Let $n$ be a positive integer and $W_n$ be the linear subspace of the real vector space $\mathbb{R}[X_1,\cdots,X_n]$ generated by the following set $$S_n=\{X_1^{i_1}\cdots X_n^{i_n}:i_1+\cdots+i_n=n\ \text{and}\ i_k\in \{0,1,2\},k=1,\cdots,n\}.$$ Let $$T_n=\{(X_{i_1}-X_{i_2})(X_{i_2}-X_{i_3})\cdots(X_{i_n}-X_{i_1}):i_1,\cdots,i_n\ \text{is a permutation of}\ 1,\cdots,n\}.$$ It is easy to see that $T_n\subset W_n$.

When $n$ is odd, the expansion of any polynomial in $T_n$ does not contain the monomial $X_1X_2\cdots X_n$. So $T_n$ can not generate $W_n$.

When $n=2m\geq 8$ is even, we have $$|S_n|=\sum\limits_{k=0}^m\binom{n}{2k}\binom{2k}{k}< \dfrac{1}{2}(n-1)!=|T_n|.$$ So I want to ask can $W_n$ generated by $T_n$ when $n\geq 8$ is even?


$\let\eps\varepsilon$Substitute $X_k\leftarrow\eps_k:=\exp(2\pi ki/n)$. Notice that $\arg(\eps_k-\eps_\ell)\equiv (\pi+\arg \eps_k+\arg\eps_\ell)/2\pmod{\pi}$. Summing up, we get that the argument of every element of $T_n$ is congruent to a constant modulo $\pi$; so all their real lnear combinations also share this property. On the other hand, the elements of $W_n$ may have $n$ distinct arguments. This works for all $n>2$, regardless of the parity.

NB. Surely, the fact is still true over $\mathbb C$, since the elements of $S_n$ are necessarily rational linear combinations of those in $T_n$, provided that they are complex such combinations.

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