PART 1
First, we begin with the circle $(C): x^2 + y^2 = 1$ over a finite field $F_q$, whose characteristic is not $2$. And the question we are interested in is that how many points lie on this circle?
If $-1$ is a square in $F_q$, we assume that $-1=a^2$, then $x^2+y^2=x^2-a^2y^2=(x-ay)(x+ay)=1$. Hence, $(x,y)\in (C)$ is equivalent $x-ay = b, x+ay = b^{-1}$. And from this $x = (b+b^-1)/2$, and $y = (b-b^{-1})/2a$. An EXERCISE is that counting the number of points in this case.
If $-1$ is not square in $F_q$, we consider the quadratic extensions $F_q(i)/F_q$, where $i$ is the square root of $-1$ in $\overline{F_q}$. The field extension is Galois, and there exists two elements of the Galois group $\{id, x\mapsto x^q\}$, or equivalent $\{id, x+iy \mapsto x-iy\}$.
Because of the definition of norm, let $u=x+iy$, we can compute its norm by two ways: $Norm(u) = x^2 + y^2 = u^{q+1}$. And hence, points on $(C)$ are exactly the kernel of the norm map.
Because $F_q(i)^\times$ is cyclic, whose order $q^2-1$ and generator $g$, the kernel of norm is the cyclic group generated by $g^{q-1}$, and has order $q+1$. That means, $(C)$ will have $(q+1)$ points.
(NOTE: at this stage, we will also conclude that Norm is surjective, because $\#Norm(F_q(i)^\times)=|F_q(i)^\times|/|Ker(Norm)|=q-1=|F_q^\times|$)
Note that there is a 1-1 corresponding between points on $(C)$ and $Ker(Norm)$. Via this, we can define the group law for points in $(C)$ (See my earlier comment on the card named "Pairing on Linear Varieties and Curves of Genus 0").
An interesting corollary of NOTE is that: the norm map is surjective, and hence, any element in $F_q$ can be represented under the form $x^2+y^2$. And we obtain the following
Theorem 1. Let $F_q$ be a finite field with characteristic not 2, and $\alpha,\beta,\gamma$ are not zero in $F_q$. Then the equation $\alpha x^2+\beta y^2+ \gamma z^2$ always has non-trivial solutions over $F_q$.
(Or equivalent statement by another language: Any quadratic form of rank $\ge 3$ is isotropic over $F_q$)
We NOTE that the equation $\alpha x^2 + \beta y^2 = 0$ does not always has non-trivial solution over $F_q$, since in this case $\frac{-\alpha}{\beta}$ must be a square in $F_q^\times$.
The interesting story now begins. One may ask how is the case of cubic forms? That means, for what $n$, we can ensure that the equation $a_1x_1^3+a_2x_2^3+...+a_nx_n^3$ always has non-trivial solution? And more generally, $d$-forms?
PART 2
The aim of this part is to prove the theorem of Chevalley, conjectured by E. Artin.
Theorem 1. Let $f(x_1,...,x_n) \in F_q[x_1,...,x_n]$ be a multi-variables of degree $d$, where $n>d$, and $f(0,0,..,0)=0$, then $f$ has at least two solutions.
Two solutions are mentioned in the book of K. Ireland and M. Rosen "A Classical Introduction to Modern Number Theory", Chapter X, and we just explain one of them, with some further comments.
(NOTE: that the remaining solution is very interesting too, and close to the method for supersingular elliptic curves discussed in the book of L. C. Washington about EC, Chapter IV.)
Assume that $f$ has no other solutions. That means, $f(a_1,...,a_n)\ne 0$, for all $(a_1,...,a_n)\ne (0,...,0)$, that means $1-f^{q-1}$ takes value 1 at $(0,..,0)$, and 0 otherwise. It can be seen that, in fact, it is "equivalent" (this will be defined later) to the polynomial $g(x_1,...,x_n)=(1-x_1^{q-1})...(1-x_n^{q-1})$. because $g$ takes value 1 at $(0,...,0)$, and 0 otherwise (*).
The term "equivalent" can be explained as follows.
Definition 2. The two polynomial $f,g$ over $F_q[x_1,...,x_n]$ is called equivalent if $f(a_1,...,a_n)=g(a_1,...,a_n)$ for all $a_i\in F_q$.
The properties Frobenius automorphism allows us to prove that $x^q$ and $x$ are equivalent. And hence, $x^{aq+r}$ is equivalent to $x^r$. And we can extend this observation to the following
Proposition 3. Let $f\in F_q[x_1,...,x_n]$, then $f$ is equivalent to $g$, where any monomial term of $g$ has the form $x_1^{a_1}...x_n^{a_n}$, and $a_i<q$.
Such a polynomial $g$ is called reduced polynomial of $f$, and it is easy to see that the reduced polynomial of a given polynomial is unique.
Proposition 4. The reduced form of a polynomial is unique.
Proof. It is equivalent to prove that if $f$ and $g$ are reduced forms, i.e. their monomial term has the forms $x_1^{a_1}...x_n^{a_n}$, where $a_i<q$, and identical at all values in $F_q^n$, then they are identical. Let $h=f-g$, and fix the value $a_2,...,a_n\in F_q$, we can see $k(x_1)=h(x_1,a_2,...,a_n)$ is the polynomial of variable $x_1$, with the degree does not exceed $q-1$, and it takes $q$ zeros. Hence $k(x_1)$ is identical zero. Extend this method, it is obvious to see that $h$ is zero, and hence $f\equiv g$ (Q.E.D).
Back to (*), we can see that actually, $g$ is a reduced polynomial of $f$, and hence, $\deg(1-f^{q-1})=(q-1)\deg(f) \ge n(q-1)$. This implies $d\ge n$, a contradiction with the hypothesis of our theorem. Hence, $f$ must have another solution other than zero. And the theorem of Chevalley is now completely proved (Q.E.D).
As a consequence, we can see in any positive characteristic, the equation $\alpha x_1^2 + \beta x_2^2 + \gamma x_3^2$, where $\alpha, \beta, \gamma \ne 0$ always has non-trivial solution (NOTE that in the first part, we implied this in the case characteristic is not 2). And for $d$-form, the number of variables must be greater than $d$ to ensure that our equation always has non trivial solution.
How about in the case $n=d$? A very interesting counter example is built in the EXERCISE section of this book (ex. 17). We will conclude this part by proving the following
FACT 5. There is a $d$-form in $d$ variables such that it has no solution other than $(0,...,0)$.
Proof. We now need some Galois theory to point out this. Consider the extension $F_{q^d}/F_q$. The Galois group in this case is cyclic of order $d$, and generated by the Frobenius automorphism $x \rightarrow x^q$, and all of its elements is $\{id, x\rightarrow x^{q}, ..., x\rightarrow x^{q^{d-1}}\}$. Let $a_1,...,a_d$ is the basis for the vector space $F_{q^d}$ over $F_q$. Consider the polynomial
$$f(x_1,...,x_m)=\prod_{i=0}^{d-1}(a_1^{q^i}x_1+...+a_d^{q^i}x_d)$$
Now, $f$ is invariant under the action from the Galois group (CHECK THIS). Hence, it is defined in $F_q[x_1,...,x_d]$. And the basic facts for the basis of a vector space implies that $f=0$ iff $x_1=...=x_d=0$. And this is our counter example. (Q.E.D)
The book gave a hint for the polynomial and the solution is mine. I don't know how the idea for this counter example comes from. And I didn't think that the Galois theory helps in this case. But actually, it does!
PART 3
IMPORTANT: Fields in this note have characteristic different from 2.
For a long time (maybe 2 weeks), I did not take note on anything, there is many things here (and, of course, quite far from Math), and it makes me feel a little bit of tired. I want to write something about the orthogonal groups over finite fields today.
It should be first noted that we need to know something about quadratic forms first, Witt's theorems (Cancelation Theorem and Extension Theorem). It is not difficult, but really interesting, because it reflects the geometric intuition, and one may be curious how to define "geometry" over arbitrarily fields.
I took a note in the past, but today I look it back, and find some stupid mistakes. I did not bring the other notes with me. But I will edit the mistakes, and then make an outline to compute the order of the orthogonal group over finite fields (associated with a given quadratic form).
But does it have anything to do with the equations over finite fields? Well, as we will see later, one of the most important steps is: compute the number of solutions for the equation $a_1x_1^2+a_2x_2^2+...+a_nx_n^2 = 1$ over $F_q$. And this can be done with the help of Jacobi and Gauss's sum over finite fields.
No comments:
Post a Comment