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Statment

Let $q$ the power of a prime number $p$ and $f(x,y)\in F_q[x,y]$ a non constant polynomial, then $K=F_q^2\setminus V(f)$ is a Kakeya set.

My attempt. I think that it is False, for it let consider the line $L_{v,a}=\lbrace a+kv\mid k\in F_q\rbrace$ be the line that pass throught by $a$ in the direction of $v\neq 0$.

Now let $f(x,y)=y-x$ which vanishes in $y=x$ and chose $v=(3,3)$, $k\in F_q$ and $a=(a_1,a_1)$ then $L_{v,a}=\lbrace (a_1+3k,a_1+3k)\rbrace $ but $a_1+3k-a_1-3k=0$ and hence $L\not \subset K$.

Hence $K$ isn´t Kakeya set.

I think that is fine, but I have some problems to understand how the line looks.

I understand that in general $F_q^2$ is like a chess board where I have all the squares with a point, but I don´t get which means that the direction of $v\neq 0$ and how looks a line in any direction of some $v$. Thanks any hint or comment was useful.

Alan Jr
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    What is the definition of a Kakeya set in $\Bbb{F}_q^2$? – Jyrki Lahtonen Sep 05 '21 at 21:20
  • A line is determined when we know a single point $a$ on it (any will do) and its direction (given by a non-zero vector $v$)- just like in $\Bbb{R}^2$! The description of the set $L_{v,a}$ serves as a definition. I don't know what is bothering you about it? – Jyrki Lahtonen Sep 05 '21 at 21:24
  • The line will not change if we replace $a$ with any other point on the line, or if we replace the vector $v$ with its non-zero scalar multiple. – Jyrki Lahtonen Sep 05 '21 at 21:26
  • I'm fairly sure that you have somehow misunderstood the definition of a Kakeya set. I have never worked with them, but surely the definition is that to each $v$ there exists an $a$ such that $L_{v,a}$ is contained in the set. It simply cannot be that every line $L_{v,a}$ must be contained. For then the only possible Kakeya set would be the entire space, and the concept would be totally uninteresting. – Jyrki Lahtonen Sep 09 '21 at 04:06

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Because you have a torus of side $q$ you generally wrap around. It is a wrapped around discrete line.

The $q$ prime case is as below:

For $q=7$ and $a=1,v=1$, write the torus as $\{0,1,\ldots,6\}^2$:

$$ 0~1~0~0~0~0~0\\ 0~0~1~0~0~0~0\\ 0~0~0~0~1~0~0\\ 0~0~0~0~0~1~0\\ 0~0~0~0~0~0~1\\ 1~0~0~0~0~0~0\\ $$

For $q=7$ and $a=5,v=5$ we have $\{5+5k:k\in F_7\}=\{5,3,1,6,4,2,0\}$ which gives $$ 0~0~0~0~0~1~0\\ 0~0~0~1~0~0~0\\ 0~1~0~0~0~0~0\\ 0~0~0~0~0~0~1\\ 0~0~0~0~1~0~0\\ 0~0~1~0~0~0~0\\ 1~0~0~0~0~0~0\\ $$

Thanks to Jyrki Lahtonen for catching my mistake, the comment I had before applied only to the prime case:

The nonprime case is equivalent to this, if the elements of $F_q$ are reordered as $\{0,1,g,g^2,\ldots,g^{q-2}\}$ where $g$ is a primitive element (which always exists) and the multiplication by the nonzero elements is just multiplication by powers of $g.$

kodlu
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    The notation $\Bbb{F}_q$ surely means the finite field of $q$ elements, so I don't understand what you mean by divisors of $q$ as elements of $\Bbb{F}_q$. – Jyrki Lahtonen Sep 06 '21 at 04:35