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Given a single-variable polynomial, we all know that the number of its roots is bounded in terms of its degree. A polynomial here is a polynomial with integer coeffecients, and a root of a polynomial is a real root.

Given a multivariate polynomial, this phenomenon, in its general form, stops working. However, by a well-known theorem by Bezout, we still can say that either it has an infinite number of roots, or it has only a finite number which is bounded above in terms of its degree.

My questions is about the number of integer roots: Does there exist a functions $B=B_n:\mathbb N\rightarrow \mathbb N$ such that given a polynomial $P(x_1,...,x_n)$ of degree $d$, we can say that the set of INTEGER roots of $P$ contains at most $B(d)$ elements, or otherwise it is infinite?

Thanks in advance.

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  • $\begingroup$ Sorry, what's the specific reference for that theorem of Bezout? I'm not familiar with that version. $\endgroup$
    – KReiser
    Jun 25 at 18:11
  • $\begingroup$ It considers a system of polynomial equatinons, and claims that the number of solutions, if finite, then bounded in terms of the degrees. Bézout's theorem $\endgroup$
    – Al-Hasan Ibn Al-Hasan
    Jun 25 at 18:46
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    $\begingroup$ I'm familiar with that form of Bezout, but I don't see how you're getting the claim that a multivariable polynomial has either infinitely many integer roots or finitely many integer roots and the number of such roots is bounded above by a function of the degree. $\endgroup$
    – KReiser
    Jun 25 at 18:50
  • $\begingroup$ Given a multivariate polynomial $P(x_1,...,x_n)$ of degree $d$ and with a finite number $m$ of zeros, I applied Bezout's theorem on the system consisting of $n$ copies of the equation $P(x_1,...,x_n)=0$. Since the solutions of this system are those of $P$, I concluded that $m\leq d^n$. But I realized now that this is actually a wrong use of the theorem, because the solutions of the system has to be counted over the algebraic closure, which is the complex numbers in this case. But then, $P$ must have an infinite number of zeros (assuming $n>1$). $\endgroup$
    – Al-Hasan Ibn Al-Hasan
    Jul 1 at 5:29
  • $\begingroup$ There's more wrong with your attempt to use Bezout there than just needing to work over the algebraic closure - taking $n$ copies of $P=0$ doesn't do anything new past considering $P=0$! Bezout tells you information about nontrivial intersections, but the scenario you describe is far from a nontrivial intersection. And yes, it is well-known that $P$ must have infinitely many zeroes (over an algebraically closed field) when $n>1$ - the interesting, tricky part is how many of those are integral (or belong to some smaller field, etc). Glad you got a good answer below! $\endgroup$
    – KReiser
    Jul 1 at 5:42

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There is no such bound.

Given an equation $X$ over $\mathbf{Q}$ with infinitely many rational points, you can always scale the coefficients so that any finite subset of the rational points all become integral points. (Put them all under a common denominator $N$, then replace each variable $x$ by $x/N$.) Take an elliptic curve $E$ with positive rank, that is, $E(\mathbf{Q})$ is infinite. (They exist: for example, $E: y^2 = x^3 - 2$.) After scaling, you can find an elliptic curve with as many integral points as you like. (You can take $y^2 = x^3 - 2 N^6$ for the greatest common denominator $N$ of the finite set of rational points.) If such a bound in your question existed, then, since these curves all have the same degree, at some point these scaled elliptic curve would have to have infinitely many integral points. But $E(\mathbf{Z})$ is always finite for any elliptic curve by a theorem of Siegel.

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