Let $GF(q)$ be a finite field of order $q=p^m$, where $p$ is a prime number. Given an element $\beta \in GF(q)$ and the integers $k, d \in \mathbb{N}$, Let $$I_{k} = \{ x \in GF(q) \mid x = \alpha^k, \alpha \in GF(q) \}$$ and $$F_{d}(\beta) = \{ x \in GF(q) \mid \exists f \in GF(p)_{d-1}[x], f(0) = 1, x = \beta \cdot f(\beta)\}$$ where $$GF(p)_{d}[x] = \{ f \in GF(p)[x] \mid \mathrm{deg}(f)\leq d\}$$ How can we calcuate or estimate $|I_{k} \cap F_{d}(\beta)|$
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@JyrkiLahtonen Yes, you are right, the coefficients belong to $GF(p)$ – Blanco Jul 29 '20 at 04:23
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Oops, I missed the condition $f(0)=1$. Need a bit more of tinkering. I expect the first attempt to give an estimate of $p^{d-1}/\ell+$ an error term of size $\approx \sqrt q$, $\ell=\gcd(k,q-1)$. Which is not at all useful, when $d\le m/2$. – Jyrki Lahtonen Jul 29 '20 at 09:54
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@JyrkiLahtonen Also, I am interested in the question without the condition $f(0) = 1$. Can I keep it if you are willing to share your ideas about both of these two questions. – Blanco Jul 29 '20 at 12:20
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If you don't make any assumptions about $f(0)$, then the dimension of the subspace $V$ in my answer increases by one, and we don't need the coset. That may simplify the continuation of the analysis (and cancellation of the Gauss sums). Are you familiar with the basic properties of characters and character sums? – Jyrki Lahtonen Jul 29 '20 at 12:24
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@JyrkiLahtonen Not very much:( – Blanco Jul 29 '20 at 12:29
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How/why did you run into this question? – Jyrki Lahtonen Jul 30 '20 at 04:27
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The original question is about algebraic coding. And it seems that the precise size has been calculated in my team, but I have not known it yet. – Blanco Jul 30 '20 at 06:25
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I am a bit curious about whether you have accumulated data on how far from the "expected value" $p^{d-1}/\ell$ the size of the intersection varies in practice? All my old code is for DOS, and Microsoft is no longer supporting that :-( – Jyrki Lahtonen Jul 30 '20 at 06:33
1 Answers
Trying to say something non-trivial about a generic case of the edited version of the question. The old answer can be viewed in the edit history of this answer. Trust me, you aren't missing anything worthwhile :-)
Because the multiplicative group $GF(q)^*$ is cyclic of order $q-1$, a non-zero $x\in GF(q)$ is a $k$th power if and only if it is an $\ell$the power, where $\ell=\gcd(k,q-1)$. I will write everything using $\ell$ as a reminder, but you may want equally well think that $k\mid q-1$ when $\ell=k$.
The set $F_d(\beta)$ is the coset $\beta+V$, where $V$ is the $GF(p)$-span of the elements $\beta^2,\beta^3,\ldots,\beta^d$. In what I call the generic case the space $V$ has dimension $d-1$. If $\beta$ belongs to a subfield $GF(p^t), t<d-1$, then the space $V$ has a lower dimension (at most $t$). Those were not excluded and I may come back to that case later. Anyway, the set $F_d(\beta)$ has $p^{d-1}$ elements.
By the properties of cyclic groups, a random element of $GF(q)^*$ is an $\ell$th power with probability $1/\ell$, so we expect the intersection $I_k\cap F_d(\beta)$ to have approximately $p^{d-1}/\ell$ elements. Character sums form a relatively universal tool for estimating the accuracy of such estimates. In the present case we need both the additive characters (to get a handle on the additive subspace $V$) as well as the multiplicative characters of $GF(q)$ (to get a handle on the subset of $\ell$th powers).
Let $\operatorname{tr}:GF(q)\to GF(p)$ be the trace function, $\operatorname{tr}(x)=\sum_{i=0}^{m-1}x^{p^i}$. Let's denote the canonical additive character of $GF(q)$ by $\psi$, so $$ \psi(x):=e^{2\pi i \operatorname{tr}(x)/p}. $$ Let's denote $$ V^\perp=\{y\in GF(q)\mid tr(xy)=0\ \forall\,x\in V\}. $$ Because the trace-form is non-degenerate, $\dim_{GF(p)}V^{\perp}=m-(d-1).$ By the well-known properties of character sums we have, for all $x\in GF(q)$, $$ \sum_{y\in V^{\perp}}\psi(yx)= \begin{cases} p^{m-d+1},&\ \text{if $x\in V$, and}\\ 0,&\ \text{otherwise.} \end{cases} $$ Consequently $$ \sum_{y\in V^{\perp}}\psi(y[x-\beta])= \begin{cases} p^{m-d+1},&\ \text{if $x\in F_d(\beta)$, and}\\ 0,&\ \text{otherwise.} \end{cases} $$ We can use this as a test of membership in $F_d(\beta)$.
Similarly, let $\chi_1:GF(q)\to\Bbb{C}^*$ be a multiplicative character of order $\ell$ (so $\chi_1(GF(q)^*)=\mu_\ell$). Let $G$ be the group of characters generated by $\chi_1$, so $G\simeq C_\ell$. Let $\chi_0$ be the principal character (= the constant map to $1$). Again, a standard fact is that $$ \sum_{\chi\in G}\chi(x)=\begin{cases} \ell,&\ \text{if $x=z^\ell$ for some $z\in GF(q)^*$,}\\ 1,&\ \text{if $x=0$, assuming the convention $\chi_0(0)=1$ and $\chi(0)=0$, when $\chi\neq\chi_0$,}\\ 0,&\ \text{otherwise.} \end{cases} $$ This is our membership test in the set of $\ell$th powers.
In the generic case $0\notin F_d(\beta)$, and we can ignore any special treatment of $x=0$.
Putting all of the above together we get a formula for the size of the intersection in terms of character sums
$$ |I_k\cap F_d(\beta)|=\frac1{\ell\cdot p^{m-d+1}} \sum_{y\in V}\sum_{\chi\in G}\sum_{x\in GF(q)^*}\psi(y[x-\beta])\chi(x).\qquad(*) $$ Let's take a close look at the innermost sum $$ \sum_{x\in GF(q)^*}\psi(y[x-\beta])\chi(x)=\overline{\psi(y\beta)} \sum_{x\in GF(q)^*}\psi(yx)\chi(x). $$ The last sum is the well studied Gauss sum $S(y,\chi)=\sum_{x\in GF(q)^*}\psi(yx)\chi(x)$. We know that $$ S(y,\chi)= \begin{cases} q-1,&\ \text{when $y=0$ and $\chi=\chi_0$,}\\ -1,&\ \text{when $y\neq0$ and $\chi=\chi_0$,}\\ 0,&\ \text{when $y=0$ and $\chi\neq\chi_0$,}\\ \text{has absolute value $\sqrt q$},&\ \text{when $y\neq0$ and $\chi\neq\chi_0$.} \end{cases} $$ In the sum $(*)$ we identify the main term ($y=0,\chi=\chi_0$) of size $\approx p^{d-1}/\ell$, and crudely apply the triangle inequality to the rest of them. Observing that there are $(\ell-1)(|V^\perp|-1)$ terms of magnitude $\sqrt q$ and smaller terms, we arrive at the approximate bound $$ \left\vert|I_k\cap F_d(\beta)|-\frac{p^{d-1}}\ell\right\vert\le\sqrt q. $$ This gives something non-trivial only when $|V|=p^{d-1}<\sqrt q$, or when $d>m/2$.
The disappointing features of this estimate are
- One would expect the Gauss sums to have random phases, and hence that there should be a lot of cancellation in the error term. There are several relations that the Gauss sums related to different choices of $y$ satisfy. Yet, the factor $\overline{\psi(y\beta)}$ make it non-obvious how to proceed here.
- This bound works for an arbitrary subspace $V$ of dimension $d-1$ just the same. The special description of $F_d(\beta)$ is not used at all. In the generic case there is a recipe for the dual (w.r.t. the trace form) of the monomial basis $\{1,\beta,\ldots,\beta^{m-1}\}$. Exploiting that structure may open a route forward, but exploring that would take more time than I can put into this question at this time.
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