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

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?

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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 $x,y\mapsto y$.

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$f$ 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.

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Yes, I meant $(x-y)$ to be the ideal generated by $x-y$. What do you mean by the other answer? Do you mean anon's answer? –  math-visitor Jun 14 '12 at 6:55
Yes, after posting I saw anon answer with the different interpretation of (x-y). –  zyx Jun 14 '12 at 6:55