I encountered the present question when investigating that other recent question of mine.

Let $x_1,x_2, \ldots, x_8$ be indeterminates. Let $s_1,s_2, \ldots s_n$ denote the elementary symmetric polynomials (so that $s_1=\sum x_i, s_2=\sum_{i<j}x_ix_j$ etc. Let us consider also

$$ \begin{align} g_1 &= x_1x_2+x_3x_4+x_5x_6+x_7x_8 \\ g_2 &= x_1x_3+x_1x_7+x_2x_4+x_2x_8+x_3x_5+x_5x_7+x_4x_6+x_6x_8 \\ g_3 &= x_1x_4+x_1x_8+x_2x_3+x_2x_7+x_4x_5+x_5x_8+x_3x_6+x_6x_7 \\ g_4 &= x_1x_5+x_2x_6+x_3x_7+x_4x_8 \\ g_5 &= x_1x_6+x_2x_5+x_3x_8+x_4x_7 \end{align} $$

If we make act the group $S$ of permutations of $\lbrace x_1,x_2, \ldots, x_8\rbrace$ on polynomials in the usual way, one can compute the subgroup $T$ of permutations fixing $g_1$ and $g_2$ (in particular, it has 16 elements). One can also check that any element of $T$ fixes $g_3$. So by the Galois correspondence, we have a polynomial $A$ with rational coefficients such that

$$ g_3=A(s_1,s_2, \ldots ,s_n,g_1,g_2) $$

How can we compute $A$ ?

  • $\begingroup$ So $x_2 x_8$ has a coefficient of $2$ in $g_2$? Or was the second appearence of $x_2x_8$ meant to be $x_6x_8$? $\endgroup$ – Jeff Tolliver Nov 14 '12 at 7:23
  • $\begingroup$ @JeffTolliver : we have $s_2=g_1+g_2+g_3+g_4$, of course. Typo corrected, thanks. $\endgroup$ – Ewan Delanoy Nov 14 '12 at 8:15
  • $\begingroup$ I think you forgot about $g_5$ $\endgroup$ – mercio Nov 14 '12 at 8:32
  • $\begingroup$ @mercio : indeed, but that’s not really important, as $g_3,g_4$ and $g_5$ are not needed in the question. $\endgroup$ – Ewan Delanoy Nov 14 '12 at 9:08

A short useless answer : try to write $g_3 = P/Q$ with $P$ homogeneous of degree $d+2$ and $Q$ of degree $d$ for $d=1$. If that fails, try again with $d=2$, and so on.

Or you can try to directly guess a good degree so that it will be solvable. Let $R = K[x_1 \ldots x_8]$, denote $R_d$ the subspace of $R$ of homogeneous polynomials of degree $d$, let $R^G$ the subring of $R$ of elements fixed by a subgroup of $S_8$, $R^G_d = R^G \cap R_d$, and $R^G[y]$ be the subring generated by $R^G$ and $y$, and $R^G[y]_d = R^G[y] \cap R_d$.

Let $G$ be the subgroup of order $16$ fixing $g_1$ and $g_2$. Then $R^{S_8}[g_1,g_2] \subset R^G$. For small $d$, the inclusion is strict but Galois theory implies that for some $d$ large enough you will have $R^{S_8}[g_1,g_2]_d = R^G_d$.

The goal is to find two nonzero elements $P \in R^{S_8}[g_1,g_2]_d, Q \in R^{S_8}[g_1,g_2]_{d+2}$ such that $P g_3 = Q$. We know $g_3 \in R^G_2$ thus $P g_3 \in R^G_{d+2}$. The multiplication by $g_3$ map $R^{S_8}[g_1,g_2]_d \to R^G_{d+2}$ is always injective, so its image is a subspace of dimension $\dim_K(R^{S_8}[g_1,g_2]_d)$. If $\dim_K(R^{S_8}[g_1,g_2]_d) + \dim_K(R^{S_8}[g_1,g_2]_{d+2}) > dim_K(R^G_{d+2})$, the intersection of those two is guaranteed to be nonzero, so you will obtain nonzero elements $P,Q$ as wanted.

So first you have to compute all the numbers $\dim_K (R^G_d)$ and $\dim_K (R^{S_8}[g_1,g_2]_d)$, find the smallest integer $d$ such that the inequality holds, then solve an extremely big system of linear equations. Since those dimensions grow like $d^8/8!16$ you should try to pick the smallest $d$ possible. Of course you may not get the smallest polynomials possible, it may happen that there was a linear combination for a smaller degree, but you can't know a priori that it exists.

Finding $\dim_K (R^G_d)$ is a combinatorics problem. It shouldn't be too difficult here, just count the orbits of $R_d$. For $\dim_K (R^{S_8}_d)$ since it is generated by $n$ polynomials, you get it easily with generating functions. For $\dim_K (R^{S_8}[g_1,g_2]_d)$, I'm not certain what the best way is, it doesn't seem too friendly. You don't need to compute it exactly, a good enough lower bound will work too.

You can compute $\dim_K (R^{S_8}[g_1]_d)$. $g_1$ is the root of a degree $105$ polynomial with coefficients in $R^{S_8}$, thus $\dim_K (R^{S_8}[g_1]_d) = \sum_{0 \le i < 105} \dim_K (R^{S_8}_{d-i})$. Next, $g_2$ is a root of a degree $24$ polynomial with coefficients in the fraction field of $R^{S_8}[g_1]$, which gives you a nice enough lower bound for $\dim_K (R^{S_8}[g_1,g_2]_d)$ (you should obtain something equivalent to $d^8/8! 16$)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.