I have a question concerning §3 of this article by Soergel (German). Consider a polynomial ring $R=k[x,y]$ where the symmetric group $S_2=\langle s\rangle$ acts by exchanging indeterminates
$$s\colon x\rightleftarrows y.$$
Denote the invariants by $R^s=k[x+y, xy]$ and let $\alpha_s = x-y. $ It is known that $R\cong R^s\oplus \alpha_s R^s$ as $R^s$-module. Consider the modules $$\begin{aligned}R⊗_{R^s} R &= R⊗_k R / (a⊗1-1⊗a)_{a∈R^s},\\ D_s &= R⊗_k R /\underbrace{(sa⊗1-1⊗a)_{a∈R}}_{=: I_s}\:\underbrace{(a⊗1-1⊗a)_{a∈R}}_{=: I_e}.\end{aligned}$$
Claim: There is an isomorphism $R⊗_{R^s} R \to D_s$; supposedly a homogeneous map.
Hints:
- The above decomposition $R\cong R^s\oplus \alpha_s R^s$,
- the short exact sequence $$ 0\to R⊗_k R /I_s \xrightarrow{\alpha_s⊗1-1⊗\alpha_s} R⊗_k R /I_sI_e\to R⊗_k R /I_e \to 0,$$
- If one sees that the desired map is surjective, the ismorphism follows from a graded rank argument.
Question: I do not know how to see that there is a map at all. What is it?