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Let $A_k$ be a set of polinomials $f(x_1,x_2,\ldots,x_n)$ in $\mathbb{Q}[x_1,x_2,\ldots,x_n]$ symmetric on $k<n$ variables: $$f(x_{\sigma(1)},\ldots, x_{\sigma(k)},x_{k+1}, \ldots,x_n)=f(x_1,x_2,\ldots,x_n),\,\, \sigma \in S_k$$

$A_k$ is a $\mathbb{Q}-$ vector space. What is the basis or dimension of $A_k$?

Silvio Sandro
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  • $A_k$ is simply $S_k \otimes \mathbb{Q}[x_{k+1} \dots, x_n]$, where $S_k\subset \mathbb{Q}[x_1, \dots, x_k]$$ is the ring of symmetric polynomials. It therefore has a infinite dimension as a vector space, but the relation above gives you the dimension of each graded part. – anomaly Jun 23 '14 at 15:37

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The dimension is $\infty$. Note e.g. that $\sum_{j=1}^k x_j^p$ is in $A_k$ for all positive integers $p$.

But perhaps you meant to consider polynomials of fixed degree?

Robert Israel
  • 448,999
  • I'm sorry. Every polynomial in $A_k$ have fixed degree $n$. I think, $dim_{\mathbb{Q}}A_{k}$ at least $p(n)$, the number of partitions of $n$. – Silvio Sandro Jun 23 '14 at 15:42