Try
For three indeterminates x,y,z, consider the polynomial
f = x²y+x²z+xy²+xz²+y²z+yz²+5xyz.
Swapping any two variables leaves f unchanged. Instead of remembering seven terms, can you recover it from the three elementary symmetric polynomials
e₁=x+y+z, e₂=xy+xz+yz, e₃=xyz?
Expand e₁e₂ before reading further. Each of the six terms with exponents 2,1,0 appears once, while xyz appears three times. Thus f=e₁e₂+2e₃. This is a formal polynomial identity: it remains valid after substituting elements of any commutative ring, including a ring in which 2=0. Agreement at a few numerical inputs alone would not prove it.
Our goal is a procedure that works for every symmetric polynomial and a proof that its answer is unique. We will also relate the elementary polynomials to power sums, while keeping track of when division is allowed.
Predict
A symmetric polynomial in n variables is a formal polynomial unchanged by every permutation of those variables. Work over a commutative ring R with identity, initially nonzero, with n≥1. Coefficients may be zero divisors. Polynomial equality means equality of coefficients; it is not equality of functions on a finite set.
For 1≤j≤n, eⱼ is the sum of all products of j distinct variables. Put e₀=1 and eⱼ=0 for j>n. The power sum pₖ is x₁ᵏ+⋯+xₙᵏ for k≥1. Integer coefficients such as k mean repeated addition of 1 in R.
Predict the expressions for p₂ and for x²y²+x²z²+y²z² in terms of e₁,e₂,e₃. Then ask a different question: if only p₁ and p₂ are supplied, can you always recover e₂ by dividing e₁²−p₂ by 2? Keep the identity and the proposed division separate.
Experiment
Use the table as a calculation sheet. Expand the proposed product, compare every monomial, and only then subtract. For a systematic choice, compare exponent vectors in lexicographic order: first compare the exponent of x, then y, then z. The larger exponent at the first disagreement determines the larger monomial.
| Current polynomial | Largest exponent vector | Product with the same leading monomial | Remainder after subtraction |
|---|
| f from Try | (2,1,0) | e₁e₂ | 2xyz |
| 2xyz, when 2≠0 in R | (1,1,1) | 2e₃ | 0 |
| p₂=x²+y²+z² | (2,0,0) | e₁² | −2e₂ |
| x²y²+x²z²+y²z² | (2,2,0) | e₂² | −2e₁e₃ |
In characteristic two, the second row is already zero and does not require a further cancellation. The other displayed identities still hold, with the integer coefficients interpreted in the ring. Notice that subtraction of a matching leading coefficient is available even when that coefficient has no inverse.
Change the example to g=x²y²+x²z²+y²z²+4xyz(x+y+z). Predict its expression before opening the check.
Check the changed polynomial
The final row gives the first three terms as e₂²−2e₁e₃. The added term is 4e₁e₃, so g=e₂²+2e₁e₃. This calculation uses no division and remains true in characteristic two, where the answer reduces to e₂².
Understand
Why leading exponents give an algorithm
Let a nonzero symmetric polynomial have leading monomial x₁ᵃ¹⋯xₙᵃⁿ with nonzero coefficient c. Its exponents satisfy a₁≥a₂≥⋯≥aₙ. Otherwise some adjacent pair has aᵢ<aᵢ₊₁. Swapping those two variables produces another monomial with the same nonzero coefficient, but with a larger lexicographic exponent vector, contradicting the choice of the leading monomial.
The leading monomial of eⱼ is x₁⋯xⱼ, with coefficient 1. Hence the product
E = e₁^(a₁−a₂) e₂^(a₂−a₃) ⋯ eₙ₋₁^(aₙ₋₁−aₙ) eₙ^aₙ
has leading exponent vector (a₁,…,aₙ) and leading coefficient 1. Lexicographic order is preserved by adding exponent vectors, which proves this assertion about products. The coefficient 1 ensures that zero divisors cannot destroy the leading term of cE.
Subtract cE. The leading monomial cancels, the remainder stays symmetric, and its leading exponent vector strictly decreases unless it is zero. To justify termination, let D be the original total degree. Every subtracted E has total degree a₁+⋯+aₙ≤D, so all remainders use only the finitely many exponent vectors of total degree at most D. Strict descent within this finite set must stop. Do not appeal to an unspecified infinite lexicographic descent argument.
Recording all subtractions constructs a polynomial Q in formal variables T₁,…,Tₙ such that f=Q(e₁,…,eₙ). Constants are handled by the all-zero exponent vector. No coefficient division, domain assumption or characteristic restriction has been used.
Why the expression is unique
The substitution map R[T₁,…,Tₙ]→R[x₁,…,xₙ] sending Tⱼ to eⱼ is injective. Indeed, a monomial T₁ᵇ¹⋯Tₙᵇⁿ maps to a product whose leading exponent vector is
(b₁+⋯+bₙ, b₂+⋯+bₙ, …, bₙ).
Distinct b vectors give distinct leading vectors: subtract consecutive entries to recover b. For a nonzero Q, select the largest leading vector contributed by its finitely many nonzero terms. No other substituted term can cancel its leading monomial, since all monomials in that term are no larger than its own strictly smaller leading monomial. Its coefficient remains the original nonzero coefficient of Q. Thus Q(e₁,…,eₙ)≠0.
Existence and injectivity together say that the ring of symmetric polynomials is exactly a polynomial ring in the elementary symmetric polynomials. This is about formal polynomials, including over finite coefficient rings. For the zero ring both polynomial rings have one element, so the statement holds without a nonzero-leading-term argument. With no variables, both rings are simply R and there is nothing to eliminate.
Newton identities without division
Introduce a separate formal variable u and the polynomial
E(u)=∏ᵢ(1−xᵢu)=Σⱼ₌₀ⁿ (−1)ʲ eⱼuʲ.
Formal power series are coefficient lists with the usual addition and convolution product; each coefficient of a product uses only finitely many terms. The factor 1−xᵢu has the formal inverse Σₖ₌₀∞ xᵢᵏuᵏ, because multiplication cancels all its positive-degree coefficients. This is an algebraic identity, not a claim about numerical convergence.
Differentiate E formally by the product rule and multiply by −u. Using these inverses gives
−u E′(u) = E(u) Σᵢ xᵢu/(1−xᵢu) = E(u) Σₖ₌₁∞ pₖuᵏ.
The fraction denotes multiplication by the displayed formal inverse, not division by an arbitrary ring element. Compare coefficients of uᵏ. For 1≤k≤n the result is
pₖ − e₁pₖ₋₁ + e₂pₖ₋₂ − ⋯ + (−1)^(k−1)eₖ₋₁p₁ + (−1)ᵏ k eₖ = 0.
For k>n it is instead
pₖ − e₁pₖ₋₁ + e₂pₖ₋₂ − ⋯ + (−1)ⁿ eₙpₖ₋ₙ = 0.
There is no p₀ term in either displayed formula. In the first case the final term comes from −uE′; in the second case that coefficient is zero because E has degree at most n. The coefficient comparison proves every identity over every commutative coefficient ring, including positive characteristic.
For example, p₁=e₁, p₂=e₁p₁−2e₂, and p₃=e₁p₂−e₂p₁+3e₃. Given elementary coefficients, these relations compute all power sums without division. Conversely the kth identity determines k eₖ from earlier elementary coefficients and power sums. Solving for eₖ by multiplying by 1/k is valid if k is a unit in R. In a characteristic-zero field this is always possible for positive k; in characteristic p it fails at k=p. In ℤ, k is generally not a unit, but known divisibility and injectivity of multiplication by k can still justify a particular exact recovery. “Not a unit” does not mean every recovery problem has multiple answers.
What power sums can forget
In characteristic two, the pairs (0,0) and (t,t), with entries in F₂[t], have identical power sums for every k≥1: tᵏ+tᵏ=0. Their e₁ values both vanish, but their e₂ values are 0 and t², respectively. Therefore even all power sums need not determine elementary symmetric values. This does not contradict uniqueness of expressing a symmetric formal polynomial in the eⱼ: that theorem concerns a different substitution map.
Apply
For three variables, derive p₄ from the Newton identities, expressing it only in e₁,e₂,e₃. Test the result at (x,y,z)=(1,2,−1). Then reduce the formula in characteristic two and explain why that reduction cannot be reversed to recover e₂ or e₃ from p₄.
Check the fourth power and the characteristic boundary
The first three identities give p₂=e₁²−2e₂ and p₃=e₁³−3e₁e₂+3e₃. Since 4>3, the next identity has no 4e₄ term: p₄=e₁p₃−e₂p₂+e₃p₁. Substitution gives
p₄=e₁⁴−4e₁²e₂+2e₂²+4e₁e₃.
At (1,2,−1), the elementary values are (2,−1,−2). The right side is 16+16+2−16=18, equal to 1+16+1. In characteristic two the formula becomes p₄=e₁⁴, which supplies no information about e₂ or e₃. The full polynomial identity remains valid; the lost coefficients prevent the proposed recovery.
Continue
Reuse polynomial division and formal equality when distinguishing coefficient identities from finite numerical checks. The quartic constructibility example already derives the three root-pairing identities explicitly; use those as an application rather than assuming that a particular expansion proves the general symmetric-polynomial theorem.
Continue to discriminants and repeated roots, where a symmetric product of root differences becomes a coefficient certificate. The following discriminants and Galois groups lesson will distinguish invariance of its square from the sign of the root-difference product itself.
