## 数学代写|编码理论代写Coding theory代考|MTH 3018

2022年7月2日

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## 数学代写|编码理论代写Coding theory代考|DETERMINING THE DEGREES OF THE IRREDUCIBLE

Since every element of order $n$ has the same number of conjugates with respect to $\mathrm{GF}(q)$, every irreducible factor of the cyclotomic polynomial $Q^{(n)}(x)$ has the same degree over $\mathrm{GF}(q)$. This degree is $m$, which is the multiplicative order of $q \bmod n$. Thus, although the explicit factors of $Q^{(n)}(x)$ may be determined only by polynomial manipulations using Euclid’s algorithm, the degrees of these irreducible factors may be computed by appropriate manipulations with integers.

The most straightforward approach for determining the order of $q \bmod n$ is to compute the successive residues, $q, q^{2}, q^{3}, \ldots, \bmod n$, until one finds the least $m$ such that $q^{m} \equiv 1 \bmod n$. However, if $n$ is large, this procedure becomes tedious. Fortunately, the calculations may be considerably shortened by the use of several simple number theoretic results which we now present.

Theorem 6.51 If $n=\prod_{i} n_{i}{ }{i}, n{i}$ distinct primes, then the order of $q \bmod n$ is the least common multiple of the orders of $q \bmod n_{i}^{j_{i}}$.

EXAMPLE Suppose $n=45, q=2$. Since the order of $2 \bmod 5$ is 4 and the order of $2 \bmod 9$ is 6 , the order of $2 \bmod 45$ is 12 .

Proof of Theorem 6.51 $q^{m} \equiv 1 \bmod n$ iff $q^{m} \equiv 1 \bmod n_{i}^{j^{3}}$ for every i. $\quad q^{m} \equiv 1 \bmod n_{i}^{j_{i}}$ iff $m$ is a multiple of the order of $q \bmod n_{i}^{j_{i}}$.

## 数学代写|编码理论代写Coding theory代考|IS THE NUMBER OF IRREDUCIBLE

Although it appears unlikely that any general method for factoring polynomials over finite fields can improve substantially on the algorithm presented in Sec. 6.1, other methods sometimes enable one to obtain certain information about the factors with even less effort. For example, every high-school algebra student knows that the roots of the real quadratic equation $a x^{2}+b x+c=0$ are $x=\left(-b \pm \sqrt{b^{2}-4 a c}\right) / 2 a$. If $q$ is a power of an odd prime which does not divide $a$, then this formula remains valid over $\operatorname{GF}(q)$. Thus, the question of whether or not a quadratic equation over $\mathrm{GF}(q)$ has two roots in $\mathrm{GF}(q)$ depends on whether or not its discriminant, $D=b^{2}-4 a c$, is a perfect square in $\mathrm{GF}(q)$. In particular, we see that the original quadratic equation has an even number (two) of irreducible factors over $G F(q)$ if its discriminant is a square, but it has an odd number (one) of irreducible factors over $\mathrm{GF}(q)$ if its discriminant is a nonsquare. This is the simplest example of a general result called Stickelberger’s theorem. In order to state the general result, we must define the discriminant of a polynomial of arbitrary degree.

Stickelberger’s theorem $\mathbf{6 . 6 8}$ (odd characteristic) Let $q$ be a power of an odd prime, let $f(x)$ be a monic polynomial of degree $m$ over GF $(q)$, with discriminant $D(f) \neq 0$. Let $r$ be the number of irreducible factors of $f(x)$ over $\mathrm{GF}(q)$. Then $r \equiv m \bmod 2$ iff $D(f)$ is a square in $\mathrm{GF}(q)$

Given an integer $a$ and a prime $p$, the Legendre symbol $(a / p)$ is defined by Definition 6.71.
Definition $6.71$
$$(a / p)= \begin{cases}1 & \text { if } x^{2}=a \text { has nonzero solutions in } \mathrm{GF}(p) \ -1 & \text { if } x^{2}=a \text { does not have solutions in } \mathrm{GF}(p) \ 0 & \text { if } p \text { divides } a\end{cases}$$
The Legendre symbol arises in applications of Stickelberger’s theorem, in the determination of the multiplicative order of $a \bmod p$, in the determination of the degrees of the irreducible factors of $Q^{(a)}(x)$ over $\mathrm{GF}(p)$, in the study of quadratic-residue codes, which will be defined in Sec. $15.2$, and in many other places in number theory. If the equation $x^{2}=a$ has nonzero solutions in $\mathrm{GF}(p), a$ is said to be a quadratic residue $\bmod p$; if the equation $x^{2}=a$ has no solution in $\mathrm{GF}(p), a$ is said to be a quadratic nonresidue.

If $a$ is a quadratic nonresidue, then $x^{2}-a$ is irreducible over $\mathrm{GF}(p)$, and $\sqrt{a} \in \mathrm{GF}\left(p^{2}\right)$; if $a$ is a quadratic residue, then $x^{2}-a$ factors over $\mathrm{GF}(p)$, and $\sqrt{a} \in \mathrm{GF}(p)$. In either case, $\sqrt{a} \in \mathrm{GF}\left(p^{2}\right)$. The element $\sqrt{a}$ is in the subfield $\mathrm{GF}(p)$ iff $\sqrt{a^{p}}=\sqrt{a}$. Since $\left(a^{(p-1) / 2}\right)^{2}=1$ and the only two roots of the equation $x^{2}-1=0$ are $x=\pm 1, a^{(p-1) / 2}=$ $\pm 1 \bmod p$. Therefore,
$$(a / p) \equiv a^{(p-1) / 2} \bmod p$$ This congruence is called Euler’s criterion. Lemmas $6.721$ to $6.723$ follow immediately,

## 数学代写|编码理论代写Coding theory代考|IS THE NUMBER OF IRREDUCIBLE

$x=\left(-b \pm \sqrt{b^{2}-4 a c}\right) / 2 a$. 如果 $q$ 是不除的奇榡数的草 $a$,那么这个公式在 $\mathrm{GF}(q)$. 因 此，一个二次方程是否超过 $\mathrm{GF}(q)$ 有两个根源 $\mathrm{GF}(q)$ 取决于它的判别式是否，
$D=b^{2}-4 a c$ ，是一个完美的正方形 $\mathrm{GF}(q)$. 特别是，我们看到原始二次方程有偶数（两 个) 不可约因子 $G F(q)$ 如果它的判别式是一个正方形，但它有奇数个 (一个) 不可约因数 $\mathrm{GF}(q)$ 如果它的判别式是非正方形的。这是称为斯蒂克尔伯格定理的一般结果的最简单示 例。为了说明一般结果，我们必须定义任意次数多项式的判别式。

$(a / p)=\left{1 \quad\right.$ if $x^{2}=a$ has nonzero solutions in $\mathrm{GF}(p)-1 \quad$ if $x^{2}=a$ does not have solutions in $\mathrm{GF}(p) 0$
Legendre 符号出现在 Stickelberger 定理的应用中，用于确定 $a \bmod p$, 在确定不可约因子

$\sqrt{a}$ 在子字段中 $\mathrm{GF}(p)$ 当且当 $\sqrt{a^{p}}=\sqrt{a}$. 自从 $\left(a^{(p-1) / 2}\right)^{2}=1$ 和方程仅有的两个根
$x^{2}-1=0$ 是 $x=\pm 1, a^{(p-1) / 2}=\pm 1 \bmod p$. 所以，
$$(a / p) \equiv a^{(p-1) / 2} \bmod p$$

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## MATLAB代写

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