This is regarding theorem 3 in this article. My problems begin after the equalities denoted by (5). My problems aren't so much about theory really, I think. I'm disregarding the authors' notation a bit and write $A^{S_n}:=k[X_1,...,X_n]^{S_n}$.
note that $\mathscr{E}=E(\Psi_1,...,\Psi_e)$ for a suitable polynomial $E \in A^{S_n}[\Psi][T_2,...,T_e]$, symmetric in the variables $T_2,...,T_e$.
My first question is really just this: How can a polynomial in $e-1$ variables be mapped to the $e$ "values" $\Psi_i$, $i=1,...,e$? If we're saying that $E$ is in a ring to which is adjoined $\Psi=\Psi_1$, I'd say $\mathscr{E}=E(\Psi_2,...,\Psi_e)$ for a suitable $E \in A^{S_n}[\Psi][T_2,...,T_e]$. My suspicion is that what's in the article is a typo or some such. What do you think?
Second, I'm afraid I don't see at all how $E(T_2,...,T_e)$ so surely is symmetric; at least $\mathscr{E}=E(\Psi_2,...,\Psi_e)$ isn't (try $e=3$).
By the main theorem on symmetric polynomials, denoting $S_1,...,S_{e-1}$ the elementary symmetric polynomials in $\Psi_2,...,\Psi_e$, we get a polynomial $F\in A^{S_n}[\Psi][Y_1,...,Y_{e-1}]$ such that $E(\Psi_2,...,\Psi_e)=F(S_1,...,S_{e-1})$.
By said main theorem, if $E$ is symmetric in the $T_i$ ($\Psi_i$) then it can be written in terms of the elementary symmetric polynomials in the $T_i$ ($\Psi_i$). But what happens here? I have no clue. If, say, $f \in B[x_1,x_2]$ happens to be symmetric I'd say $f \in B[x_1,x_2]^{S_2}=B[x_1+x_2,x_1 x_2]$, but here they introduce yet another bunch of variables $Y_i$ and... I'm completely lost what they're doing. Could someone clarify what's going on?
(couldn't think of a more precise title)
Progress edit: When checking if $\mathscr{E}$ was symmetric in the $\Psi_i$ I didn't include the factor $\prod (\Psi_i - \Psi_j)$. I tried that now for $e=3,4,5$ and then $\mathscr{E}$ is indeed symmetric. I'll look into why that factor changes the picture, can't say I have any ideas on how to do that though...
Progress edit 2: The assertion is that $\mathscr{E}$ is symmetric in the $\Psi_i$. Now, the factor $\prod (\Psi_i - \Psi_j)$ is alternating, i.e. even permutations stabilize it. I then worked with the assumption that $\mathscr{D}$ ($\mathscr{E}=\delta_\Psi \mathscr{D}$) is alternating as well; if the assertion were true then $\sigma \mathscr{E}=\pm \delta_\Psi \sigma (\mathscr{D})=\mathscr{E}$, so $\mathscr{D}$ would have to be alternating.
I picked up my old linear algebra textbook but it wasn't immediately obvious to me what was key. But after a while, after many unnecessary calculations, I realized: Applying permutations to the determinant (i.e. $\mathscr{D}$) amounts to moving the columns around. One such move changes the sign (a result I'd forgotten), in other words, odd permutations change the sign, even permutations preserve it. Hence $\sigma \mathscr{E}=\sigma (\delta_\Psi) \sigma (\mathscr{D}) = sgn(\sigma) \delta_\Psi \cdot sgn(\sigma) \mathscr{D}=\delta_\Psi \mathscr{D}=\mathscr{E}$, qed.
Progress edit 3: I'm pretty sure my problem with the last sentence I quoted has to do with what I would say is something completely unnecessary; I think the introduction of the $Y_i$ is done simply to get the indices "right". I see it this way: $E$ is symmetric in the $\Psi_i$, $i=2,...,e$. Then $E$ can be expressed in terms of the elementary symmetric polynomials in the $\Psi_i$, which we denote by $S_i$, $i=1,...,e-1$. That the $i$'s assume different "values" is not confusing at all, we simply have $F\in A^{S_n}[\Psi][T_2,...,T_e]^{S_{e-1}}$, where $S_{e-1}=Sym(\{2,...,e\})$ rather than $Sym([e-1])$. We then go on to show that the $S_i$ lie in such and such and we're done.
Of course, I'm not 100% sure of this, so some input would be appreciated here.
