# $n$-forms representing zero (versus division rings)

• 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 $Q_p$.

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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.

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+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.

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