Constructing the sequence: $0\rightarrow (x-y)^{S_2} \stackrel{f}{\rightarrow} k[x+y,xy]\stackrel{g}{\rightarrow} k[y]$

Question

Let $S_2$, a group of two elements, act on $k[x,y]$ by permuting $x$ and $y$.

It is clear that $$ 0\rightarrow (x-y) \rightarrow k[x,y]\rightarrow \dfrac{k[x,y]}{(x-y)}\cong k[y] \rightarrow 0 $$ is exact.

Taking its invariant subrings, we obtain $$ 0\rightarrow (x-y)^{S_2} \rightarrow k[x,y]^{S_2}\rightarrow k[y]^{S_2},$$ which simplifies as

$$ 0\rightarrow (x-y)^{S_2} \stackrel{f}{\rightarrow} k[x+y,xy]\stackrel{g}{\rightarrow} k[y]. $$

What are $f$ and $g$ concretely?

Answer 1

The elements of the ideal $(x-y)$ are of the form $u=p(x,y)(x-y)$, and $p$ is antisymmetric if and only if the element $u$ is $S_2$-invariant. As such, $u$ being a symmetric polynomial, it can be written as a polynomial in the elementary symmetric polynomials $e_1(x,y)=x+y$ and $e_2(x,y)=xy$, hence we can let $f$ simply be the inclusion map. Just as before, $g$ is the evaluation map $x,y\mapsto y$.

Answer 2

$f$ is symmetrization, taking $P(x,y)$ to $P(x,y) + P(y,x)$ (or the same divided by 2).

$g$ is restriction to the diagonal, taking $Q(x,y)$ to $Q(x,x)$.

Edit: if you mean $(x-y)$ as an ideal and not the polynomial $f(x,y)=x-y$, then the other answer is what you want.