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I came across the following system of polynomial equations on $X_1,\dots,X_{m-2}$: $$ \begin{cases} 2X_{2s}+\sum\limits_{t=1}^{2s-1}(-1)^tX_tX_{2s-t}=0,\quad s=1,\dots,\frac{m}{2}-1,\\ X_sX_{m-s}+(-1)^s=0,\quad s=2,\dots,\frac{m}{2},\\ X_1X_{\frac{m}{2}}-2X_{\frac{m}{2}+1}=0, \end{cases} $$ where $m\ge10$ is an even integer.

By Groebner basis computation, I verified up to $m=20$ that $1$ is in the ideal generated by the above polynomials in $\mathbb{Q}[X_1,\dots,X_{m-2}]$, whence the system has no solution in $\mathbb{C}^{m-2}$. Also, the number of equations is one more than the number of variables. Thus it is reasonable to conjecture that the system has no solution in $\mathbb{C}^{m-2}$ for any even $m\ge10$, and so the question arises naturally as how to prove this.

Remark: For a given value (not too lagre) of $m$, there are ways such as Groebner basis and various kinds of multipolynomial resultants to show that the system has no solution. But it seems to me that these algorithmic ways give not much insight for general $m$.

A further question: If we have known that the system has no solution in $\mathbb{C}^{m-2}$, we immediately deduce that it has no solution in $\mathbb{F}_p^{m-2}$ for all sufficiently large prime $p$. However, how can I then get a bound $N$ such that the system has no solution in $\mathbb{F}_p^{m-2}$ for prime $p>N$? I guess a bound like $p>m$ holds, but achieving this would be ad hoc to the equations.

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  • $\begingroup$ In your first equation, do you mean $2X_{2s}$? $\endgroup$
    – Walter Neff
    Commented Sep 20, 2014 at 17:59
  • $\begingroup$ @B.Wellington The first equation is $2X_2-X_1^2=0$, so $X_2=\frac{X_1^2}{2}$. In fact, the first line of equations shows that $X_{2t}$ is a polynomial of $X_1,\dots,X_{2t-1}$ for $1\le t\le\frac{m}{2}-1$. $\endgroup$
    – Binzhou Xia
    Commented Sep 21, 2014 at 2:19
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    $\begingroup$ Call the variety $V(m)$. Have you tried looking for morphisms $V(m) \to V(d)$ for divisors $d$ of $m$? Maybe $m$ will have to be shifted to $m - 1$ or $(m - 1)/2$ or something. What's the story behind this variety? Sometimes that gives rise to natural morphisms. $\endgroup$
    – James Weigandt
    Commented Sep 21, 2014 at 3:04
  • $\begingroup$ @B.Wellington Thanks to your comment. I've changed $2X_{2t}$ to $2X_{2s}$ now. $\endgroup$
    – Binzhou Xia
    Commented Sep 21, 2014 at 3:33
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    $\begingroup$ For the "further question", even $p > m^2$ is likely way too optimistic. One can easily write linear equations in $m$ variables, with no coefficient larger than $2$, that have no solution in $\bf C$ but do have a solution modulo some prime $p \sim 2^m$. (Use the binary expansion of $p$ to build up to $x_m=p$, and then add the equation $x_m=0$.) $\endgroup$
    – Noam D. Elkies
    Commented Sep 21, 2014 at 4:22

1 Answer 1

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For the Further question, See Pascal Koiran's paper (which, I believe, is the last word in the subject, even if almost 20 years old).

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    $\begingroup$ For a given $m$, provided we have know that the system has no solution over $\mathbb{C}$, this paper says that the number of primes $p$ such that the system has solution over $\mathbb{F}_p$ is bounded. But does this provide a concrete bound $N$ such that the system has no solution whenever $p>N$? $\endgroup$
    – Binzhou Xia
    Commented Sep 24, 2014 at 11:15
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    $\begingroup$ Hi, this link is broken as of 2024 $\endgroup$
    – FShrike
    Commented Sep 10 at 9:06

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