# 数学代写|数论作业代写number theory代考|MATH1001

## 数学代写|数论作业代写number theory代考|The Multiplicative Structure

One surprising fact about finite fields is that their multiplicative structure is as simple as possible, in a sense that we will now make precise. In order to do this, we need a little more language from abstract algebra. We know that in general a field $F$ possesses both an addition and a multiplication which must obey certain rules, for example we know that every non-zero element of $F$ must have a multiplicative inverse in $F$. In abstract algebra we say that the elements of $F$ form a “group” under addition, and that the non-zero elements of $F$ (often denoted $F^*$ ) form a group under multiplication. If we now focus in on the case $F=\mathbb{F}_q$ (where $q$ is a prime power), we know that the additive group has order $q$ and the multiplicative group has order $q-1$. If $q=p^e$ ( $p$ prime), we also know that the additive structure is simply addition modulo $p$ if $e=$ 1 and is polynomial addition modulo $p$ if $e>1$. The multiplicative structure is multiplication modulo $p$ when $e=1$, but we saw in the previous section that when $e>1$, multiplication seems more complicated. It turns out, however and as we said above, that even though doing multiplication is tricky, the underlying multiplicative structure is not complicated.

One of the fundamental and beautiful theorems of the finite group theory is Lagrange’s Theorem, which says quite simply that the order of a subgroup of a finite group must divide the order of the group. Here we wish to apply this result to to a certain type of subgroup called a cyclic subgroup, as follows.

Suppose $a \neq 1$ is an element of $\mathbb{F}_q^$ (which is a multiplicative group of order $q-1$ ) and consider the set $\left{a, a^2, a^3, \ldots\right}$ of powers of $a$. Since $\mathbb{F}_q^$ is finite, eventually some power $a^m$ will be the first to have the same value as a previous power $a^n$. But $a^m=a^n$ tells us that $a^{m-n}=1$. Writing $k=m-n$, we say that $k$ is the order of the element $a$ in the multiplicative group $\mathbb{F}_q^$; i.e., $k$ is the smallest positive integer such that $a^k=1$. Moreover, the set of $k$ elements $\left{1, a, a^2, \ldots, a^{k-1}\right}$ form a subgroup of $\mathbb{F}_q^$. This subgroup is called the cyclic subgroup generated by $a$, and is often denoted $\langle a\rangle$. We may also denote the order of $a$ by $|a|$.

Now applying Lagrange’s Theorem, to the cyclic subgroup $<$ $a>$ inside the group $\mathbb{F}_q^$, we must have that $k$ (the multiplicative order of $a$ and hence of $\langle a\rangle)$ must divide $q-1$, say $q-1=s k$ for some integer $s$. Hence we have that $a^{q-1}=a^{s k}=\left(a^k\right)^s=1^s=1$; that is, we have shown that for every element $a$ of $\mathbb{F}_q^, a^{q-1}=1$, and the order $|a|$ of $a$ divides $q-1$.

## 数学代写|数论作业代写number theory代考|Counting Irreducible Polynomials

In Section 10.4, while discussing a procedure for constructing the finite field $\mathbb{F}_{p^n}$, we asserted that, given any prime $p$ and any positive integer exponent $e$, there always exists a monic irreducible polynomial $P(\theta)$ of degree $n$ with coefficients in the prime field $\mathbb{F}_p$. In this section we not only show that this is true, but we also show that we can count exactly how many such polynomials there are. It turns out that the Möbius Inversion Formula (Theorem 8.9) comes into play here.

Let $N_q(n)$ denote the number of monic irreducible polynomials of degree $n$ over the finite field $F_q$, i.e., with coefficients of the polynomial in the field $F_q$, where $q$ is a prime power.

Lemma 10.8. Let $T_n$ be the set of all monic irreducible polynomials over the field $F_q$ of degree dividing the positive integer $n$. Then the polynomial $\theta^{q^n}-\theta$ factors over the field $F_q$ as $\prod_{f \in T_n} f$.

Proof. A fundamental result from finite group theory in abstract algebra is that if the order of a multiplicative group $G$ is $|G|$ and if $a$ is an element of $G$, then $a^{|G|}=1$. Since the order of the multiplicative group $\mathbb{F}{q^n}^*$ is $q^n-1$, this tells us that if $a \neq 0$ is in $\mathbb{F}{q^n}$, then $a$ is a root of the polynomial $\theta^{q^n-1}-1$. Multiplying through by $\theta$ (and thus including the element 0 ) we get that the $q^n$ distinct elements of $\mathbb{F}{q^n}$ are exactly the roots of the polynomial $g(\theta)=\theta^{q^n}-\theta$. If we now factor $g(\theta)$ into its product of monic irreducible polynomials (just as we factor a positive composite integer into its product of primes), we see that every element of $\mathbb{F}{q^n}$ is a root of exactly one of these irreducibles.

A question now is, What are the possible degrees of these irreducible polynomials? The answer is: The degrees are numbers $k$ which divide $n$. The basic idea is the following: The roots of a monic irreducible of degree $k$ over $\mathbb{F}q$ all come from the set of elements of the field $\mathbb{F}{q^k}$ which do not lie in any proper subfields of $\mathbb{F}{q^k}$ (such elements are called primitive in $\mathbb{F}{q^k}$ ). However, all this is occurring inside the field $\mathbb{F}{q^n}$, so $\mathbb{F}{q^k}$ must be a subfield of $\mathbb{F}{q^n}$, and in Theorem $10.7$ we proved that $k$ must be a divisor of n. (For example, looking at the subfield lattice at the end of the previous section, the roots of a 6-th degree irreducible polynomial over $\mathbb{F}_p$ would come from those elements of $\mathbb{F}{p^6}$ which do not lie in either $\mathbb{F}{p^3}$ or $\mathbb{F}{p^2}$.) Since $g(\theta)$ factors into the product of all of these monic irreducibles, the proof is complete.

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