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It is basic that the norm map $N:\mathbf{F}_{q^n}^* \to \mathbf{F}_q^*$ is surjective for finite fields. In fact $N(x) = x^{(q^n-1)/(q-1)}$. How well does this simple fact extend to subspaces?

A basic example is an intermediate extension $\mathbf{F}_{q^d}$. On $\mathbf{F}_{q^d}^*$ we have $$N(x) = \left(x^{(q^d-1)/(q-1)}\right)^{(q^n-1)/(q^d-1)} = \left(x^{(q^d-1)/(q-1)}\right)^{n/d}$$ since the term in the brackets is in $\mathbf{F}_q^*$ and $(q^n-1)/(q^d-1) \equiv n/d \pmod {q-1}$. So $N$ is surjective on $\mathbf{F}_{q^d}^*$ if and only if $(n/d, q-1) = 1$. In particular $N$ fails to be surjective on a subspace of dimension $n/2$ whenever $n$ is even and $(n/2, q-1) > 1$.

As a sort of converse note that if $(n,q-1)=1$ then $N$ is surjective on every one-dimensional subspace.

Is it true that if $V \leq \mathbf{F}_{q^n}$ is a $\mathbf{F}_q$-rational subspace of dimension $>n/2$ then $N$ is surjective on $V$?

Equivalently, if $\dim_{\mathbf{F}_q} V > n/2$, can we always find $x^{q-1} \in V$?

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  • $\begingroup$ When you say "In particular $N$ fails to be surjective on a subspace of dimension $n/2$ whenever …" you mean "when …, there is some subspace of dimension $n/2$ on which $N$ is not surjective", right? (I.e., the quantification is existential, not universal?) $\endgroup$ – LSpice Jul 16 '18 at 2:03
  • $\begingroup$ @LSpice Yes, that's what I meant $\endgroup$ – Sean Eberhard Jul 16 '18 at 5:12
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The number of elements with norm $a$ is $$\frac{1}{q-1} \sum_{\chi: \mathbb F_q^\times \to \mathbb C^\times} \sum_{x \in V} \chi(Nx) \overline{\chi(a)}$$ The summand vanishes unless $\chi$ has order dividing $n$ so there are at most $gcd(n,q-1)$ terms. One of the terms has size $q^{\dim V}$ so it is sufficient that the other terms have size $<q^{ \dim V} / (\gcd (n,q-1)-1)$.

Each term is an average of $q^{n-\dim V}$ different Gauss sums associated to the additive characters that vanish on $V$ and hence has size $\leq q^{n/2}$.

So it suffices that $q^{ \dim V -n/2} > \gcd(n,q-1)-1$. This is automatically satisfied for $\dim V \geq n/2+1$ and is satisfied for $\dim V= (n+1)/2$ as long as $q$ is sufficiently large with respect to $n$.

This method is the same as Felipe Voloch's answer, because the eigenvalues in the curve he writes down are Gauss sums, except that we get a savings by dropping the unnecessary multiplicative characters.

It is likely possible to improve the dependence on $q$ somewhat by using etale cohomology theorem (the key thing is not to view this as a $1$-dimensional variety over $\mathbb F_{q^n}$ but as an $n$-dimensional variety over $\mathbb F_q$). But this will probably lead to some loss in $n$ so I don't know whether tit will improve the $\dim (n+1)/2$ case.

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  • $\begingroup$ Very nice! I'll leave this open for now while I think about your answer and the remaining cases (or, more likely, convince myself that I can live without them). $\endgroup$ – Sean Eberhard Jul 16 '18 at 5:51
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Here is an answer when $V$ is a hyperplane and $n \ge 4$. The method genereralizes but you'd have to do the calculation to see if it covers all the cases. Represent the hyperplane as the equation $T(bx) = 0$ for some $b\ne0$, where $T$ is the trace map. To have an element of norm $1$ in $V$ (you may need some additional hypothesis on $n$ to extend this to arbitrary norm elements), as you notice, you need $x$ with $x^{q-1} \in V$, i.e. $T(bx^{q-1})=0$. The elements of trace zero are the elements of the form $y^q-y$, so you need to solve $y^q-y =bx^{q-1}$. This defines a curve of genus $(q-1)(q-2)/2$ so the Weil bound guarantees a solution (other than the one at infinity) if $q^n > (q-1)(q-2)q^{n/2}$ which holds if $n \ge n/2 +2$ or $n \ge 4$.

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  • $\begingroup$ Nice, thank you. I regard the unit-norm and arbitrary-norm problems as equivalent, because $V$ contains an element of norm $1$ if and only if $xV$ contains an element of norm $N(x)$ for $x \neq 0$. $\endgroup$ – Sean Eberhard Jul 16 '18 at 5:45

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