Proof: general degree-5 or more polynomials not solvable with nth roots

[lpetrich]

"Galois-Free Guarantee! | The Insolubility of the Quintic" -- avoiding Galois theory and using only some properties of complex numbers, crediting mathematician Vladimir Arnold. But it does have some Galois-theory connections.

The quadratic formula is well-known: the solution of a general degree-2 polynomial requires a square root. Less well-known are solutions for cubic and quartic polynomials, degrees 3 and 4. The cubic one requires a cube root of a quadratic solution, and the quartic one a square root of a cubic solution.

But no further with addition, subtraction, multiplication, division, and nth roots.

At 04:11 the presenter has a proof of the Fundamental Theorem of Algebra, as it's called, that a degree-n polynomial in complex numbers has at n complex roots. The proof has two parts: showing that at least one complex root always exists, and using it to show that the remaining (n-1) complex roots also exist.

Consider polynomial
\( \displaystyle{ P(z) = \sum_{k=0}^{n} a_k z^k } \)

Make the independent variable z make a circle in the complex plane: \( z = z_0 + r e^{i\theta} \) for radius r and angle θ from 0 to 2π. For large r, \( P(z) \simeq a_n z^n \simeq a_n r^n e^{n i\theta} \) and it makes n loops around the origin. Shrinking r to 0, one arrives at the coordinate origin, the curve has to go through the origin, or both. That means that the polynomial has at least one root, which I will call z₁. P(z) must be divisible by (z - z₁):

\( P(z) = (z - z_1) P'(z) \)

where P'(z) has degree (n-1). Continue further, and one finds that

\( \displaystyle{ P(z) = a_n \prod_{k=1}^{n} (z - z_k) } \)

This is an existence proof as opposed to a proof by construction: it shows that the roots exist without showing how to find them. It also does not say whether or not any roots coincide, or how many sets of coincident roots that there are.

 

[lpetrich]

The presenter noted that one can interchange solutions and come up with the same polynomial. Polynomial roots have no natural order, and they can be interchanged arbitrarily.

He starts off with quadratic equations, using a simple case. I've simplified it a bit: \( z^2 = e^{i\theta} \) It is solved by two values of z, \( z_1 = e^{i\theta/2} ,\ z_2 = - e^{i\theta/2} \)

He then rotates θ in a circle, and when that is done, the two roots are interchanged: z₁ -> z₂ and z₂ -> z₁.

That means that there is no general solution for a quadratic equation using only arithmetic operations: addition, subtraction, multiplication, and division. These operations are single-valued, so they cannot give multiple solution values.

 

[lpetrich]

He then considers the effect of swapping pairs of roots, effects on arithmetic combinations of the polynomial coefficients. What phase change for going around some point?

• 1 -> 2, 2 -> 1 gives 2nπ
• 2 -> 3, 3 -> 2 gives 2mπ
• 2 -> 1, 1 -> 2 (first one in reverse) gives - 2nπ
• 3 -> 2, 2 -> 3 (second one in reverse) gives - 2mπ

Total phase change = 0. The first permutation is 213 and the second one 132. Applying the second one to the first one gives 231, the inverse of the first one gives 321, and the inverse of the second one gives 312. This is a rotation of the three points.

This is the "commutator" of permutations 213 and 132, and the presenter noted that some Rubik's-cube solution moves are essentially commutators: comm(A,B) = (inv-B).(inv-A).B.A for tfms A and B and their inverses.

comm(213,132) = 312, and not the identity permutation 123.

This means that general formulas for cubic or higher-degree polynomials cannot contain only one nth root.

 

[lpetrich]

But the cubic formula has nested roots. How does one handle them?

Taking commutators of root interchanges will restore the innermost nth root but not the ones that contain it.

So one has to repeat the commutator operation to get the next nth root out until one reaches the outermost one.

For linear equations, it is easy. There is only one length-1 permutation, 1, the identity permutation.

For quadratic equations, the group of permutations contains 12 21. The commutator of every pair of them is 12, the identity permutation.

For cubic equations, that group has 123 132 213 231 312 321. The commutator group for it has 123 231 312, and that group's commutator has 123, the identity permutation.

Permutations can be expressed as combinations of cyclic permutations, and I note the set of all those with cycle lengths n1 n2 ... as (n1 n2 ...).

Thus, length-1 permutations are 1: (1), length-2 ones are 2: (11) (2), with commutator 1: (11), and length-3 ones are 6: (111) (12) (3), with commutator sequence 3: (111) (3) and 1: (111).

Length-4 ones are 24: (1111) (112) (13) (22) (4) with commutator sequence 12: (1111) (13) (22) and 4: (1111) (22) and 1: (1111).

 

[lpetrich]

There is an additional curiosity that the presenter didn't go into: "quotient groups". I will illustrate with length-3 permutations, since it is nontrivial without being too small.

The set of length-3 permutations S is 123 231 312 132 213 321 -- (111) has 123, (3) has 231 312, (12) has 132 213 321

Take the commutator group of it. Its set S0 is 123 231 312 -- (111), (3). Multiply every element of S0 by some element of S. one gets either S0 or another set, S1: 132 213 321 -- (12). If one multiplies elements from S0 or S1, one gets members of S0 or S1, according to this table:

S0*S0 = S0, S0*S1 = S1, S1*S0 = S1, and S1*S1 = S0

This is equivalent to doing addition modulo 2, the group Z2. So the quotient group of S and S0 is Z2.

But the innermost nth root of the cubic formula is a square root, what one expects for Z2. The permutation group S0 is equivalent to addition modulo 3, the group Z3. The outermost nth root is a cube root, which fits.

The quotient-group sequences are

1. (none)
2. Z2
3. Z2 - Z3
4. Z2 - Z3 - Z2*Z2

 

[lpetrich]

Let's introduce some more features of permutation groups. The set of all permutations with length n forms the "symmetric group", Sym(n.). The set of all "even permutations" forms the "alternating group", Alt(n.). The n. is to keep XF from making a thumbs-down emoji.

Even permutations? These are all permutations that can be formed with an even number of interchanges or length-2 permutations. There also exist "odd permutations", formed from an odd number of interchanges.

Whether even or odd is "parity". The parity of a cyclic permutation of length n can be shown to be the parity of (n-1). Parities add for subpermutations of a permutation, thus the parity of a permutation is the parity of the number of even-length cycles in it.

Thus:

1. Even: (1)
2. Even: (11), Odd: (2)
3. Even: (111) (3), Odd: (12)
4. Even: (1111) (13) (22), Odd: (112) (4)
5. Even: (11111) (113) (122) (5) Odd: (1112) (14) (23)

The parity of a permutation of a permutation is the sum of the parities of the two permutations:

Even * Even = Even _ Even * Odd = Odd _ Odd * Even = Odd _ Odd * Odd = Even

The group Z2 again.

Thus, Sym(n.) -Z2-> Alt(n.) for all n. With e = identity group or trivial group, and with (group) -(quotient-group)-> subgroup,

1. (Sym(1) = e)
2. Sym(2) -Z2-> (Alt(2) = e)
3. Sym(3) -Z2-> (Alt(3) = Z3) -Z3-> e
4. Sym(4) -Z2-> Alt(4) -Z3-> (Z2*Z2) -(Z2*Z2)-> e

Can we continue?

Before that, I note that the Z2*Z2 permutation group has elements 1234 2143 3412 4321

 

[lpetrich]

Back to the video. The presenter then explained what I did about length-3 permutations, noting that a sequence of commutator groups is a "derived series". He left it to us to prove that length-4 permutations have a commutator series that ends with the identity.

If you have some programming ability, it should be easy to write a program that finds a group's commutator series by brute force.

I note what one needs to generate the symmetric and alternating groups of permutations:

• Symmetric: permutations with an interchange, a cycle of 2 symbols
• Alternating: permutations with a cycle of 3 symbols

Also, their order or number of elements:

• Symmetric: n!
• Alternating: n!/2 for n > 1, 1 for n = 1

The numbers of even permutations and odd permutations are equal whenever there are any odd permutations.

 

[lpetrich]

Then degree-5 polynomials. They require length-5 permutations. One can do Sym(5) -Z2-> Alt(5) with them, but can one proceed further?

One can get a permutation with a length-3 cycle from two permutations with such a cycle if one has 5 symbols.

23145
12453 * - = 23451
31245 * - = 42351
12534 * - = 42135

2 and 5 fixed, 1 -> 4 -> 3 -> 1 -- a 3-cycle. One can make any 3-cycle permutation with an appropriate selection of two of them.

This means that the commutator group of Alt(n.) is Alt(n.) itself if n >= 5.

So it's

1. (Sym(1) = Alt(1) = e)
2. (Sym(2) = Z2) -Z2-> (Alt(1) = e)
3. Sym(3) -Z2-> (Alt(3) = Z3) -Z3-> e
4. Sym(4) -Z2-> Alt(4) -Z3-> (Z2*Z2) -(Z2*Z2)-> e
5. Sym(5) -Z2-> Alt(5)

If a group's derived series (commuator series) ends in the identity group, then the group is "solvable". If a group is equal to its commutator group, then it is "perfect".

The quotient group of a group and its commutator group is abelian (commutative) and the maximal (largest possible) one.

All finite abelian groups are the identity group, cyclic groups, and products of cyclic groups.

 

[lpetrich]

As to  Galois theory it requires some abstract algebra. It features a generalization of the rational numbers and their supersets: an algebraic field.

It features a set S, an analog of addition, +, and an analog of multiplication, *, both binary operations on S: (S,+,*) with S + S -> S and S * S -> S. The * is distributive over the +, the + forms an abelian group over S, and the * forms an abelian group over all of S but the + identity: 0. The * identity is 1.

Every field's members are the rational numbers, Q, the integers modulo some prime p, Z(p) ~ Z/pZ, or some superset of any of these.

Consider a single-variable polynomial P of some variable x whose coefficients are in field F. The field generated by F and the polynomial's roots is the "splitting field" of P over F. Some examples:

Take P(x) = x - 2. The splitting field over the rational numbers, Q(P), is generated by Q and 2. But since 2 is in Q, the splitting field is Q.

Now take P(x) = x^2 - 2. Though all the coefficients are in Q, the roots are not in Q. The splitting field Q(P) is generated by Q and x for x satisfying P(x) = 0, and its elements are a + b*x for a and b in Q.

The solution of P(x) is +- sqrt(2), and its splitting field is often called Q(sqrt(2)). One can choose the other root and make Q(-sqrt(2)), and one can go between definitions with a -> a and b -> -b. This is an automorphism of Q(sqrt(2)) and combined with the identity one, they form the group Z2.

The automorphisms of field F that leave subfield E field are called the  Galois group of F / E. Here, the Galois group of Q(sqrt(2)) / Q is Z2.

Here is a less trivial example of a Galois group. Solve x^4 - 5*x^2 + 6 = 0 for x. One finds
sqrt(2) + sqrt(3), sqrt(2) - sqrt(3), - sqrt(2) + sqrt(3), - sqrt(2) - sqrt(3)
Reversing the signs on the square roots makes permutations of these roots. One must reverse all the signs of each sqrt(2) together, and also of each sqrt(3). One find permutations 1234, 2143, 3412, 4321 with group Z2 * Z2, the Viergruppe or four-group.

 

[lpetrich]

One can show that a polynomial can be solved with radicals if its Galois group is solvable -- can be decomposed step by step with going down by a cyclic group at each step. The orders of these cyclic groups are related to the inverse powers in the nth roots. Z2 is related to square roots, Z3 to cube roots, etc.

That means that one can find whether a polynomial can be solved with a ruler-and-compass construction -- a polynomial whose roots can be found with basic arithmetic and square roots. If one uses a marked ruler ( Neusis), one can also do cube roots.