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Chevalley's theorem says that if $k$ is a finite field and $f(X_1,...,X_n)$ is a form (homogeneous polynomial) of degree $d < n$, then the equation $f(X_1,...,X_n) = 0$ has a non-trivial solution in $k^n$.

It is known that this result is optimal, in the sense that for each $n$ there exists a form $f(X_1,...,X_n)$ - coming from a norm - of degree $d = n$ which has only the trivial zero. See Brian Conrad's answer below.

I am interested in each form with this property, i.e. each form $f(X_1,...,X_n)$ of degree $d = n$ which has only the trivial zero. What are the known examples/classes of such forms? It would be very nice if we could describe/classify them all...

I am in particular interested in the case of quartic forms.

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I think that the Fermat quartic in $\mathbb{P}^3$ is an example of a form without zero with $n=d=4$ having no zeros over $\mathbb{F}_5$. If I remember correctly, it is the only diagonal quartic over a finite field admitting no solution. If this is correct, then Martin Bright told me this, otherwise I am wrong! Note that the Fermat quartic is non-singular, and hence it is geometrically irreducible. – damiano Mar 17 '10 at 9:53

It is optimal for every $d > 0$. One way to make examples is using extension fields. For any $d > 0$, let $k'/k$ be an extension of degree $d$ and consider the norm map $N:k' \rightarrow k$. Choose a $k$-basis of $k'$, this is expressed as a homogeneous polynomial $f$ of degree $d$ in $d$ variables over $k$. More specifically, if $A$ is any $k$-algebra then the norm map $N_A:k' \otimes_k A \rightarrow A$ is given by evaluation of $f$ on $d$-tuples from $A$. This has no nonzero $k$-rational zeros, since nonzero elements of $k'$ have nonzero norm in $k$.

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Thanks for the reply. Yes, I know that it is optimal for every $d > 0$, and norms are the obvious examples to show this - I should have mentioned that in the question. But I'm rather interested in "other types" of examples, whatever that may mean. – Wanderer Mar 17 '10 at 1:20
You're asking that the projective hypersurface $(f=0)$ of degree $d$ in projective $(d-1)$-space has no $k$-rational points. If $k$ has sufficiently large size purely in terms of $d$ then the RH bound provides rational points if this hypersurface is smooth and geometrically connected. So if we restrict attention to irreducible $f$ over $k$ then it seems inevitable for it to be a $k^{\times}$-multiple of a norm of a lower-degree homogeneous form over a nontrivial finite extension of degree dividing $d$, or that $(f=0)$ has singularities. – BCnrd Mar 17 '10 at 2:02
@Brian: inevitable only if $\# k \gg d$, you mean, right? Certainly there are smooth counterexamples when $d$ is large compared to $\# k$ and $n$. – Pete L. Clark Mar 17 '10 at 3:45
Yes Pete, I was only considering d=n as in the question, and so I was only pointing out that if we want to make constructions that include most finite fields and use a fixed d then blah-blah-blah... – – BCnrd Mar 17 '10 at 5:40
@Brian: not to belabor the point, but you're not suggesting that this is the case even for $n = d \gg \# k$, right? I ask because (i) this is relevant to the original question and (ii) I am interested myself. – Pete L. Clark Mar 17 '10 at 13:28

Perhaps it is obvious to most readers, but about a year ago I spent several days trying to determine for which pairs (d,n) there existed an anisotropic degree d form in n variables over a finite field $\mathbb{F}_q$. The question was motivated by Exercise 10.16 in Ireland and Rosen's classic number theory text: "Show by explicit calculation that every cubic form in two variables over $\mathbb{F}_2$ has a nontrivial zero."

As many students have discovered over the years, this is false: e.g. take

$f(x_1,x_2) = x_1^3 + x_1^2 x_2 + x_2^3$.

I knew about the existence and anisotropy of norm hypersurfaces for all $n = d$. But what about $n < d$? I confess that I spent some time proving this result in several special cases and even dragged a postdoc into it. Here is a copy of the sheepish email I sent out (in particular to Michael Rosen) later on:

If K is a field, and f(x_1,...,x_n) is an anisotropic form of degree d in n variables, then f(x_1,...,x_{n-1},0) is an anisotropic form of degree d in n-1 variables.

So let K be any field which admits field extensions of every positive degree d. Then for all d there is an anisotropic norm form N in d variables of degree d. For any n < d, setting (d-n) of the variables equal to 0 gives an anisotropic form of degree d in n variables. In particular, this proves "the converse of Chevalley-Warning".

So, not so fascinating after all, then.

I think it is still nontrivial to ask what happens if the hypersurface f is required to be geometrically irreducible. For instance, despite the fact that (q,3,3) is anisotropic, every geometrically irreducible cubic curve over a finite field has a rational point.

AS's question about classifying anisotropic hypersurfaces with $d = n$ is interesting. It may also be interesting to look at the case $d < n$. It is certainly not clear to me that all such anistropic hypersurfaces come from intersecting a norm hypersurface of larger dimension with a linear subspace.

I also want to add that the following generalization seemed less trivial to me (and I still don't know the answer): Chevalley-Warning is also true for sytems of polynomial equations $f_1(x_1,\ldots,x_n) = \ldots = f_r(x_1,\ldots,x_n)$ so long as the sum of the degrees of the $f_i$'s is strictly less than $n$. What kind of counterexamples can we construct here when $d = d_1 + \ldots + d_r \geq n$?

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The third term of your cubic form should be x_2^3. – Franz Lemmermeyer Mar 17 '10 at 13:37
Thanks. $ $ – Pete L. Clark Mar 17 '10 at 14:14
Your last question is interesting, but doesn't the classical norm example still work? Just take norm forms - in disjoint sets of variables - for extensions of degree $d_1,d_2,\,\cdots,d_r$ where $d_1 + d_2 + \cdots + d_r = n$. – Wanderer Mar 17 '10 at 14:19
@AS: Yes, it does. Thanks for pointing this out. Amusingly the case with $d_i = 1$ for all $i$ is given in (and is the only anisotropic example given there), but it did not inspire me to think of the generalization to simultaneous norm equations. Again, looking for geometrically integral examples seems more interesting... – Pete L. Clark Mar 17 '10 at 20:27

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