Symmetric Polynomials and the Fundamental Theorem of Symmetric Functions

Every variable permutation fixes the constant polynomials $0$ and $1$, so both are symmetric.

For every permutation $\sigma$, substitution of permuted variables commutes with the ring operations, so $\sigma (f+g)=\sigma (f)+\sigma (g)=f+g$, $\sigma (-f)=-\sigma (f)=-f$, and $\sigma (fg)=\sigma (f)\sigma (g)=fg$.

Thus the symmetric polynomials contain $1$ and are closed under addition, additive inverses, and multiplication, so they form a subring. When $n=0$ every polynomial is a coefficient in $R$ and the assertion gives $R$ itself.

In expanding the product, choose either $t$ or $-x_i$ from each factor. A choice of $-x_i$ from exactly the indices in a subset $S$ of size $k$ contributes $(-1{)}^k({\prod }_{i\in S}{x}_i)t^{n-k}$.

Summing the contributions with $\mathrm{\mid }S\mathrm{\mid }=k$ gives $(-1{)}^k{e}_k{t}^{n-k}$ by the definition of $e_k$.

Summing over $0\le k\le n$ accounts for every term in the expansion exactly once and proves the identity. If $n=0$, the sole term is $e_0{t}^0=1$.

Substitute $x_i={\alpha }_i$ in [L1] inside $S[t]$ to obtain $f(t)={\sum }_{k=0}^n(-1{)}^k{e}_k({\alpha }_1,\dots ,{\alpha }_n)t^{n-k}$.

Equality of polynomials is coefficientwise, so comparison with $f(t)=t^n+a_1{t}^{n-1}+\cdots +a_n$ gives the displayed formula for every $k$. If $n=0$, the comparison has no positive index.

Variable permutations partition the monomials into disjoint orbits, and every orbit contains exactly one weakly decreasing exponent tuple $\lambda$.

If $f$ is symmetric, the coefficients of two monomials in the same orbit are equal, because a variable permutation carries either monomial to the other and fixes $f$. Since $f$ has finite support, it is therefore a finite $R$-linear combination of the corresponding orbit sums $m_{\lambda }$.

Distinct $m_{\lambda }$ have disjoint monomial supports. Hence a finite relation ${\sum }_{\lambda }{c}_{\lambda }{m}_{\lambda }=0$ has $c_{\lambda }=0$ for every $\lambda$, by comparing the coefficient of any monomial in the orbit of $\lambda$.

Steps 1.2 and 1.3 give spanning and linear independence, respectively, so the $m_{\lambda }$ form an $R$-basis.

Suppose, for contradiction, that $a_i<a_{i+1}$ for some $i<n$.

Interchanging $x_i$ and $x_{i+1}$ fixes $f$, so the tuple $a^{\mathrm{\prime }}=(a_1,\dots ,a_{i-1},a_{i+1},a_i,a_{i+2},\dots ,a_n)$ occurs in $f$ with the same nonzero coefficient as $a$.

The tuples agree before coordinate $i$ and $a_i^{\mathrm{\prime }}=a_{i+1}>a_i$, so $a^{\mathrm{\prime }}{>}_{\mathrm{l}\mathrm{e}\mathrm{x}}a$, contradicting the maximality in [L2]. Therefore no such $i$ exists and the tuple is weakly decreasing.

The lexicographically leading monomial of $e_k$ is $x_1\cdots x_k$, and its coefficient is $1$: at the first omitted variable, any other squarefree degree-$k$ monomial has exponent $0$ where this one has exponent $1$.

If $a{>}_{\mathrm{l}\mathrm{e}\mathrm{x}}b$ first differs at coordinate $r$, then for every $c$ the tuples $a+c$ and $b+c$ still first differ at $r$, with the former coordinate larger. Thus, among the products formed from copies of the $e_k$, the unique largest term is obtained by choosing the leading monomial from every factor. Its coefficient is $1$.

Applying step 2.1 to $e_1^{b_1}\cdots e_n^{b_n}$ gives exponent $b_k+\cdots +b_n$ on $x_k$ and coefficient $1$.

The displayed cumulative sums determine $b_n$ and then successively $b_{n-1},\dots ,b_1$ by adjacent subtraction, so the map from $b$ to the leading multidegree is injective.

The assertion is immediate for $f=0$ and for $n=0$, when the symmetric-polynomial ring is $R$. Assume now that $n>0$ and $f\mathrm{\ne }0$; then some coefficient of $f$ is nonzero, so $1\mathrm{\ne }0$ in $R$ and [L2] applies.

Let $a=(a_1,\dots ,a_n)$ be the leading multidegree of $f$ and let $c$ be its leading coefficient. By [L1], set $b_i=a_i-a_{i+1}$ for $i<n$ and $b_n=a_n$, all natural numbers.

By [L2], the polynomial $c{e}_1^{b_1}\cdots e_n^{b_n}$ has the same leading term as $f$, so their difference $f_1$ is either zero or has strictly smaller leading multidegree. It remains symmetric by [L3].

Repeat step 2.1 while the remainder is nonzero. The process terminates because all exponent tuples encountered are bounded coordinatewise by the finite support box of the original polynomial and strictly decrease lexicographically at each subtraction.

Summing the finitely many subtracted monomials in the $e_i$ gives a polynomial $Q$ with $f=Q(e_1,\dots ,e_n)$.

Suppose, for contradiction, that $Q\mathrm{\ne }0$ but $Q(e_1,\dots ,e_n)=0$.

Since $Q\mathrm{\ne }0$, some coefficient $c_b$ is nonzero by [L2], so $1\mathrm{\ne }0$ in $R$ and [L1] applies. Among the finitely many monomials $c_b{T}_1^{b_1}\cdots T_n^{b_n}$ of $Q$ with $c_b\mathrm{\ne }0$, choose one whose substituted leading multidegree is greatest.

By [L1], no other substituted monomial has that leading multidegree, and the chosen substituted monomial has leading coefficient $c_b\mathrm{\ne }0$. Hence this term cannot cancel in $Q(e_1,\dots ,e_n)$, even if $R$ has zero divisors.

This contradicts $Q(e_1,\dots ,e_n)=0$, whose every coefficient is zero by [L2]. Therefore $Q=0$.

Substitution $T_k\mapsto e_k$ defines an $R$-algebra homomorphism whose image lies in the symmetric-polynomial subring.

The map is surjective by [L1].

Its kernel is zero by [L2], so it is injective.

The substitution map is therefore an isomorphism. Surjectivity gives existence of an expression, and injectivity gives its uniqueness.

Use [L1] to write $P=Q(e_1,\dots ,e_n)$ for a unique $Q$.

Evaluate at the roots and apply [L2] in each coordinate to obtain $P({\alpha }_1,\dots ,{\alpha }_n)=Q(-a_1,a_2,\dots ,(-1{)}^n{a}_n)$.

The right side is computed from coefficients in $R$, and symmetry makes it unchanged when the roots are reordered. The asserted uniqueness is the uniqueness in [L1] of the $Q$ representing $P$ as an identity of polynomials, and it is not uniqueness of a $Q$ satisfying the evaluated equality alone: when $n\ge 1$ and $1\mathrm{\ne }0$ in $R$, the polynomial $T_1+a_1$ has $T_1$-coefficient $1$ and is therefore nonzero, while substituting the tuple $(-a_1,a_2,\dots ,(-1{)}^n{a}_n)$ sends it to $-a_1+a_1=0$, so $Q$ and $Q+(T_1+a_1)$ take the same value there.

Expand ${\prod }_{j=1}^n(1-x_{j}t)$ by choosing either $1$ or $-x_{j}t$ from each factor. The choices taking $-x_{j}t$ at exactly the indices of an $i$-element set $S$ contribute $(-1{)}^i({\prod }_{j\in S}{x}_j)t^i$, so summing over $\mathrm{\mid }S\mathrm{\mid }=i$ and then over $i$ gives ${\prod }_{j=1}^n(1-x_{j}t)={\sum }_{i=0}^n(-1{)}^i{e}_i{t}^i$ by [L1], and both displayed descriptions of $E(-t)$ therefore agree.

For one variable, $(1-x_{j}t)(1+x_{j}t+x_j^2{t}^2+\cdots )=1$ coefficientwise as a formal power series.

Multiplying the one-variable geometric series over $j=1,\dots ,n$ gives ${\prod }_j(1-x_{j}t{)}^{-1}$, whose coefficient of $t^k$ is the sum of $x_1^{a_1}\cdots x_n^{a_n}$ over $a_1+\cdots +a_n=k$, namely $h_k$.

Thus $H(t)={\prod }_j(1-x_{j}t{)}^{-1}$, which is $E(-t{)}^{-1}$ by step 1.1, so $E(-t)H(t)=1$. Comparing the coefficient of $t^k$ gives the displayed recurrence, including $k=0$ as $e_0{h}_0=1$.

The recurrence in [L1] expresses $h_k$ as $(-1{)}^{k-1}{e}_k$ plus a polynomial in $e_1,\dots ,e_{k-1}$, and also expresses $e_k$ as $(-1{)}^{k-1}{h}_k$ plus a polynomial in $h_1,\dots ,h_{k-1}$.

Recursion on $k$ therefore gives mutually inverse triangular substitutions between $R[e_1,\dots ,e_n]$ and $R[h_1,\dots ,h_n]$; every diagonal coefficient is $1$ or $-1$, hence a unit in $R$.

Composing either triangular isomorphism with [L2] shows that $U_k\mapsto h_k$ is an $R$-algebra isomorphism onto the symmetric-polynomial ring.

Write $G(t):=E(-t)={\prod }_{j=1}^n(1-x_{j}t)$, which by [L2] is a polynomial in $t$ of degree at most $n$ over $R[x_1,\dots ,x_n]$, so [L3] and [L4] apply to it. Iterating the Leibniz rule of [L4] over the $n$ factors, and using $(1-x_{j}t{)}^{\mathrm{\prime }}=-x_j$ from [L3], gives $G^{\mathrm{\prime }}(t)=-{\sum }_j{x}_j{\prod }_{\mathrm{\ell }\mathrm{\ne }j}(1-x_{\mathrm{\ell }}t)$.

Multiply by the power series $H(t)=G(t{)}^{-1}$ from [L2]. Then $-G^{\mathrm{\prime }}(t)H(t)={\sum }_j{x}_j\mathrm{/}(1-x_{j}t)={\sum }_{i\ge 1}{p}_i{t}^{i-1}$, where the last equality is coefficientwise geometric expansion.

Multiply step 2.1 by $G(t)$ and use $G(t)H(t)=1$ from [L2]; no derivative of the infinite series $H$ is taken. This gives $-G^{\mathrm{\prime }}(t)=G(t){\sum }_{i\ge 1}{p}_i{t}^{i-1}$. Since $G(t)={\sum }_{i=0}^n(-1{)}^i{e}_i{t}^i$, [L3] evaluates the left side as ${\sum }_{k\ge 1}(-1{)}^{k-1}k{e}_k{t}^{k-1}$. Comparing the coefficient of $t^{k-1}$ yields $(-1{)}^{k-1}k{e}_k={\sum }_{i=1}^k(-1{)}^{k-i}{e}_{k-i}{p}_i$.

Multiplying the identity in step 3.1 by $(-1{)}^{k-1}$ proves the displayed Newton identity. When $k>n$, the term $k{e}_k$ is zero and reindexing gives the stated recurrence for $p_k$.

For each $1\le k\le n$, the element $k{1}_R$ is a unit: the product of $k{1}_R$ with the images of all the other factors in $n!$ is the unit $n!1_R$, and a factor of a unit in a commutative ring is a unit.

Using the inverse of $k{1}_R$, [L1] recursively expresses $e_k$ as a polynomial in $p_1,\dots ,p_k$. Conversely [L1] expresses $p_k$ as $(-1{)}^{k-1}k{e}_k$ plus a polynomial in $e_1,\dots ,e_{k-1}$.

These mutually inverse triangular substitutions have unit diagonal coefficients, so they give an isomorphism $R[e_1,\dots ,e_n]\cong R[p_1,\dots ,p_n]$. Composing with [L2] proves free generation.

In a field, a positive integer image is a unit exactly when it is nonzero. Thus all of $1,\dots ,n$ are nonzero exactly in characteristic zero or characteristic greater than $n$, which is equivalent to the factorial image being nonzero and hence invertible.

A variable permutation bijects the unordered pairs $\{i,j\}$ with themselves. For each pair, it sends $(x_i-x_j{)}^2$ to either $(x_{\sigma (i)}-x_{\sigma (j)}{)}^2$ or the same factor with its two terms reversed.

Reversing a difference has no effect after squaring, since $(u-v{)}^2=(v-u{)}^2$ in every commutative ring, including characteristic two. Hence the permutation merely reorders the factors of ${\mathrm{\Delta }}_n^2$.

Every variable permutation fixes ${\mathrm{\Delta }}_n^2$, so it is symmetric by [L2]. For $n=0,1$, the product is $1$ and the same conclusion holds.