# Lethear\'s Blog

Just another WordPress.com weblog

## So the number theory is all about tricks and more tricks?

I was reading Jean-Pierre Serre’s A Course in Arithmetic right now. Thought group theory and Galois theory were not hard for me, but I had had almost no training in number theory. So I guessed it may take time to get familiar with it.

I still think number theory is awful because of the tricks so my study has been no obvious progression. Well, at least I was getting to know how the group theory is applied to specific objects such as ${\mathbb F}_p^*$ . So I was to jot down what I learned today.

Considering Brother Zeng is constantly browsing this blog, I should explain the concepts a bit, although some of them is rather basic or trivial.

DEFINITIONS: For conveniency, we always write ${\mathbb F}_p={\mathbb Z}/p{\mathbb Z}$, and denote the group $F-\{ 0\}$ with the operation of multiplication as $F^*$. Given a field $F$, $char(F)$ equals to the order of its identity under addition, but when the order is infinite, we simply write $char(F)=0$. For eg., $char({\mathbb Q})=0, char({\mathbb F}_p)=p$.

PROPOSITIONS: Given a field $F$, $char(F)$ is either a prime or 0. When $char(F)=0$, we can always imbed ${\mathbb Q}$ into $F$. When $char(F)=p$, we have ${\forall}x,y{\in}F, (x+y)^p=x^p+y^p$, which indicates that the mapping:$f:F{\rightarrow}F, f(x)=x^p$ is a ring homomorphism. Besides, when $F$ is finite, $|F|=p^n$ for some n because $F$ can be seen as a vector space over the field ${\mathbb F}_p$.

THEOREM: All $|F|=p^n$ with the same n and p are isomorphic

This thm. is usually proved by the Isomorphism Extension Theorem, but JPS gives us another. Denote the algebraic closure of ${\mathbb F}_p$ as $\Omega$. By the theory of polynomials over fields, the number of roots in $\Omega$ of above polynomial is at most $p^n$. Because the elements of the subfield in $\Omega$ with order $p^n$ also satisfies the above polynomial, we get to know that if it exist, the subfield is unique. To construct it, consider the endomorphism:$f:F{\rightarrow}F, f(x)=x^{p^n}$, and the set $K=\{ x|f(x)=x\}$. Of course $K$ is a field with $p^n$ elements. Because we can trivially imbed a field with order $p^n$ into $\Omega$, we get to know it is isomorphic with $K$.

THEOREM: ${\mathbb F}_q^*$ is cyclic. (when $q=p^n$, we denote ${\mathbb F}_q$ as the unique field of order q)

After decompositing the abelian group, the conclusion is trivially followed. But JPS gives another proof. Let’s consider the Euler-fai function. ${\phi}(x)$ denotes the number of integers from 1 to x that are coprime with x. We noted ${\sum}_{d|n}{\phi}(d)=n$.

LEMMA: Given a finite group $H$ of order n. $\forall d, Suppose the number of elements with order d is at most d. Then $H$ has to be cyclic.

${\forall}h{\in}H$, there is no second cyclic group with the same order as (h), and there are ${\phi}(|h|)$ generating elements of (h). Let’s summed up the numbers of generating elem. of different cyclic subgroups. Since every elem. generates a cyclic group, the sum has to be bigger than n. But we know ${\sum}_{d|n}{\phi}(d)=n$ exactly. Since n|n, there has to be an elem. with order n. Thus H is a cyclic group generated by this one.

Now apply the lemma to ${\mathbb F}_q^*$.

======Above is the basic introduction. Now we came to the nonbasic part======

LEMMA 1: Field $F$. Denote $S(f)={\sum}_{x{\in}F}f(x)$ where f is a function on $F$. When $u{\geqslant}1$ and $(q-1)|u$, $S(x^u)=-1$. Otherwise $S(x^u)=0$.

It’s easy to verify the trivial cases. And note the fact that ${\forall}y{\in}F, {\sum}_{x{\in}F}x^u={\sum}_{x{\in}F}y^ux^u$.

THEOREM 1(Chevalley-Warning): let $f_{\alpha}{\in}F[x_1,x_2,...,x_n]. {\sum}_{\alpha}deg f_{\alpha}. The number of the common zeros of $f_{\alpha}$ can be divided by p

Consider $P(x)={\prod}_{\alpha}(1-f_{\alpha}^{q-1}), x{\in}F^n$. Prove that when x is the common zero, P(x)=0. Otherwise P(x)=1. So the number of common zeros is $S(P)$ where S is the same as the former lemma. Since P is a polynomial, prove that $S(x_1^{u_1}...x_n^{u_n})=0$. Because ${\sum}_{i=1}^nu_i, and the fact of the former lemma, the thm. holds.

(IT'S TRICKY! IT SEEMS TO ME WITHOUT TRICKS, THERE IS NO DEEP RESULT.)

REMARK: It’s an important case when $n{\geqslant}3$ and $f_{\alpha}$ are the quadratic forms. Since the 0 is a common zero, the theorem indicates there are nontrivial zeros.

I didn’t know why JPS remark this. He said a quadratic form is a conic hyper surface(?). I admitted I knew nothing about it.

THEOREM 2: $|F|=q$. When $char(F)=2$, every unit is a square. Otherwise, all the square unit in $F$ forms a subgroup of $F^*$, which is the kernel of the mapping:$f:F{\rightarrow}\{ 1,-1\} , f(x)=x^{(q-1)/2}$.

Note that the roots of $x^2=1$ is 1 and -1.

This theorem is an example of how to use the language of group theory to intepret the concepts in number theory, such as the square units. And this kind of translation offers a new way to look into it.(Well, we were just trivially considering the kernel of a mapping from an ABELIAN group to itself…Yes, it’s not exciting enough)

The remaining part of quadratic reciprocity law is not exciting except for another trick:

DEFINITION: $p{\neq}2, x{\in}{\mathbb F}_p^*. (x/p)=x^{(p-1)/2}$

PROBLEM: Calculate $(2/p)$

Consider the primitive root $\alpha$ of $x^8=1$. Let $y={\alpha}+{\alpha}^{-1}$. Because ${\alpha}^4=-1$, $y^2=2.$. $2^{(p-1)/2}=y^{p-1}$. Simple calculation implies that if $p{\equiv}{\pm}1(mod 8), y^{p-1}=1$, otherwise $y^{p-1}=-1$.

(OH TRICK…)

When we looked back we may figure out we had only learned one deep theorem, i.e., Chevalley theorem. The rest of it were translations of trivial facts to the modern language. They were both enlightening.

But why Chivalley or Warning had tried to contruct such a P(x)?…

Written by lethear

February 11, 2010 at 4:02 pm

Posted in Old things

Tagged with