I believe the question below should be fairly standard in invariant theory ; I hope someone more familiar with it than me can explain a bit more or point to a reference.

Let $F$ be polynomial field $F={\mathbb R}(X_1,X_2,X_3, \ldots ,X_8)$. Denote by $S$ the group of permutations on $\lbrace X_1,X_2,X_3, \ldots ,X_8\rbrace$, which acts on $F$ in the usual way. Consider the two permutations $\gamma$ and $\tau$ in $S$ defined by

$$ \gamma=(X_1,X_3,X_5,X_7) (X_2,X_4,X_6,X_8) , \ \ \ \tau=(X_1,X_2) (X_3,X_4) (X_5,X_6) (X_7,X_8) $$

These two commute and thus generate a subgroup of $S$ isomorphic to $\frac{\mathbb Z}{2\mathbb Z} \times \frac{\mathbb Z}{4\mathbb Z}$. Let us denote this subgroup by $D$. Let $I$ be the subfield of invariants under the action of $D$ in $F$. Clearly, $I$ contains the fully symmetric polynomials and also contains the polynomials of the form :

$$ T_f=\sum_{k=1}^{4}f(X_{2k-1},X_{2k})+f(X_{2k},X_{2k-1}) $$ (where $f$ is any bivariate polynomial).

Now, do the fully symmetric polynomials plus all the $T_f$'s suffice to generate the whole of $I$ ?

  • 1
    $\begingroup$ The OP initially claimed mistakenly that $D$ was a dihedral group, I corrected that. $\endgroup$ – Ewan Delanoy Nov 14 '12 at 6:17

The answer is NO. If we denote by $I’$ the field generated by the fully symmetric polynomials plus all the $T_f$’s, then $I’$ is invariant by the transposition $\theta=(X_7,X_8)$. If we consider

$$ g=X_1X_5+X_2X_6+X_3X_7+X_4X_8 $$ then $g$ is fixed by $\gamma,\tau$ but not by $\theta$, so $g\in I \setminus I'$ and hence $I \neq I'$.


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