2026-08-21
The degree of a polynomial as the number of active sheets of a split geometry
What is the difference between equations of the 5th, 4th, 3rd and 2nd degree?
In the works "From Euler's formula to split geometry" and "Two-periodic regime multi-level idempotent splitting" the unit was successively revealed, first into two and then into four mutually orthogonal components. Each component received its own complex plane, and the degrees of commuting hyperbolic units made it possible to independently control the phases of different levels.
A natural question arises: is it possible to relate the number of active idempotent components to the degree of an algebraic equation? In other words, is it possible to consider the roots of a polynomial not only as a set of numbers, but also as projections of one multi-level element onto mutually orthogonal sheets of split geometry?
It will be shown below that such an interpretation is possible. Moreover, for a polynomial with different roots, the corresponding idempotents do not need to be introduced arbitrarily: they are canonically constructed using the Lagrange interpolation formula. Decreasing the degree from five to four, three and two takes on the geometric meaning of the sequential transition of the root sheets to the zero state.
The main result of this work concerns the geometric and idempotent representation of roots. This representation should be distinguished from the calculation of unknown roots from coefficients. For powers two, three and four, there are general formulas using radicals or trigonometric functions. For a general equation of the fifth degree there is no similar formula in radicals.
1. Polynomial degree and number of root branches
Let a normalized polynomial of degree \(n\) be given over the field of complex numbers:
\[\tag{1} P_n(x) =x^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0. \] According to the main theorem of algebra, it has exactly \(n\) complex roots, taking into account their multiplicity:
\[\tag{2} P_n(x)=\prod_{k=1}^{n}(x-r_k). \] This allows us to formulate the original geometric principle:
\[\tag{3} \boxed{ \deg P_n = \text{number of active idempotent root leaves} } \] The phrase “one idempotent corresponds to one power of \(x\)” can only be used as a preliminary intuition. In the strict formulation, idempotent corresponds not to the monomial \(x^k\), but to one root branch. A polynomial of degree \(n\) has \(n+1\) monomials, including the free term, but only \(n\) roots, taking into account multiplicity.
Introduce \(n\) pairs of orthogonal projectors:
\[\tag{4} E_k^2=E_k, \qquad E_kE_l=0\quad(k\ne l), \qquad \sum_{k=1}^{n}E_k=I_n. \] Then all roots are combined into one root element:
\[\tag{5} \boxed{ \mathcal R_n =\sum_{k=1}^{n}r_kE_k. } \] Projection \(E_k\mathcal R_n\) selects exactly one branch:
\[\tag{6} E_k\mathcal R_n=r_kE_k. \] Therefore, \(\mathcal R_n\) is not a new numerical root, but a single diagonal element, the spectrum of which consists of all the roots of the polynomial.
2. Canonical idempotents constructed from the polynomial
itself Until now, projectors \(E_k\) have been introduced formally. Let us now show that for different roots they arise directly from the polynomial. Consider the factor algebra
\[\tag{7} \mathcal A_n =\mathbb C[x]/(P_n). \] Two polynomials in this algebra are considered the same if their difference is divisible by \(P_n(x)\). Let us denote by \([x]\) the class of the variable \(x\). By definition of factor algebra
\[\tag{8} P_n([x])=0. \] Let us first assume that all roots \(r_1,\ldots,r_n\) are different. For each root, we introduce the fundamental Lagrange polynomial:
\[\tag{9} \boxed{ E_k(x) =\prod_{\scriptstyle 1\le l\le n,\;l\ne k} \frac{x-r_l}{r_k-r_l} = \frac{P_n(x)}{(x-r_k)P_n'(r_k)}. } \] At the root points it takes values
\[\tag{10} E_k(r_l)=\delta_{kl}. \] That is, \(E_k\) is equal to one on its own root branch and zero on all others. Hence the idempotent properties in quotient algebra:
\[\tag{11} E_k^2\equiv E_k\pmod{P_n}, \qquad E_kE_l\equiv0\pmod{P_n}\quad(k\ne l). \] The sum of all fundamental polynomials is equal to one:
\[\tag{12} \sum_{k=1}^{n}E_k(x) \equiv1\pmod{P_n}. \] Finally, the interpolation formula for the function \(f(x)=x\) gives
\[\tag{13} \boxed{ [x] =\sum_{k=1}^{n}r_kE_k. } \] Thus, the root element \(\mathcal R_n\) in formula (5) can be identified with the class of the variable \([x]\) in the factor algebra. For any function correctly defined on a finite set of roots, the componentwise rule holds
\[\tag{14} \boxed{ f(\mathcal R_n) =\sum_{k=1}^{n}f(r_k)E_k. } \] In particular,
\[\tag{15} P_n(\mathcal R_n) =\sum_{k=1}^{n}P_n(r_k)E_k =0. \] From an algebraic point of view, this is a special case of the Chinese remainder theorem. For distinct roots, the linear ideals \((x-r_k)\) are pairwise coprime, therefore
\[\tag{16} \boxed{ \mathbb C[x]/(P_n) \cong \bigoplus_{k=1}^{n}\mathbb C. } \] Split geometry then receives a standard algebraic foundation: each direct component corresponds to one orthogonal idempotent and one root.
3. A Visual Example of Constructing Three Projectors
Consider a Simple Polynomial
\[\tag{17} P_3(x)=(x-1)(x-2)(x-3). \] Its fundamental idempotents are
\[\tag{18} \begin{aligned} E_1(x)&=\frac{(x-2)(x-3)}{(1-2)(1-3)} =\frac{(x-2)(x-3)}{2},\\ E_2(x)&=\frac{(x-1)(x-3)}{(2-1)(2-3)} =-(x-1)(x-3),\\ E_3(x)&=\frac{(x-1)(x-2)}{(3-1)(3-2)} =\frac{(x-1)(x-2)}{2}. \end{aligned} \] In the quotient algebra modulo \(P_3\), we have
\[\tag{19} E_1+E_2+E_3=1, \qquad E_iE_j=\delta_{ij}E_i. \] The root element takes the form
\[\tag{20} \mathcal R_3 =E_1+2E_2+3E_3. \] After simplification, the right-hand side coincides with \(x\) modulo \(P_3(x)\). Three ordinary numbers \(1,2,3\) turn out to be three projections of a single element \(\mathcal R_3\).
4. Coefficients as symmetric invariants of sheets
Idempotents correspond to roots, but not to individual coefficients. The coefficients of the polynomial are elementary symmetric functions of all root coordinates:
\[\tag{21} \begin{aligned} a_{n-1}&=-\sum_k r_k,\\ a_{n-2}&=\sum_{i < j}r_ir_j,\\ a_{n-3}&=-\sum_{i < j < l}r_ir_jr_l,\\ &\, \vdots\\ a_0&=(-1)^n\prod_{k=1}^{n}r_k. \end{aligned} \] Geometrically, an individual root is a coordinate of one leaf, while the coefficient is an invariant of the entire configuration. For example, the sum of the roots describes the collective displacement of all the sheets, the pairwise sum of the products describes their second-order relations, and the free coefficient describes the product of all the root coordinates.
The last invariant is especially important:
\[\tag{22} a_0=(-1)^n r_1r_2\cdots r_n. \] If \(a_0=0\), then at least one coordinate \(r_k\) is zero. The corresponding idempotent does not cease to exist algebraically, but its contribution to the root element vanishes:
\[\tag{23} E_kr_k=E_k\cdot0=0. \] This state will henceforth be called the zeroth, or inactive, root leaf.
5. Two levels of splitting and four root leaves
The first idempotent level is defined by the hyperbolic unit \(\j\):
\[\tag{24} \ep=\frac{1+\j}{2}, \qquad \em=\frac{1-\j}{2}, \qquad \j=\ep-\em. \] The second independent level is defined by the unit \(\jp\):
\[\tag{25} \pmp=\frac{1+\jp}{2}, \qquad \pme=\frac{1-\jp}{2}, \qquad \jp=\pmp-\pme. \] If both units commute, four minimal projectors arise:
\[\tag{26} \boxed{ E_1=\ep\pmp, \quad E_2=\ep\pme, \quad E_3=\em\pmp, \quad E_4=\em\pme. } \] Their sum is equal to one:
\[\tag{27} E_1+E_2+E_3+E_4 =(\ep+\em)(\pmp+\pme)=1. \] Therefore, the general quartic equation naturally fits into the balanced two-level geometry \(2\times2\):
\[\tag{28} \mathcal R_4 =\ep\pmp r_1 +\ep\pme r_2 +\em\pmp r_3 +\em\pme r_4. \] It is important that the four sheets are not four different algebras. They represent four mutually orthogonal components of a single unit.
6. How to Get Five Sheets Without Completely Doubling the Space
Two binary levels yield four minimal channels. A general fifth-degree equation requires five independent root branches. A third global binary level could be introduced to obtain eight channels at once, but for five roots, this is excessive.
A more economical construction consists of locally splitting only one of the four sheets. We denote the last leaf by \(E_4=\em\pme\) and introduce within the corresponding ideal a relative pair of idempotents \(\qmp,\qme\):
\[\tag{29} \qmp^2=\qmp, \qquad \qme^2=\qme, \qquad \qmp\qme=0, \qquad \qmp+\qme=E_4=\em\pme. \] Both new projectors are contained within \(E_4\), so
\[\tag{30} E_4\qmp=\qmp, \qquad E_4\qme=\qme, \qquad E_4=\qmp+\qme. \] As a result, we obtain five primitive projectors:
\[\tag{31} \boxed{ \begin{aligned} F_1&=\ep\pmp,\ F_2&=\ep\pme,\ F_3&=\em\pmp,\ F_4&=\qmp,\ F_5&=\qme. \end{aligned} } \] They are pairwise orthogonal and form a complete unit:
\[\tag{32} F_iF_j=\delta_{ij}F_i, \qquad F_1+F_2+F_3+F_4+F_5=1. \] Thus, a multi-level splitting does not have to double all existing channels simultaneously. It can form a branched tree in which only the selected terminal leaf is split at each step.
Local splitting of one leaf should be understood as constructing a refined direct sum. Unlike the global third pair with a sum equal to one, here \(\qmp+\qme=\em\pme\). Therefore, the tertiary level exists only within the selected ideal, and the other three channels do not receive an extra index.
7. General fifth-degree equation
Consider the normalized equation
\[\tag{33} \boxed{ P_5(x) =x^5+Ax^4+Bx^3+Cx^2+Dx+E=0. } \] Let its roots be \(r_1,\ldots,r_5\). The five-sheet root element is written as
\[\tag{34} \boxed{ \begin{aligned} \mathcal R_5={}& \ep\pmp r_1 +\ep\pme r_2 +\em\pmp r_3\\ &+\qmp r_4 +\qme r_5. \end{aligned} } \] Its five projections give five roots:
\[\tag{35} F_k\mathcal R_5=r_kF_k, \qquad k=1,\ldots,5. \] The coefficients of the equation are related to the roots by Vieta's formulas:
\[\tag{36} \begin{aligned} A&=-\sum_{k=1}^{5}r_k,\\ B&=\sum_{i < j}r_ir_j,\\ C&=-\sum_{i < j < l}r_ir_jr_l,\\ D&=\sum_{i < j < l < m}r_ir_jr_lr_m,\\ E&=-r_1r_2r_3r_4r_5. \end{aligned} \] The five-sheet notation is an exact representation of the root space, but by itself does not yet calculate \(r_k\) through the coefficients \(A,\ldots,E\). This difference becomes especially significant for the fifth power.
8. Reduction from the Fifth to the Fourth Power
Set the free coefficient equal to zero:
\[\tag{37} E=0. \] Since \(E=-r_1r_2r_3r_4r_5\), at least one root must vanish. Without loss of generality, we choose
\[\tag{38} r_5=0. \] Then the polynomial is factored:
\[\tag{39} \begin{aligned} P_5(x) &=x^5+Ax^4+Bx^3+Cx^2+Dx\ &=x\left(x^4+Ax^3+Bx^2+Cx+D\right)\ &=xP_4(x). \end{aligned} \] The remaining factor
\[\tag{40} \boxed{ P_4(x)=x^4+Ax^3+Bx^2+Cx+D } \] is a general normalized quartic equation. In the root element, the last leaf becomes zero:
\[\tag{41} \mathcal R_5 =F_1r_1+F_2r_2+F_3r_3+F_4r_4+F_5\cdot0. \] The active part is as follows
\[\tag{42} \boxed{ \mathcal R_4 =F_1r_1+F_2r_2+F_3r_3+F_4r_4. } \] The unit of the active four-sheet subgeometry becomes
\[\tag{43} I_4=F_1+F_2+F_3+F_4=1-F_5. \] The idempotent \(F_5\) is not destroyed: it remains part of the full extended algebra, but its root amplitude is zero. Therefore, it is geometrically more correct to speak not of the leaf disappearing, but of its contraction to the origin and exclusion from the active root space.
9. Reduction from the fourth power to the third power
After the condition \(E=0\), the free coefficient of the remaining quartic is \(D\). Let
\[\tag{44} D=0. \] Then another root becomes zero. Let us choose \(r_4=0\). The original fifth-degree polynomial takes the form
\[\tag{45} \begin{aligned} P_5(x) &=x^5+Ax^4+Bx^3+Cx^2\ &=x^2\left(x^3+Ax^2+Bx+C\right)\ &=x^2P_3(x). \end{aligned} \] We obtain the general cubic equation
\[\tag{46} \boxed{ P_3(x)=x^3+Ax^2+Bx+C=0. } \] Its three roots form an active element
\[\tag{47} \boxed{ \mathcal R_3 =F_1r_1+F_2r_2+F_3r_3, } \] with an active unit
\[\tag{48} I_3=F_1+F_2+F_3=I_4-F_4. \] The geometry of the three roots can be viewed in two ways. The first is as a degenerate multi-level geometry of five sheets, two of which have zero amplitude. The second is as a proper three-phase orbit in which the three roots are related by phase shifts of \(2\pi/3\). The first representation better illustrates the reduction of powers, while the second is more convenient for calculating roots in the trigonometric case.
10. Reduction from the Third to the Second Power
The free coefficient of a cubic factor is \(C\). When
\[\tag{49} C=0 \] the third root also becomes zero. We obtain
\[\tag{50} \begin{aligned} P_5(x) &=x^5+Ax^4+Bx^3\ &=x^3\left(x^2+Ax+B\right)\ &=x^3P_2(x). \end{aligned} \] The remaining factor is a general quadratic equation:
\[\tag{51} \boxed{ P_2(x)=x^2+Ax+B=0. } \] The root element is reduced to two active components:
\[\tag{52} \boxed{ \mathcal R_2 =F_1r_1+F_2r_2, } \] and the active unit is
\[\tag{53} I_2=F_1+F_2=I_3-F_3. \] In the selected numberingBoth remaining sheets lie inside the primary component \(\ep\):
\[\tag{54} \mathcal R_2 =\ep\left(\pmp r_1+\pme r_2\right). \] Inside the ideal \(\ep\mathcal A\), the element \(\ep\) plays the role of its own unit. Therefore, after three successive contractions, the pentagonal geometry naturally returns to the usual binary decomposition.
11. A unified algebraic and geometric sequence
All transitions can be collected into a single formula:
\[\tag{55} \boxed{ \begin{aligned} P_5(x) &=x^5+Ax^4+Bx^3+Cx^2+Dx+E,\\ E=0 &\Longrightarrow P_5(x)=xP_4(x),\\ D=0 &\Longrightarrow P_5(x)=x^2P_3(x),\\ C=0 &\Longrightarrow P_5(x)=x^3P_2(x). \end{aligned} } \] At the root level:
\[\tag{56} \boxed{ \begin{aligned} \{r_1,r_2,r_3,r_4,r_5\} &\longrightarrow \{r_1,r_2,r_3,r_4,0\}\\ &\longrightarrow \{r_1,r_2,r_3,0,0\}\\ &\longrightarrow \{r_1,r_2,0,0,0\}. \end{aligned} } \] At the level of active units:
\[\tag{57} \boxed{ I_5=1 \longrightarrow I_4=1-F_5 \longrightarrow I_3=1-F_5-F_4 \longrightarrow I_2=1-F_5-F_4-F_3. } \] Thus, lowering the degree is not an arbitrary deletion of part of the formula. It corresponds to an exact algebraic transition to an ideal whose unit is the sum of the remaining active projectors.
12. How are four roots calculated within two-level geometry
The five-sheet diagram describes the general geometry of roots. For degree four, we can further show how the classical Ferrari solution can be packed into two binary idempotent levels.
After the substitution
\[\tag{58} x=X-\frac A4 \] the equation \(P_4(x)=0\) is reduced to the form
\[\tag{59} X^4+pX^2+qX+r=0, \] where
\[\tag{60} \begin{aligned} p&=B-\frac{3A^2}{8},\\ q&=C-\frac{AB}{2}+\frac{A^3}{8},\\ r&=D-\frac{AC}{4}+\frac{A^2B}{16}-\frac{3A^4}{256}. \end{aligned} \] Represent the left side as the difference of squares:
\[\tag{61} (X^2+\alpha)^2-(\beta X+\gamma)^2=0. \] Comparing the coefficients yields
\[\tag{62} p=2\alpha-\beta^2, \qquad q=-2\beta\gamma, \qquad r=\alpha^2-\gamma^2. \] If \(z=\beta^2\), then \(z\) is determined by the cubic resolvent
\[\tag{63} \boxed{ z^3+2pz^2+(p^2-4r)z-q^2=0. } \] After choosing a suitable root of the resolvent
\[\tag{64} \beta=\sqrt z, \qquad \alpha=\frac{p+z}{2}, \qquad \gamma=-\frac{q}{2\beta}. \] Introduce two sign units
\[\tag{65} \j=\ep-\em, \qquad \jp=\pmp-\pme. \] Then all four roots of the reduced quartic are combined in one operator:
\[\tag{66} \boxed{ \mathcal X_4 =-\frac{\beta}{2}\j +\frac12\jp \sqrt{\beta^2-4\alpha-4\gamma\j}. } \] The square root is taken component-wise at the first level:
\[\tag{67} \sqrt{\beta^2-4\alpha-4\gamma\j} =\ep\sqrt{\Delta_+} +\em\sqrt{\Delta_-}, \] where
\[\tag{68} \Delta_+=\beta^2-4\alpha-4\gamma, \qquad \Delta_-=\beta^2-4\alpha+4\gamma. \] After expanding on four sheets, we get
\[\tag{69} \begin{aligned} \mathcal X_4={}& \ep\pmp\frac{-\beta+\sqrt{\Delta_+}}2 +\ep\pme\frac{-\beta-\sqrt{\Delta_+}}2\\ &+\em\pmp\frac{\beta+\sqrt{\Delta_-}}2 +\em\pme\frac{\beta-\sqrt{\Delta_-}}2. \end{aligned} \] The first level selects one of two quadratic equations, and the second one of two roots within the selected equation. Thus, the classical nested structure of the Ferrari solution exactly corresponds to the \(2\times2\) geometry.
13. Cubic Equation: Singular Leaves and Eigenphase Geometry
The cubic equation can be obtained from the general scheme as a five-leaf configuration with two zero leaves. However, three active branches do not form a balanced binary tree. Therefore, to calculate the three real roots, it is more natural to use cyclic phase geometry of order three.
For the equation
\[\tag{70} x^3+Ax^2+Bx+C=0 \] Introduce
\[\tag{71} x=y-\frac A3, \qquad y^3+py+q=0, \] Where
\[\tag{72} p=B-\frac{A^2}{3}, \qquad q=\frac{2A^3}{27}-\frac{AB}{3}+C. \] For three real roots, we set
\[\tag{73} \rho=\sqrt{-\frac p3}, \qquad \cos(3\theta)=-\frac{q}{2\rho^3}. \] Then
\[\tag{74} \boxed{ x_k =-\frac A3 +2\rho\cos\left(\theta+\frac{2\pi k}{3}\right), \qquad k=0,1,2. } \] This formula arises from a conjugate pair of powers
\[\tag{75} \j^s=\ep+\em e^{i\pi s}, \qquad \j^{-s}=\ep+\em e^{-i\pi s}, \] since the projectionTheir sum on the plane \(\em\) is equal to
\[\tag{76} \operatorname{P}_{\em} \left(\j^s+\j^{-s}\right) =2\cos(\pi s). \] The three roots correspond to shifts
\[\tag{77} s_k=s+\frac{2k}{3}. \] Therefore, the cubic simultaneously admits two geometric interpretations: three orthogonal root projections in the quotient algebra and three phase positions of one conjugate pair. The first construction is applicable to any distinct roots, the second directly calculates the roots only in the corresponding trigonometric mode.
14. A quadratic equation as a single binary level
For the equation
\[\tag{78} ax^2+bx+c=0 \] one idempotent splitting is sufficient. Let's introduce
\[\tag{79} \jp=\pmp-\pme. \] Two roots are combined in the operator
\[\tag{80} \boxed{ \mathcal R_2 =-\frac{b}{2a} +\frac{\jp}{2a}\sqrt{b^2-4ac}. } \] Expanding it across two sheets, we obtain
\[\tag{81} \mathcal R_2 =\pmp\frac{-b+\sqrt{b^2-4ac}}{2a} +\pme\frac{-b-\sqrt{b^2-4ac}}{2a}. \] In this case, the algebraic and geometric structures are completely identical: one binary level contains exactly two projections and exactly two roots.
15. Special Quintic and the Boundary of the General Method
One conjugate phase pair allows us to obtain a special fifth-degree equation. Let
\[\tag{82} y=2\rho\cos\theta. \] From the identity for the Chebyshev polynomial of degree five it follows
\[\tag{83} \boxed{ y^5-5\rho^2y^3+5\rho^4y -2\rho^5\cos(5\theta)=0. } \] All five roots of this special equation are of the form
\[\tag{84} \boxed{ y_k =2\rho\cos\left(\theta+\frac{2\pi k}{5}\right), \qquad k=0,1,2,3,4. } \] This is a de Moivre quintic. Its coefficients are bound by additional conditions, so it is not a general quintic. The five-phase orbit here actually calculates the roots, whereas the five-sheeted element \(\mathcal R_5\) represents the roots of an arbitrary fifth-degree equation, but does not automatically determine them through the coefficients.
Abel's theorem states that there is no general formula expressing the roots of an arbitrary equation of degree five or higher through a finite combination of arithmetic operations and radicals. Idempotent splitting does not remove this restriction: if the roots of a general quintic are unknown, constructing their numerical coordinates remains an independent problem.
Therefore, the correct result of the new construction consists not in "solving a general quintic by radicals," but in constructing its five-sheeted root space and in a geometric description of the power reduction.
16. Multiple Roots and Nilpotent Directions
Formula (9) contains the factor \(P_n'(r_k)\) in the denominator. If the root is multiple, then \(P_n'(r_k)=0\), and the usual Lagrange projection is not defined in this form. This isn't a technical detail, but a change in the structure of the quotient algebra.
For example, for
\[\tag{85} P(x)=(x-r)^m \] the quotient algebra has the form
\[\tag{86} \mathbb C[x]/\bigl((x-r)^m\bigr). \] It introduces a nilpotent element
\[\tag{87} \nu=[x-r], \qquad \nu^m=0. \] Therefore, the multiplicity of the root is encoded not by several canonical orthogonal idempotents with the same coordinate, but by a local nilpotent structure around a single root leaf. For a polynomial
\[\tag{88} P(x)=\prod_{j=1}^{s}(x-r_j)^{m_j} \] The Chinese remainder theorem gives the factorization
\[\tag{89} \boxed{ \mathbb C[x]/(P) \cong \bigoplus_{j=1}^{s} \mathbb C[x]/\bigl((x-r_j)^{m_j}\bigr). } \] The number of orthogonal root leaves here is equal to the number of distinct roots \(s\), and the total degree
\[\tag{90} n=m_1+\cdots+m_s \] additionally takes into account the nilpotent depth of each leaf. Therefore, the main thesis of the paper needs to be clarified:
\[\tag{91} \boxed{ \begin{aligned} \text{distinct root} &\longleftrightarrow \text{idempotent leaf},\\ \text{root multiplicity} &\longleftrightarrow \text{nilpotent leaf depth}. \end{aligned} } \] For polynomials without multiple roots, the nilpotent part is absent, and the degree indeed coincides with the number of active idempotent leaves. For multiple roots, formula (3) remains valid only when counting leaves with formal multiplicity; canonical algebraic geometry distinguishes between the number of leaves and their internal depth.
17. What is known, and what does split geometry add?
Polynomial factorization, Lagrange interpolation polynomials, the Chinese remainder theorem, and the Cardano and Ferrari formulas are classical results. The proposed developmentIt consists in combining them with multilevel split geometry:
\[\tag{92} \boxed{ \begin{aligned} \text{polynomial degree} &\longrightarrow \text{number of active root channels},\\ \text{root} &\longrightarrow \text{coordinate of an idempotent leaf},\\ \text{coefficients} &\longrightarrow \text{symmetric configuration invariants},\\ \text{zero root} &\longrightarrow \text{contracted inactive leaf},\\ \text{multiplicity} &\longrightarrow \text{nilpotent depth}. \end{aligned} } \]
This reading connects the phase operators of previous works, deep idempotent splitting, and classical polynomial algebra into a unified geometric picture.
Conclusion
A polynomial of degree \(n\) with distinct roots canonically generates \(n\) mutually orthogonal idempotents. Each idempotent distinguishes one root branch, and the class of a variable in the quotient algebra decomposes into a root element \(\mathcal R_n=\sum r_kE_k\). Therefore, the degree of a polynomial receives a geometric interpretation as the number of active sheets of its root space.
Two levels of binary splitting naturally create four sheets and coincide with the structure of the solution of a general quartic equation. The fifth sheet can be obtained by locally splitting one of the four channels, without doubling the entire space. Successively vanishing the free coefficients leads to an exact chain.
\[\tag{93} \boxed{ P_5 \longrightarrow xP_4 \longrightarrow x^2P_3 \longrightarrow x^3P_2. } \] Algebraically, an additional zero root appears at each step. Geometrically, one of the sheets is contracted to the origin, and the unit of active space is reduced by the corresponding projection. Thus, equations of the fifth, fourth, third, and second degree turn out to be different active sections of a single multilevel root geometry.
However, the idempotent representation should not be confused with the universal method for calculating roots. It accurately reveals the structure of the root space, but preserves the classical limitations of algebra: a general quintic is not solvable by radicals, and multiple roots require the inclusion of nilpotent directions.
Materials used
- Encyclopedia of Mathematics. Idempotent.
- Encyclopedia of Mathematics. Interpolation formula.
- The Stacks Project. Lemma 10.15.4: Chinese remainder.
- Encyclopedia of Mathematics. Galois theory.
- Encyclopedia of Mathematics. Abel theorem on algebraic equations.