Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let me start with two observations.

  • In the classification of quadratic forms with rational coefficients, one has the following statement: a quadratic form in five indeterminates represents $0$ over $\mathbb Q$ if and only if it represents $0$ over $\mathbb R$.
  • For every $n\ge2$, there exists a division ring $R_n$ of dimension $n^2$ over its center $\mathbb Q$. There is a 'norm' over $R_n$, which is an $n$-form (homogeneous polynomial of degree $n$) with rational coefficients, multiplicative over $R_n$. It is an example of $n$-form in $n^2$ indeterminates that represents $0$ over $\mathbb R$ but does not over $\mathbb Q$.

Question. Is it true that for every $n\ge2$, an $n$-form with rational coefficients in $n^2+1$ indeterminates represents $0$ over $\mathbb Q$ if and only if it represents $0$ over $\mathbb R$ ?

A positive answer would say that the norm in a division ring is a maximal (in terms of the number of indeterminates) example of an $n$-form representing zero over $\mathbb R$ but not over $\mathbb Q$.

Edit. After Laurent's answer, I see that my question was close to Artin's conjecture, that Terjanian's example disproves. However, for fixed degree $n$, Artin's conjecture is true for almost every prime $p$: every $n$-form in $n^2+1$ indeterminates represents zero over $Q_p$.

share|improve this question

2 Answers 2

up vote 10 down vote accepted

Take Terjanian's example of a form of degree 4 in 18 variables over $\mathbb{Q}$ without nontrivial zero over $\mathbb{Q}_2$ (hence over $\mathbb{Q}$), as explained e.g. in Serre, Cours d'arithmétique, chap. 4. It is easy to see that this form represents zero over $\mathbb{R}$: in fact it is negative at the point (1,0,0,...) and positive at (1,1,1,0,0,...). Hence you get a 17-variable example by setting the last variable equal to zero.

share|improve this answer
1  
+1. Formidable. –  Chandan Singh Dalawat Jan 3 '11 at 12:59

You are asking about a venerable and active area in Diophantine equations lying at the border of arithmetic geometry and analytic number theory. Some keywords are forms in many variables and circle method.

Let $P(x_1,\ldots,x_n) \in \mathbb{Z}[x_1,\ldots,x_n]$ be a form (i.e., homogeneous polynomial) of degree $d$. As you say, if $n \leq d^2$, then there are very natural examples of forms without nontrivial integral solutions: e.g., for each $d \geq 2$, take the reduced norm form on a central division algebra of degree $d^2$. (One might express this in shorthand by saying that $\mathbb{Q}$ is not $C_2(d)$ for any $d \geq 2$.)

Note that when $d$ is odd, solutions over $\mathbb{R}$ are guaranteed, so one may simply ask whether taking $n \gg d$ is enough to guarantee the existence of nontrivial $\mathbb{Z}$-solutions. The answer is a resounding yes: this is a celebrated $1957$ theorem of B.J. Birch. Analogous results are known with $\mathbb{Q}$ replaced by any number field $K$: if $K$ has no real places, then the condition that $d$ is odd may be omitted.

So the problem becomes a quantitative one: for (say) odd $d$, just how large must $n$ be compared to $d$ to ensure that a form $P$ necessarily has a nontrivial $\mathbb{Z}$-solution? For instance, taking $d = 3$, we know that $n$ must be at least $10$ and in fact it is conjectured that this is the sharp answer. But we are very far from being able to prove this: in celebrated work of the 1960's, Davenport showed that one may take $n \geq 16$. In a 2007 Inventiones paper, Heath-Brown improved this to $n \geq 14$.

The classical reference for this material is M.J. Greenberg's little book Lectures on Forms in Many Variables. But this was written in the late 1960's and a lot of work has been done since then. A bit of googling finds some nice survey papers, e.g. this one by Trevor Wooley, who is one the current leaders in the field. Note in particular that on page $4$ of this document Wooley relates an example of Cassels and Guy which shows that when $d = 6$, there are forms with arbitrarily large $n$ and points over $\mathbb{R}$ (and also over $\mathbb{Q}_p$ for all $p$) but no nontrivial $\mathbb{Q}$-points. So the case of even $d$ is really different.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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