How do I prove that

any symmetric polynomial P is given by an expression involving only additions and multiplication of constants and elementary symmetric polynomials.

I have no clue of where to start, I just know the basic definition:

The polynomial $P(x_1,x_2,...,x_n)$ is symmetric if for any permutation $\sigma$ of $\{x_1,x_2,...,x_n\}$, $$ P(x_1,x_2,...,x_n)=P(x_{\sigma_1},x_{\sigma_2},...,x_{\sigma_n}) $$ The elementary symmetric polynomials for a polynomial consists of $n$ variables, $\{x_1,x_2,...,x_n\}$, and are defined as $\{e_1,e_2,...,e_n\}$: $$ e_0=1\\ e_1=\sum_{i}x_i\\ e_2=\sum_{i,j}x_ix_j\\ e_3=\sum_{i,j,k}x_ix_jx_k\\ ....................\\ e_n=x_1x_2...x_n $$

Can I use induction, if possible where do I start ?


1 Answer 1


Proof for symmetric polynomials with two variables X, Y:

Let $f(X,Y)$ be a symmetric polynomial with degree n. By definition, $f(X,Y)=f(Y,X)$. Now, $$f(X,Y)= \sum_{0 \le i\le n,0\le j \le n,i+j\le n} a_{ij}X^iY^j$$ for some $a_{ij}$, and $a_{ij}=a_{ji} \;\forall{0\le i \le n,0\le j\le n}.\;\;\;\;(*)$

If $\alpha,\beta \in{\mathbb R}, c$ are some arbitrary constants, and $f(X,Y), g(X,Y)$ are two symmetric polynomials, then $\alpha f(X,Y)+\beta g(X,Y)+c$ is also a symmetric polynomial. This obviously extends to cases with multiple symmetric functions, i.e.

$$\sum_{j=1}^{p} \alpha_j f_j(X,Y)+c \; \text{is symmetric} \;\forall{\alpha_j \in{\mathbb R}, f_j \;\text{are all symmetric}} \;\;\;\;(**)$$

If n is even, then let $n=2m$ $$(X+Y)^n=X^n+\binom n 1 X^{n-1}Y+ \;...\;+\binom n mX^mY^m+ \;...\;+\binom n {n-1} XY^{n-1}+Y^n$$

If n is odd, then let $n=2m+1$ $$(X+Y)^n=X^n+\binom n 1 X^{n-1}Y+ \;...\;+\binom n mX^{m+1}Y^m+\binom n {m+1}X^mY^{m+1}+ \;...\;+\binom n {n-1} XY^{n-1}+Y^n$$

It is clear that $(X+Y)^n$ is symmetric $\forall{n\in {\mathbb N}}$. Moreover, $\forall 0\le k\le n,$ the $k^{th}$ term is the same as the $(n+2-k)^{th}$ term.

By $(*)$ and $(**)$ , it suffices to prove that every symmetric polynomial $f_0(X,Y)$ with at most 2 non-constant terms can be generated by $X+Y$ and $XY$, and that all polynomials with at most 2 non-constant terms generated by $X+Y$ and $XY$ are symmetric.

Suppose $\mu X^iY^j$ is symmetric, where $i\ge 0, j\ge 0$ are not both $0$. Then $X^iY^j=X^jY^i,$ which implies $i=j$.

Now, let us consider the polynomial $\;\eta_1 X^iY^j + \eta_2 X^rY^s$, where $i+j \ne r+s, i+j \ne 0, r+s \ne 0$. If it is symmetric, then $$\;\eta_1 X^iY^j + \eta_2 X^rY^s=\;\eta_1 X^jY^i + \eta_2 X^sY^r$$

This means either $X^iY^j=X^jY^i,X^rY^s=X^sY^r$ or $\eta_1=\eta_2, X^iY^j=X^sY^r, X^jY^i=X^rY^s$. Therefore, in the first case, $i=j, r=s$, and in the second case, $i=s,j=r$. Hence, all symmetric polynomials with two non-constant terms are of the form $\eta X^iY^i+\omega X^jY^j+c \;\text{or}\; \gamma (X^iY^j+X^jY^i)+c$.

Now, $\mu X^iY^i$ and $\eta X^iY^i+\omega X^jY^j$ can clearly be generated by $XY$, so we only have to check $\gamma (X^iY^j+X^jY^i)\;,$ where $i\ne j$.

Without loss of generality, suppose that $i\gt j$, so we have $$\gamma (X^iY^j+X^jY^i)=\gamma X^jY^j(X^{i-j}+Y^{i-j})$$. Now we only have to check if $$X^{i-j}+Y^{i-j}$$ can be generated by $X+Y$ and $XY$.

Let $i-j=q$. When $q=1$, $X^q+Y^q$ can trivially be generated by $X+Y$, since it's itself $X+Y$. When $q=2, X^q+Y^q=X^2+Y^2=(X+Y)^2-2XY,$ so $X^q+Y^q$ can be generated by $X+Y$ and $XY$ as well. Notice that


Therefore, $X^{q+1}+Y^{q+1}$ can be generated by $X+Y,XY,X^{q-1}+Y^{q-1},$ and $X^q+Y^q$. Hence, by induction,$\forall{q\in {\mathbb N}}, X^q+Y^q$ can be generated by $X+Y$ and $XY$. This means

$$\gamma(X^iY^j+X^jY^i)$$ can be generated by $X+Y$ and $XY$, hence all symmetric polynomials $f(X,Y)$ can be generated by $X+Y$ and $XY$.

All polynomials of degree n generated by $X+Y$ and $XY$ are of the form $$\sum_{0 \le i'\le n,0\le j' \le n,i'+j'\le n} \kappa_{i'j'}(X+Y)^{i'}(XY)^{j'}$$

Each term of this sum is symmetric, so the sum is symmetric as well. Hence, every polynomial generated by $X+Y$ and $XY$ is symmetric, and every symmetric polynomial can be generated by $X+Y$ and $XY$.


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