数学代写|现代代数代写Modern Algebra代考|MATH067

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数学代写|现代代数代写Modern Algebra代考|The Field of Quotients of an Integral Domain

The example of an integral domain that is most familiar to us is the set $\mathbf{Z}$ of all integers, and the most familiar example of a field is the set of all rational numbers. There is a very natural and intimate relationship between these two systems. In fact, a rational number is by definition a quotient $a / b$ of integers $a$ and $b$, with $b \neq 0$; that is, the set of rational numbers is the set of all quotients of integers with nonzero denominators. For this reason, the set of rational numbers is frequently referred to as “the quotient field of the integers.” In this section, we shall see that an analogous field of quotients can be constructed for an arbitrary integral domain.

Before we present this construction, let us review the basic definitions of equality, addition, and multiplication in the rational numbers. We recall that for rational numbers $\frac{a}{b}$ and $\frac{c}{d}$,
\begin{aligned} & \frac{a}{b}=\frac{c}{d} \text { if and only if } a d=b c, \ & \frac{a}{b}+\frac{c}{d}=\frac{a d+b c}{b d}, \ & \frac{a}{b} \cdot \frac{c}{d}=\frac{a c}{b d} . \end{aligned}
Note that the definitions of equality, addition, and multiplication for rational numbers are based on the corresponding definitions for the integers. These definitions guide our construction of the quotient field for an arbitrary integral domain $D$.
Our first step in this construction is the following definition.

A Relation on Ordered Pairs
Let $D$ be an integral domain and let $S$ be the set of all ordered pairs $(a, b)$ of elements of $D$ with $b \neq 0$ :
$$S={(a, b) \mid a, b \in D \text { and } b \neq 0}$$

The relation $\sim$ is defined on $S$ by
$(a, b) \sim(c, d)$ if and only if $a d=b c$.
The relation $\sim$ is an obvious imitation of the equality of rational numbers, and we can show that it is indeed an equivalence relation on $S$.

数学代写|现代代数代写Modern Algebra代考|The Equivalence Relation

The relation $\sim$ in Definition 5.21 is an equivalence relation on $S$.
Proof We shall show that $\sim$ is reflexive, symmetric, and transitive. Let $(a, b),(c, d)$, and $(f, g)$ be arbitrary elements of $S$.

$(a, b) \sim(a, b)$, since the commutative multiplication in $D$ implies that $a b=b a$.

$(a, b) \sim(c, d) \Rightarrow a d=b c$
by definition of
$\Rightarrow d a=c b$ or $c b=d a$
since multiplication is commutative in $D$
$\Rightarrow(c, d) \sim(a, b)$
by definition of $\sim$.

Assume that $(a, b) \sim(c, d)$ and $(c, d) \sim(f, g)$.
$$\left.\begin{array}{l} (a, b) \sim(c, d) \Rightarrow a d=b c \Rightarrow a d g=b c g \ (c, d) \sim(f, g) \Rightarrow c g=d f \Rightarrow b c g=b d f \end{array}\right} \Rightarrow a d g=b d f$$
Using the commutative property of multiplication in $D$ once again, we have ${ }^{\dagger}$
$$d a g=d b f$$
where $d \neq 0$, and therefore
$$a g=b f$$
by Theorem 5.16. According to Definition 5.21, this implies that $(a, b) \sim(f, g)$.
Thus $\sim$ is an equivalence relation on $S$.
The next definition reveals the basic plan for our construction of the quotient field of $D$.

Let $D, S$, and $\sim$ be the same as in Definition 5.21 and Lemma 5.22. For each $(a, b)$ in $S$, let $[a, b]$ denote the equivalence class in $S$ that contains $(a, b)$, and let $Q$ denote the set of all equivalence classes $[a, b]$ :
$$Q={[a, b] \mid(a, b) \in S} .$$
The set $Q$ is called the set of quotients for $D$.

We shall at times need the fact that for any $x \neq 0$ in $D$ and any $[a, b]$ in $Q$,
$$[a, b]=[a x, b x] .$$
This follows at once from the equality $a(b x)=b(a x)$ in the integral domain $D$.

现代代数代考

数学代写|现代代数代写Modern Algebra代考|The Field of Quotients of an Integral Domain

\begin{aligned} & \frac{a}{b}=\frac{c}{d} \text { if and only if } a d=b c, \ & \frac{a}{b}+\frac{c}{d}=\frac{a d+b c}{b d}, \ & \frac{a}{b} \cdot \frac{c}{d}=\frac{a c}{b d} . \end{aligned}

$$S={(a, b) \mid a, b \in D \text { and } b \neq 0}$$

$(a, b) \sim(c, d)$当且仅当$a d=b c$。

数学代写|现代代数代写Modern Algebra代考|The Equivalence Relation

$(a, b) \sim(a, b)$，因为$D$中的交换乘法意味着$a b=b a$。

$(a, b) \sim(c, d) \Rightarrow a d=b c$

$\Rightarrow d a=c b$或$c b=d a$

$\Rightarrow(c, d) \sim(a, b)$

$$\left.\begin{array}{l} (a, b) \sim(c, d) \Rightarrow a d=b c \Rightarrow a d g=b c g \ (c, d) \sim(f, g) \Rightarrow c g=d f \Rightarrow b c g=b d f \end{array}\right} \Rightarrow a d g=b d f$$

$$d a g=d b f$$

$$a g=b f$$

$$Q={[a, b] \mid(a, b) \in S} .$$

$$[a, b]=[a x, b x] .$$

数学代写|现代代数代写Modern Algebra代考|MTH350

statistics-lab™ 为您的留学生涯保驾护航 在代写现代代数Modern Algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写现代代数Modern Algebra代写方面经验极为丰富，各种代写现代代数Modern Algebra相关的作业也就用不着说。

数学代写|现代代数代写Modern Algebra代考|Integral Domains and Fields

In the preceding section we defined the terms ring with unity, commutative ring, and zero divisors. All three of these terms are used in defining an integral domain.
Integral Domain
Let $D$ be a ring. Then $D$ is an integral domain provided these conditions hold:

1. $D$ is a commutative ring.
2. $D$ has a unity $e$, and $e \neq 0$.
3. $D$ has no zero divisors.
Note that the requirement $e \neq 0$ means that an integral domain must have at least two elements.

Example 1 The ring $\mathbf{Z}$ of all integers is an integral domain, but the ring $\mathbf{E}$ of all even integers is not an integral domain, because it does not contain a unity. As familiar examples of integral domains, we can list the set of all rational numbers, the set of all real numbers, and the set of all complex numbers-all of these with their usual operations.

Example 2 The ring $\mathbf{Z}{10}$ is a commutative ring with a unity, but the presence of zero divisors such as [2] and [5] prevents $\mathbf{Z}{10}$ from being an integral domain. Considered as a possible integral domain, the ring $M_2(\mathbf{R})$ of all $2 \times 2$ matrices with real numbers as elements fails on two counts: Multiplication is not commutative, and it has zero divisors.

In Example 4 of Section 5.1, we saw that $\mathbf{Z}_n$ is a ring for every value of $n>1$. Moreover, $\mathbf{Z}_n$ is a commutative ring since
$$[a] \cdot[b]=[a b]=[b a]=[b] \cdot[a]$$
for all $[a],[b]$ in $\mathbf{Z}_n$. Since $\mathbf{Z}_n$ has $[1]$ as the unity, $\mathbf{Z}_n$ is an integral domain if and only if has no zero divisors. The following theorem characterizes the $\mathbf{Z}_n$ with no zero divisors, providing us with a large class of finite integral domains (that is, integral domains that have a finite number of elements).

数学代写|现代代数代写Modern Algebra代考|The Integral Domain Zn When n Is a Prime

For $n>1, \mathbf{Z}_n$ is an integral domain if and only if $n$ is a prime.
Proof From the previous discussion, it is clear that we need to only prove that $\mathbf{Z}_n$ has no zero divisors if and only if $n$ is a prime.

Suppose first that $n$ is a prime. Let $[a] \neq[0]$ in $\mathbf{Z}_n$, and suppose $[a][b]=[0]$ for some $[b]$ in $\mathbf{Z}_n$. Now $[a][b]=[0]$ implies that $[a b]=[0]$, and therefore, $n \mid a b$. However, $[a] \neq[0]$ means that $n \nmid a$. Thus $n \mid a b$ and $n \nmid a$. Since $n$ is a prime, this implies that $n \mid b$, by Theorem 2.16 (Euclid’s Lemma); that is, $[b]=[0]$. We have shown that if $[a] \neq[0]$, the only way that $[a][b]$ can be $[0]$ is for $[b]$ to be $[0]$. Therefore, $\mathbf{Z}_n$ has no zero divisors and is an integral domain.
Suppose now that $n$ is not a prime. Then $n$ has divisors other than \pm 1 and $\pm n$, so there are integers $a$ and $b$ such that
$$n=a b \text { where } 1<a<n \text { and } 1<b<n .$$
This means that $[a] \neq[0],[b] \neq[0]$, but
$$[a][b]=[a b]=[n]=[0] .$$
Therefore, $[a]$ is a zero divisor in $\mathbf{Z}_n$, and $\mathbf{Z}_n$ is not an integral domain.
Combining the two cases, we see that $n$ is a prime if and only if $\mathbf{Z}_n$ is an integral domain.

现代代数代考

数学代写|现代代数代写Modern Algebra代考|Integral Domains and Fields

$D$ 是一个交换环。

$D$ 有一个统一的$e$，和$e \neq 0$。

$D$ 没有零因子。

$$[a] \cdot[b]=[a b]=[b a]=[b] \cdot[a]$$

数学代写|现代代数代写Modern Algebra代考|The Integral Domain Zn When n Is a Prime

$$n=a b \text { where } 1<a<n \text { and } 1<b<n .$$

$$[a][b]=[a b]=[n]=[0] .$$

数学代写|现代代数代写Modern Algebra代考|MATH310

statistics-lab™ 为您的留学生涯保驾护航 在代写现代代数Modern Algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写现代代数Modern Algebra代写方面经验极为丰富，各种代写现代代数Modern Algebra相关的作业也就用不着说。

数学代写|现代代数代写Modern Algebra代考|Sylow p-Subgroup

If $G$ is a finite abelian group and $p$ is a prime such that $p$ divides $|G|$, then $G_p$ is a Sylow $p$-subgroup.

Proof Assume that $G$ is a finite abelian group such that $p^m$ divides $|G|$ but $p^{m+1}$ does not divide $|G|$. Then $|G|=p^m k$, where $p$ and $k$ are relatively prime. We need to prove that $G_p$ has order $p^m$.

We first argue that $\left|G_p\right|$ is a power of $p$. If $\left|G_p\right|$ had a prime factor $q$ different from $p$, then $G_p$ would have to contain an element of order $q$, according to Cauchy’s Theorem. This would contradict the very definition of $G_p$, so we conclude that $\left|G_p\right|$ is a power of $p$. Let $\left|G_p\right|=p^t$.

Suppose now that $\left|G_p\right|<p^m$-that is, that $t<m$. Then the quotient group $G / G_p$ has order $p^m k / p^t=p^{m-t} k$, which is divisible by $p$. Hence $G / G_p$ contains an element $a+G_p$ of order $p$, by Theorem 4.41 . Then
$$G_p=p\left(a+G_p\right)=p a+G_p,$$
and this implies that $p a \in G_p$. Thus $p a$ has order that is a power of $p$. This implies that $a$ has order a power of $p$, and therefore $a \in G_p$; that is, $a+G_p=G_p$. This is a contradiction to the fact that $a+G_p$ has order $p$. Therefore, $\left|G_p\right|=p^m$, and $G_p$ is a Sylow $p$-subgroup of $G$.

The next theorem shows the true significance of the Sylow $p$-subgroups in the structure of abelian groups.

Let $G$ be an abelian group of order $n=p_1^{m_1} p_2^{m_2} \cdots p_r^{m_r}$ where the $p_i$ are distinct primes and each $m_i$ is a positive integer. Then
$$G=G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}$$
where $G_{p_i}$ is the Sylow $p_i$-subgroup of $G$ that corresponds to the prime $p_i$.
Proof Assume the hypothesis of the theorem. For each prime $p_i, G_{p_i}$ is a Sylow $p$-subgroup of $G$ by Theorem 4.42. Suppose an element $a_1 \in G_{p_1}$ is also in the subgroup generated by $G_{p_2}, G_{p_3}, \ldots, G_{p_r}$. Then
$$a_1=a_2+a_3+\cdots+a_r$$
where $a_i \in G_{p_i}$. Since $G_{p_i}$ has order $p_i^{m_i}, p_i^{m_i} a_i=0$ for $i=2, \ldots, r$. Hence
$$p_2^{m_2} p_3^{m_3} \cdots p_r^{m_r} a_1=0 \text {. }$$
Since the order of any $a_1 \in G_{p_1}$ is a power of $p_1$, and $p_1$ is relatively prime to $p_2^{m_2} p_3^{m_3} \cdots p_r^{m_r}$, this requires that $a_1=0$. A similar argument shows that the intersection of any $G_{p_i}$ with the subgroup generated by the remaining subgroups
$$G_{p_1}, G_{p_2}, \ldots, G_{p_{i-1}}, G_{p_{i+1}}, \ldots, G_{p_r}$$
is the identity subgroup ${0}$. Hence the sum
$$G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}$$
is direct and has order equal to the product of the orders $p_i^{m_i}$ :
$$\left|G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}\right|=p_1^{m_1} p_2^{m_2} \cdots p_r^{m_r}=|G| .$$
Therefore,
$$G=G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}$$

数学代写|现代代数代写Modern Algebra代考|Direct Sum of Cyclic Groups

Any finitely generated abelian group $G$ (and therefore any finite abelian group) is a direct sum of cyclic groups.

Proof The proof is by induction on the rank of $G$. If $G$ has rank 1 , then $G$ is cyclic and the theorem is true.

Assume that the theorem is true for any group of rank $k-1$, and let $G$ be a group of rank $k$. We consider two cases.

Case 1 Suppose there exists a minimal generating set $\left{a_1, a_2, \ldots, a_k\right}$ for $G$ such that any relation of the form
$$z_1 a_1+z_2 a_2+\cdots+z_k a_k=0$$
with $z_i \in \mathbf{Z}$ implies that $z_1 a_1=z_2 a_2=\cdots=z_k a_k=0$. Then
$$G=\left\langle a_1\right\rangle+\left\langle a_2\right\rangle+\cdots+\left\langle a_k\right\rangle,$$
and the theorem is true for this case.

Case 2 Suppose that Case 1 does not hold. That is, for any minimal generating set $\left{a_1, a_2, \ldots, a_k\right}$ of $G$, there exists a relation of the form
$$z_1 a_1+z_2 a_2+\cdots+z_k a_k=0$$
with $z_i \in \mathbf{Z}$ such that some of the $z_i a_i \neq 0$. Among all the minimal generating sets and all the relations of this form, there exists a smallest positive integer $\bar{z}_i$ that occurs as a coefficient in one of these relations. Suppose this $\bar{z}_i$ occurs in a relation with the generating set $\left{b_1, b_2, \ldots, b_k\right}$. If necessary, the elements in $\left{b_1, b_2, \ldots, b_k\right}$ can be rearranged so that this smallest positive coefficient occurs as $\bar{z}_1$ with $b_1$ in
$$\bar{z}_1 b_1+\bar{z}_2 b_2+\cdots+\bar{z}_k b_k=0 .$$
Now let $s_1, s_2, \ldots, s_k$ be any set of integers that occur as coefficients in a relation of the form
$$s_1 b_1+s_2 b_2+\cdots+s_k b_k=0$$
with these generators $b_i$. We shall show that $\bar{z}_1$ divides $s_1$. By the Division Algorithm, $s_1=\bar{z}_1 q_1+r_1$, where $0 \leq r_1<\bar{z}_1$. Multiplying equation (1) by $q_1$ and subtracting the result from equation (2), we have
$$r_1 b_1+\left(s_2-\bar{z}_2 q_1\right) b_2+\cdots+\left(s_k-\bar{z}_k q_1\right) b_k=0 .$$

现代代数代考

数学代写|现代代数代写Modern Algebra代考|Sylow p-Subgroup

$$G_p=p\left(a+G_p\right)=p a+G_p,$$

$$G=G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}$$

$$a_1=a_2+a_3+\cdots+a_r$$

$$p_2^{m_2} p_3^{m_3} \cdots p_r^{m_r} a_1=0 \text {. }$$

$$G_{p_1}, G_{p_2}, \ldots, G_{p_{i-1}}, G_{p_{i+1}}, \ldots, G_{p_r}$$

$$G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}$$

$$\left|G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}\right|=p_1^{m_1} p_2^{m_2} \cdots p_r^{m_r}=|G| .$$

$$G=G_{p_1} \oplus G_{p_2} \oplus \cdots \oplus G_{p_r}$$

数学代写|现代代数代写Modern Algebra代考|Direct Sum of Cyclic Groups

$$z_1 a_1+z_2 a_2+\cdots+z_k a_k=0$$

$$G=\left\langle a_1\right\rangle+\left\langle a_2\right\rangle+\cdots+\left\langle a_k\right\rangle,$$

$$z_1 a_1+z_2 a_2+\cdots+z_k a_k=0$$

$$\bar{z}_1 b_1+\bar{z}_2 b_2+\cdots+\bar{z}_k b_k=0 .$$

$$s_1 b_1+s_2 b_2+\cdots+s_k b_k=0$$

$$r_1 b_1+\left(s_2-\bar{z}_2 q_1\right) b_2+\cdots+\left(s_k-\bar{z}_k q_1\right) b_k=0 .$$

数学代写|现代代数代写Modern Algebra代考|MATH611

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数学代写|现代代数代写Modern Algebra代考|Nonzero preservation and the gcd of several polynomials

In this section, we discuss the following problem: Given nonzero polynomials $f_1, \ldots, f_n \in F[x]$ over a field $F$, compute $h=\operatorname{gcd}\left(f_1, \ldots, f_n\right)$. Let $d \in \mathbb{N}$ be such that $\operatorname{deg} f_i \leq d$ for all $i$. We are particularly interested in the case where $d$ is close to $n$. A simple approach is to set $h_1=f_1$ and compute $h_i=\operatorname{gcd}\left(h_{i-1}, f_i\right)$ for $i=2, \ldots, n$. If $\operatorname{deg} h$ is fairly large, say $d / 10$, then this will take $n-1 \operatorname{gcd}$ calculations of polynomials of degree at least $d / 10$.

We now present a more efficient algorithm that uses only one gcd calculation. The basic tool for this probabilistic algorithm is the following useful lemma. It says that a nonzero polynomial is likely to take a nonzero value at a random point. In other words, random evaluations probably preserve nonzeroness.

LEMMA 6.44. Let $R$ be an integral domain, $n \in \mathbb{N}, S \subseteq R$ finite with $s=# S$ elements, and $r \in R\left[x_1, \ldots, x_n\right]$ a polynomial of total degree at most $d \in \mathbb{N}$.
(i) If $r$ is not the zero polynomial, then $r$ has at most $d s^{n-1}$ zeroes in $S^n$.
(ii) If $s>d$ and $r$ vanishes on $S^n$, then $r=0$.
Proof. (i) We prove the claim by induction on $n$. The case $n=1$ is clear, since a nonzero univariate polynomial of degree at most $d$ over an integral domain has at most $d$ zeroes (Lemma 25.4). For the induction step, we write $r$ as a polynomial in $x_n$ with coefficients in $x_1, \ldots, x_{n-1}: r=\sum_{0 \leq i \leq k} r_i x_n^i$ with $r_i \in R\left[x_1, \ldots, x_{n-1}\right]$ for $0 \leq i \leq k$ and $r_k \neq 0$. Then $\operatorname{deg} r_k \leq d-k$, and by the induction hypothesis, $r_k$ has at most $(d-k) s^{n-2}$ zeroes in $S^{n-1}$, so that there are at most $(d-k) s^{n-1}$ common zeroes of $r$ and $r_k$ in $S^n$. Furthermore, for each $a \in S^{n-1}$ with $r_k(a) \neq 0$, the univariate polynomial $r_a=\sum_{0 \leq i \leq k} r_i(a) x_n^i \in R\left[x_n\right]$ of degree $k$ has at most $k$ zeroes, so that the total number of zeroes of $r$ in $S^n$ is bounded by
$$(d-k) s^{n-1}+k s^{n-1}=d s^{n-1} .$$
(ii) follows immediately from (i).

In the example $r=\prod_{1 \leq i \leq d}\left(x_n-a_i\right)$, where $a_1, \ldots, a_d \in S$ are distinct, the bound in (i) is achieved. A typical application of (i) is in the analysis of probabilistic algorithms, rewriting it as
$$\operatorname{prob}\left{r(a)=0: a \in S^n\right} \leq \frac{d}{# S},$$
where $a$ is chosen in $S^n$ uniformly at random. In the applications, the tricky part is usually to show that $r$ is nonzero. An amazing fact is that this bound on the probability is independent of the number of variables.

数学代写|现代代数代写Modern Algebra代考|Subresultants

In this section, we extend the resultant theory-which governs the gcd-to the subresultants which cover all results of the Extended Euclidean Algorithm. As before, this leads to efficient modular methods, but now for the whole algorithm. The reader only interested in efficient gcd algorithms may skip this and proceed directly to the implementation report in Section 6.13.

So now let $F$ be an arbitrary field, and $f, g \in F[x]$ nonzero of degrees $n \geq m$, respectively. We use the notation for the results of the Extended Euclidean Algorithm, as in (1) on page 141, and $n_i=\operatorname{deg} r_i$ for $0 \leq i \leq \ell+1$, with $r_{\ell+1}=0$ and $\operatorname{deg} r_{\ell+1}=-\infty$.
THEOREM 6.47.
Let $0 \leq k \leq m \leq n$. Then $k$ does not appear in the degree sequence if and only if there exist $s, t \in F[x]$ satisfying
$$t \neq 0, \quad \operatorname{deg} s<m-k, \quad \operatorname{deg} t<n-k, \quad \operatorname{deg}(s f+t g)<k .$$
there exists an $i$ with $2 \leq i \leq \ell+1$ such that $n_i<k<n_{i-1}$. We claim that $s=s_i$ and $t=t_i$ do the job. We have $s f+t g=r_i$, and $\operatorname{deg} r_i=n_i<k$. Furthermore, from Lemma 3.15 (b) we have
\begin{aligned} \operatorname{deg} s & =m-n_{i-1}<m-k, \ 0 \leq \operatorname{deg} t & =n-n_{i-1}<n-k . \end{aligned}

现代代数代考

数学代写|现代代数代写Modern Algebra代考|Nonzero preservation and the gcd of several polynomials

(i)如果$r$不是零多项式，则$r$在$S^n$中最多有$d s^{n-1}$个零。
(ii)如果$s>d$和$r$在$S^n$上消失，那么$r=0$。

$$(d-k) s^{n-1}+k s^{n-1}=d s^{n-1} .$$
(ii)紧接(i)。

$$\operatorname{prob}\left{r(a)=0: a \in S^n\right} \leq \frac{d}{# S},$$

数学代写|现代代数代写Modern Algebra代考|Subresultants

$$t \neq 0, \quad \operatorname{deg} s<m-k, \quad \operatorname{deg} t<n-k, \quad \operatorname{deg}(s f+t g)<k .$$

\begin{aligned} \operatorname{deg} s & =m-n_{i-1}<m-k, \ 0 \leq \operatorname{deg} t & =n-n_{i-1}<n-k . \end{aligned}

数学代写|现代代数代写Modern Algebra代考|MA320

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数学代写|现代代数代写Modern Algebra代考|Modular ged algorithm in F[x, y]

In Section 6.11, we present a modular algorithm that computes all results of the Extended Euclidean Algorithm, including the gcd and the Bézout coefficients. But if just the gcd is required, there is a better way which we now describe.

In a modular algorithm, say with a big prime $p$, we need two conditions to be satisfied: $p$ has to be large enough so that the coefficients of the gcd can be recovered from their images modulo $p$, and $p$ should not divide the resultant and the leading coefficient of the gcd, so that its degree does not change modulo $p$. When both input polynomials have degree about $n$ and coefficients of length $n$, then the bound for the first condition is $O(n)$, but for the second one it is about $n^2$. The trick now is to choose $p$ randomly so that coefficient recovery is always guaranteed, but the non-divisibility condition only with high probability.

This introduces the important method of probabilistic algorithms. Such an algorithm takes an input, makes some random choices (say, chooses several times a bit, either 0 or 1 , each with equal probability), does some calculations, and returns an output. If one can prove that the probability of returning the correct output is at least some value greater than $1 / 2$, say $2 / 3$, then one can run the algorithm repeatedly, and will obtain the correct answer by a majority vote with probability arbitrarily close to 1 . This is called a Monte Carlo algorithm. In some applications, as in this chapter, one can easily test the output for correctness. Then the error probability becomes zero, and only the running time is a random variable. This is called a Las Vegas algorithm. See Notes 6.5 and Section 25.8 for discussions.

These probabilistic algorithms actually started in computer algebra, with Berlekamp’s polynomial factorization (Section 14.8) and Solovay \& Strassen’s primality test (Section 18.5). Their power and simplicity has made them a ubiquitous tool in many areas of computer science. We have seen examples of probabilistic modular testing in Section 4.1. These methods have an inherent uncertainty, but it can be made arbitrarily small, and thus they are like playing a highly attractive lottery: the stakes are only a tiny fraction of the jackpot (say, polynomial time vs. exponential time), but you are almost guaranteed to win!

数学代写|现代代数代写Modern Algebra代考|Mignotte’s factor bound and a modular gcd algorithm in Z[x]

In order to adapt Algorithm 6.28 to $\mathbb{Z}[x]$, we need an a priori bound on the coefficient size of $h$. Over $F[y]$, the bound
$$\operatorname{deg}_y h \leq \operatorname{deg}_y f$$
is trivial and quite sufficient. Over $\mathbb{Z}$, we could use the subresultant bound of Theorem 6.52 below, but we now derive a much better bound. It actually depends only on one argument of the gcd, say $f$, and is valid for all factors of $f$. We will use this again for the factorization of $f$ in Chapter 15 .

We extend the 2-norm to a complex polynomial $f=\sum_{0 \leq i \leq n} f_i x^i \in \mathbb{C}[x]$ by $|f|_2=\left(\sum_{0 \leq i \leq n}\left|f_i\right|^2\right)^{1 / 2} \in \mathbb{R}$, where $|a|=(a \cdot \bar{a})^{1 / 2} \in \mathbb{R}$ is the norm of $a \in \mathbb{C}$ and $\bar{a}$ is the complex conjugate of $a$. We will derive a bound for the norm of factors of $f$ in terms of $|f|_2$, that is, a bound $B \in \mathbb{R}$ such that any factor $h \in \mathbb{Z}[x]$ of $f$ satisfies $|h|_2 \leq B$. One might hope that we can take $B=|f|_2$, but this is not the case. For example, let $f=x^n-1$ and $h=\Phi_n \in \mathbb{Z}[x]$ be the $n$th cyclotomic polynomial (Section 14.10). Thus $\Phi_n$ divides $x^n-1$, and the direct analog of (8) would say that each coefficient of $\Phi_n$ is at most 1 in absolute value, but for example $\Phi_{105}$, of degree 48 , contains the term $-2 x^7$. In fact, the coefficients of $\Phi_n$ are unbounded in absolute value if $n \longrightarrow \infty$, and hence this is also true for $|h|_2$. Worse yet, for infinitely many integers $n, \Phi_n$ has a very large coefficient, namely larger than $\exp (\exp (\ln 2 \cdot \ln n / \ln \ln n))$, where $\ln$ is the logarithm in base $e$; such a coefficient has word length somewhat less than $n$. It is not obvious how to control the coefficients of factors at all, and it is not surprising that we have to work a little bit to establish a good bound.

现代代数代考

数学代写|现代代数代写Modern Algebra代考|Mignotte’s factor bound and a modular gcd algorithm in Z[x]

$$\operatorname{deg}_y h \leq \operatorname{deg}_y f$$

数学代写|现代代数代写Modern Algebra代考|MATH342

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数学代写|现代代数代写Modern Algebra代考|Rational number reconstruction

The integer analog of rational function reconstruction is, given integers $m>g \geq 0$ and $k \in{1, \ldots, m}$, to compute a rational number $r / t \in \mathbb{Q}$, with $r, t \in \mathbb{Z}$, such that
$$\operatorname{gcd}(t, m)=1 \text { and } r t^{-1} \equiv g \bmod m, \quad|r|<k, \quad 0 \leq t \leq \frac{m}{k},$$
where $t^{-1}$ is the inverse of $t$ modulo $m$. As in the polynomial case, we will see that the related problem
$$r \equiv t g \bmod m, \quad|r|<k, \quad 0 \leq t \leq \frac{m}{k},$$
is always solvable, while (24) need not have a solution. The uniqueness statements are a bit weaker than in the polynomial case, however. The following lemma is the integer analog of the Uniqueness Lemma 5.15.

LeMmA 5.25. Let $f, g \in \mathbb{N}$ and $r, s, t \in \mathbb{Z}$ with $r=s f+t g$, and suppose that
$$|r|<k \text { and } 0<t \leq \frac{f}{k} \text { for some } k \in{1, \ldots, f}$$
We let $r_i, s_i, t_i \in \mathbb{Z}$ for $0 \leq i \leq \ell+1$ be the results of the traditional Extended Euclidean Algorithm for $f, g$, with $r_i \geq 0$ for all $i$. Moreover, we define $j \in$ ${1, \ldots, \ell+1}$ by
$$r_j<k \leq r_{j-1}$$
and if $j \leq \ell$, we choose $q \in \mathbb{N}{\geq 1}$ such that $$r{j-1}-q r_j<k \leq r_{j-1}-(q-1) r_j$$
and let $q=0$ if $j=\ell+1$. Then there exists a nonzero $\alpha \in \mathbb{Z}$ such that
$$\text { either }(r, s, t)=\left(\alpha r_j, \alpha s_j, \alpha t_j\right) \text { or }(r, s, t)=\left(\alpha r_j^, \alpha s_j^, \alpha t_j^\right),$$ where $r_j^=r_{j-1}-q r_j, s_j^=s_{j-1}-q s_j$, and $t_j^=t_{j-1}-q t_j$.

数学代写|现代代数代写Modern Algebra代考|Partial fraction decomposition

Let $F$ be a field, $f_1, \ldots, f_r \in F[x]$ nonconstant monic and pairwise coprime polynomials, $e_1, \ldots, e_r \in \mathbb{N}$ positive integers, and $f=f_1^{e_1} \cdots f_r^{e_r}$. (We will see in Part III how to factor polynomials over finite fields and over $\mathbb{Q}$ into irreducible factors, but here we do not assume irreducibility of the $f_i$.) For another polynomial $g \in F[x]$ of degree less than $n=\operatorname{deg} f$, the partial fraction decomposition of the rational function $g / f \in F(x)$ with respect to the given factorization of the denominator $f$ is
$$\frac{g}{f}=\frac{g_{1,1}}{f_1}+\cdots+\frac{g_{1, e_1}}{f_1^{e_1}}+\cdots+\frac{g_{r, 1}}{f_r}+\cdots+\frac{g_{r, e_r}}{f_r^{e_r}},$$
with $g_{i j} \in F[x]$ of smaller degree than $f_i$, for all $i, j$. If all $f_i$ are linear polynomials, then the $g_{i j}$ are just constants.

EXAMPLE 5.28. Let $F=\mathbb{Q}, f=x^4-x^2$, and $g=x^3+4 x^2-x-2$. The partial fraction decomposition of $g / f$ with respect to the factorization $f=x^2(x-1)(x+1)$ of $f$ into linear polynomials is
$$\frac{x^3+4 x^2-x-2}{x^4-x^2}=\frac{1}{x}+\frac{2}{x^2}+\frac{1}{x-1}+\frac{-1}{x+1} .$$
The following questions pose themselves: Does a decomposition as in (31) always exist uniquely, and how can we compute it? The next lemma is a first step towards an answer.

Lemma 5.29. There exist unique polynomials $c_i \in F[x]$ with $\operatorname{deg} c_i<e_i \operatorname{deg} f_i$ for all $i$ such that
$$\frac{g}{f}=\frac{c_1}{f_1^{e_1}}+\cdots+\frac{c_r}{f_r^{e_r}} .$$
Proof. We multiply both sides in (33) by $f$ and obtain the linear equation
$$g=c_1 \prod_{j \neq 1} f_j^{e_j}+\cdots+c_r \prod_{j \neq r} f_j^{e_j}$$
with “unknowns” $c_1, \ldots, c_r$. (We have already seen in Section 4.5 how to find polynomial solutions of such equations.) For any $i \leq r$, each summand with the possible exception of the $i$ th one is divisible by $f_i^{e_i}$, whence $g \equiv c_i \prod_{j \neq i} f_j^{e_j} \bmod f_i^{e_i}$. Now each $f_j$ is coprime to $f_i$ and hence invertible modulo $f_i^{e_i}$, and we obtain
$$c_i \equiv g \prod_{j \neq i} f_j^{-e_j} \bmod f_i^{e_i},$$
which together with $\operatorname{deg} c_i<\operatorname{deg} f_i^{e_i}$ uniquely determines $c_i$.

现代代数代考

数学代写|现代代数代写Modern Algebra代考|Rational number reconstruction

$$\operatorname{gcd}(t, m)=1 \text { and } r t^{-1} \equiv g \bmod m, \quad|r|<k, \quad 0 \leq t \leq \frac{m}{k},$$

$$r \equiv t g \bmod m, \quad|r|<k, \quad 0 \leq t \leq \frac{m}{k},$$

$$|r|<k \text { and } 0<t \leq \frac{f}{k} \text { for some } k \in{1, \ldots, f}$$