to the Alternating symmetry groups. These describe the ways you can permute a set, specifially A_n describes the even permutations of a set of n elements. The even permutations of a sequence are those obtained by making an even number of transpositions ("moves"). I can make 2 swaps to go 1,2,3,4,5 -> 1,4,3,2,5 -> 1,4,5,2,3. That's an even permutation. The order (size) of A_n is the number of even permutations of n unique elements. The orders are as follows: A2: 1 A3: 3 A4: 12 A5: 60 A6: 360 A_n: n! / 2 The "solvability" of a polynomial depends on the the degrees of freedom and number of constraints. Quintic is where the number of degrees of freedom blows up and you can no longer factor out roots in such a way that you can uniquely factor the polynomial. Quadratic is basically trivial. Cubic cannot be directly factored, because you have 3 variables and 2 equations, but you can "factor out" a quadratic that you can solve, and then solve the cubic. Quartic is the same, you end up with 4 variables and 3 unique equations, you can factor out a cubic, which you can solve, etc. Quartic, you have 4 equations to solve and...something like 30 variables (or maybe it's 60. It's some way more than we can factor!). https://en.wikipedia.org/wiki/Alternating_group https://en.wikipedia.org/wiki/Alternating_group https://math.stackexchange.com/questions/550401/intuitive-reasoning-why-are-quintics-unsolvable https://math.stackexchange.com/questions/550401/intuitive-re...
- paulpauper 5y ago
  Sorta. There are obviously infinite quintics that can be solved despite this. For example any polynomial in which the roots are arranged on a circle on the real,complex plane
anon_tor_12345 5y ago
arnold's proof is the most intuitive and what made it click after years of being responsible for understanding it (several years of algebra at the undergrad and grad level): https://www.youtube.com/watch?v=zeRXVL6qPk4 https://www.youtube.com/watch?v=zeRXVL6qPk4
- mjreacher 5y ago
  Is there any difference between that video and this one?: https://www.youtube.com/watch?v=RhpVSV6iCko https://www.youtube.com/watch?v=RhpVSV6iCko seems to be the same author but that video is 10 minutes longer.
  - anon_tor_12345 5y ago
    there is some slight difference but i don't remember what it is. i wrote a blog post about abel-ruffini based on those two videos and i remember there being some things that were clearer in one rather than the other (but i don't remember which!)
paulpauper 5y ago
The original proofs are quite long. Not exacyly something you can compress to a comment or an article. I dunno
kevinventullo 5y ago
So first, consider nth roots. These correspond to particularly simple field extensions. If I take two choices of nth root, their ratio is always an nth root of unity. This more or less forces the symmetry group to be a subgroup of the nth roots of unity, i.e. Z/n. Now if you have a general expression built out of extracting nth roots for various n, that might generate some big complicated field extension. But it corresponds to a tower of field extensions which are all just extracting nth roots. Thus the intermediate symmetry groups are still all subsets of Z/n for various n. The Galois correspondence tells you that the symmetry group of the entire extension, call it G, has a corresponding decomposition: there is an increasing sequence of subgroups 1 < G_1 < … < G_k = G such that each intermediate quotient group G_{i+1}/G_i is a subgroup of Z/n for some n. Groups with such a decomposition are called solvable. Insolvability of the quintic then boils down to this: the symmetry group of any degree k polynomial must be a subgroup of S_k. For k <= 4, S_k is solvable; for k >= 5, it is not. This last sentence is the closest thing I can offer to intuition. It is a very finite and tangible fact that one can write down and computationally verify for themselves. Like, you can write down the corresponding solvable decomposition of S_4, and you can see that S_5 is not solvable by verifying that A_5 is simple which itself boils down to a small number of cases.

go_elmo 5y ago

The proof reduces it to galois-groups and shows some easy properties, which are intuitive in the galois-group domain (only divisor of S_5 is A_5 which is not of prime-order). Maybe studying the relationship / reduction to / from polynomial to galois will help

jstx1 5y ago

It's apparently the topic of 3blue1brown's next video - https://twitter.com/3blue1brown/status/1420461376896520193 https://twitter.com/3blue1brown/status/1420461376896520193

gmadsen 5y ago

I think the intuition only comes from understanding the arguments intuitively. so basically an entire algebra class, you need to understand field extensions and PIDs.

kortex 5y ago

Gonna butcher this, but it's good practice, so I'm gonna take a stab at it. The roots of polynomials are intimately related to the Alternating symmetry groups. These describe the ways you can permute a set, specifially A_n describes the even permutations of a set of n elements. The even permutations of a sequence are those obtained by making an even number of transpositions ("moves"). I can make 2 swaps to go 1,2,3,4,5 -> 1,4,3,2,5 -> 1,4,5,2,3. That's an even permutation. The order (size) of A_n is the number of even permutations of n unique elements. The orders are as follows: A2: 1 A3: 3 A4: 12 A5: 60 A6: 360 A_n: n! / 2 The "solvability" of a polynomial depends on the the degrees of freedom and number of constraints. Quintic is where the number of degrees of freedom blows up and you can no longer factor out roots in such a way that you can uniquely factor the polynomial. Quadratic is basically trivial. Cubic cannot be directly factored, because you have 3 variables and 2 equations, but you can "factor out" a quadratic that you can solve, and then solve the cubic. Quartic is the same, you end up with 4 variables and 3 unique equations, you can factor out a cubic, which you can solve, etc. Quartic, you have 4 equations to solve and...something like 30 variables (or maybe it's 60. It's some way more than we can factor!). https://en.wikipedia.org/wiki/Alternating_group https://en.wikipedia.org/wiki/Alternating_group https://math.stackexchange.com/questions/550401/intuitive-reasoning-why-are-quintics-unsolvable https://math.stackexchange.com/questions/550401/intuitive-re...

paulpauper 5y ago
Sorta. There are obviously infinite quintics that can be solved despite this. For example any polynomial in which the roots are arranged on a circle on the real,complex plane

anon_tor_12345 5y ago

arnold's proof is the most intuitive and what made it click after years of being responsible for understanding it (several years of algebra at the undergrad and grad level): https://www.youtube.com/watch?v=zeRXVL6qPk4 https://www.youtube.com/watch?v=zeRXVL6qPk4

mjreacher 5y ago
Is there any difference between that video and this one?: https://www.youtube.com/watch?v=RhpVSV6iCko https://www.youtube.com/watch?v=RhpVSV6iCko seems to be the same author but that video is 10 minutes longer.
- anon_tor_12345 5y ago
  there is some slight difference but i don't remember what it is. i wrote a blog post about abel-ruffini based on those two videos and i remember there being some things that were clearer in one rather than the other (but i don't remember which!)

paulpauper 5y ago

The original proofs are quite long. Not exacyly something you can compress to a comment or an article. I dunno

kevinventullo 5y ago

So first, consider nth roots. These correspond to particularly simple field extensions. If I take two choices of nth root, their ratio is always an nth root of unity. This more or less forces the symmetry group to be a subgroup of the nth roots of unity, i.e. Z/n. Now if you have a general expression built out of extracting nth roots for various n, that might generate some big complicated field extension. But it corresponds to a tower of field extensions which are all just extracting nth roots. Thus the intermediate symmetry groups are still all subsets of Z/n for various n. The Galois correspondence tells you that the symmetry group of the entire extension, call it G, has a corresponding decomposition: there is an increasing sequence of subgroups 1 < G_1 < … < G_k = G such that each intermediate quotient group G_{i+1}/G_i is a subgroup of Z/n for some n. Groups with such a decomposition are called solvable. Insolvability of the quintic then boils down to this: the symmetry group of any degree k polynomial must be a subgroup of S_k. For k <= 4, S_k is solvable; for k >= 5, it is not. This last sentence is the closest thing I can offer to intuition. It is a very finite and tangible fact that one can write down and computationally verify for themselves. Like, you can write down the corresponding solvable decomposition of S_4, and you can see that S_5 is not solvable by verifying that A_5 is simple which itself boils down to a small number of cases.