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 Ralgebra, 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


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.


Leave a comment

Filed under Algebra

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s