# Symmetric polynomials in two variables

Let $R$ be a commutative ring. The fundamental theorem of symmetric polynomials says that any symmetric polynomial in $R[X,Y]$ can be expressed uniquely as a polynomial in $X+Y$ and $XY$.

Recently I was thinking about this along the following lines. Let $R[X,Y]^{S_2}$ denote the set of all symmetric polynomials in $R[X,Y]$. Then the theorem above is saying that $R[X,Y]^{S_2}$ is generated by $\{X+Y,XY\}$ as an $R$algebra, i.e., $R[X,Y]^{S_2}=R[X+Y,XY]$. Being unsuccessful at utilising this I ended up with the following. (I can’t see the best way of showing that a set generates an algebra.)

Let $f(X,Y)=\sum_{i,j}a_{i,j}X^iY^j\in R[X,Y]^{S_2}$. Since $f(X,Y)=f(Y,X)$, we have $a_{i,j}=a_{j,i}$ for all $i,j$. Hence $f(X,Y)=\sum_{i\le j}a_{i,j} (XY)^i(X^{j-i}+Y^{j-i})$. So it suffices to show that $X^n+Y^n$ can be expressed as a polynomial in $X+Y$ and $XY$ for each $n\ge 0$. But this follows easily by induction and the following identity

$X^n+Y^n=(X+Y)(X^{n-1}+Y^{n-1})-XY(X^{n-2}+Y^{n-2})$.

However, this argument doesn’t generalise immediately to more variables, and I don’t particularly like any of the proofs that I’ve found so far.