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Let $\Sigma_m$ be the permutation group on $m$ letters and $R=\mathbb{C}[x_1,x_2,\dots,x_m]$. Let $\Sigma_m$ act on $R$ in the usual way. Let $R^{\Sigma_m}$ denote the ring of $\Sigma_m$ invariant polynomials and $R^{\Sigma_r \times \Sigma_{m-r}}$ denote the ring of polynomials invariant under permutation of the first $r$ and the last $m-r$ variables. I found a claim in a reference that says $R^{\Sigma_r \times \Sigma_{m-r}}$ is a free $R^{\Sigma_m}$-module of free rank $\binom{m}{r}$. The reference claims that this is "well-known" but doesn't give a specific reference or cite a theorem.

1) Can anyone explain why this is true?

2) Is there any knowledge of the specific form of the generators?

3) Can we say anything similar over more general fields?

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All three questions are answered by Proposition (1.5) in Chapter PARTL ("Partially symmetric functions") in The LLPT Notes (by Dan Laksov, Alain Lascoux, Piotr Pragacz and Anders Thorup). (The notes are lacking a few sections, but they're nevertheless self-contained.)

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The generators for this are just of the forms $ \alpha _i \beta_j$. where $ \alpha_i$ is a Symmetric Polynomial in $R_1 =\mathbb{C}[x_1,_2..x_r]$ and $\beta_j$ is a symmetric polynomial of $ R_2 =\mathbb{C}[x_{r+1}...x_m]$. Now i hope it becomes easy to check the condition of linear independency of them as because $ \alpha_i$'s are basis of $R_1^{\Sigma_r}$ and $\beta_j$ 's are a basis of $R_2^{\Sigma_{m-r}}$ and $R$ is $R_1 \otimes R_2$.

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  • $\begingroup$ But how is the rank $\binom{m}{r}$? $\endgroup$ Feb 4, 2017 at 2:21

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