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I was wondering whether there exists any result of the form

"if $f \in \mathbb{Z}[x_1, ..., x_k]$ is a polynomial (not form! I don't require homogeneity) of total degree $n$, with $k \geq \delta n$ for some explicit $\delta$, then $f(x) = 0$ satisfies the Hasse principle."

I'm particularly interested in the case $n = 4$. Thanks!

EDIT: I'm interested in integral solutions.

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    $\begingroup$ Are you interested in rational solutions or integral solutions? $\endgroup$
    – Daniel Loughran
    Commented Sep 27, 2014 at 12:26
  • $\begingroup$ Integral solutions (sorry I forgot to mention it) $\endgroup$
    – user58702
    Commented Sep 27, 2014 at 13:49
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    $\begingroup$ Ok, this makes the problem quite a bit more difficult. Let me just mention that the answer to your question as stated is no (even in the case of homogeneous forms). Here the circle method is usually used to tackle such problems, and when it works it yields a bound for $k$ which his exponential in the degree $n$. Whereas you seem to be looking for a linear bound on the degree. Such bounds are not known in general. Of course if you are only interested in the case $n=4$, then this is not so much a problem. $\endgroup$
    – Daniel Loughran
    Commented Sep 27, 2014 at 14:09
  • $\begingroup$ My naive guess is that the circle method should be able to give the result you want when $n=4$. However you are right that people normally consider homogeneous forms as they are easier, and I don't know whether the result you want has been worked out. Have you already tried looking at the circle method literature to see if you can find the result you want? $\endgroup$
    – Daniel Loughran
    Commented Sep 27, 2014 at 14:12
  • $\begingroup$ Yes, I've tried looking at the literature a bit, but all seems geared towards forms instead of just polynomials. Do you know any references which might deal with a case similar to mine (i.e. with polys instead of forms)? Thanks! $\endgroup$
    – user58702
    Commented Sep 27, 2014 at 16:39

1 Answer 1

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The integral Hasse principle can fail for polynomials of degree $2$ (or $4$), even if the number of variables is huge! For example, if $k \equiv 1 \pmod{8}$, then $$(2x_1-1)^2 + \cdots + (2x_k-1)^2 = 1$$ has the solution $(1,1/2,\ldots,1/2)$ over $\mathbf{R}$ and $\mathbf{Z}_p$ for all odd $p$, and has a solution $(a,1,\ldots,1)$ over $\mathbf{Z}_2$ for some $a$ near $1$ by Hensel's lemma, but it has no solution over $\mathbf{Z}$ for $k>1$ since each square on the left is at least $1$.

For a much more detailed study of the integral Hasse principle, see Colliot-Thélène and Xu, Brauer-Manin obstruction for integral points of homogeneous spaces and representation by integral quadratic forms, Compos. Math. 145 (2009), no. 2, 309–363.

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  • $\begingroup$ If $f(x,y)=0$ is a counterexample in two variables then $f(x_1+\ldots+x_n,y)=0$ is also. $\endgroup$
    – Felipe Voloch
    Commented Dec 29, 2014 at 0:06

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