如果你也在 怎样代写现代代数Modern Algebra 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。现代代数Modern Algebra就像数学的其他分支一样——只有从最基本的思想和例子中仔细地推导才能掌握。但这需要时间，而且有些目标在你实现之前是不明确的。

现代代数Modern Algebra这门学科的思想和方法几乎渗透到现代数学的每一个部分。此外，没有一门学科更适合培养处理抽象概念的能力，即理解和处理问题或学科的基本要素。这包括阅读数学的能力，提出正确的问题，解决问题，运用演绎推理，以及写出正确、切中要害、清晰的数学。

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|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

在本节中，我们讨论以下问题:给定域$F$上的非零多项式$f_1, \ldots, f_n \in F[x]$，计算$h=\operatorname{gcd}\left(f_1, \ldots, f_n\right)$。让$d \in \mathbb{N}$为所有$i$成为$\operatorname{deg} f_i \leq d$。我们对$d$接近$n$的情况特别感兴趣。一种简单的方法是为$i=2, \ldots, n$设置$h_1=f_1$并计算$h_i=\operatorname{gcd}\left(h_{i-1}, f_i\right)$。如果$\operatorname{deg} h$相当大，比如$d / 10$，那么这将需要$n-1 \operatorname{gcd}$次多项式的计算，次数至少为$d / 10$。

我们现在提出了一个更有效的算法，它只使用一个gcd计算。这个概率算法的基本工具是以下有用的引理。它说一个非零多项式很可能在随机点取一个非零值。换句话说，随机求值可能保持非零性。

引理6.44。设$R$为一个积分域，$n \in \mathbb{N}, S \subseteq R$为包含$s=# S$个元素的有限域，$r \in R\left[x_1, \ldots, x_n\right]$为至多$d \in \mathbb{N}$个总次的多项式。
(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$r=0$。
证明。(i)我们通过归纳法在$n$上证明了这一主张。情况$n=1$是清楚的，因为在一个积分域上次数最多为$d$的非零单变量多项式最多有$d$个零(引理25.4)。对于归纳步骤，我们将$r$写成$x_n$中的多项式，其系数为$x_1, \ldots, x_{n-1}: r=\sum_{0 \leq i \leq k} r_i x_n^i$, $0 \leq i \leq k$和$r_k \neq 0$的系数为$r_i \in R\left[x_1, \ldots, x_{n-1}\right]$。然后是$\operatorname{deg} r_k \leq d-k$，根据归纳假设，$r_k$在$S^{n-1}$中最多有$(d-k) s^{n-2}$个零，因此$r$和$r_k$在$S^n$中最多有$(d-k) s^{n-1}$个公共零。更进一步，对于每个含有$r_k(a) \neq 0$的$a \in S^{n-1}$，次为$k$的单变量多项式$r_a=\sum_{0 \leq i \leq k} r_i(a) x_n^i \in R\left[x_n\right]$最多有$k$个零，因此$S^n$中$r$的总零个数为
$$
(d-k) s^{n-1}+k s^{n-1}=d s^{n-1} .
$$
(ii)紧接(i)。

在示例$r=\prod_{1 \leq i \leq d}\left(x_n-a_i\right)$中，$a_1, \ldots, a_d \in S$是不同的，实现了(i)中的边界。(i)的典型应用是在概率算法的分析中，将其重写为
$$
\operatorname{prob}\left{r(a)=0: a \in S^n\right} \leq \frac{d}{# S},
$$
其中$S^n$均匀随机选取$a$。在应用程序中，棘手的部分通常是证明$r$是非零的。一个令人惊奇的事实是概率的界限与变量的数量无关。

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

在本节中，我们将控制gcd的结果理论扩展到涵盖扩展欧几里得算法的所有结果的子结果。和以前一样，这导致了高效的模块化方法，但现在是整个算法。对高效gcd算法感兴趣的读者可以跳过这里，直接进入6.13节的实现报告。

现在设$F$为任意域，和$f, g \in F[x]$为非零度$n \geq m$。我们对扩展欧几里得算法的结果使用表示法，如第141页(1)所示，对$0 \leq i \leq \ell+1$使用$n_i=\operatorname{deg} r_i$，其中包含$r_{\ell+1}=0$和$\operatorname{deg} r_{\ell+1}=-\infty$。
定理6.47。
让$0 \leq k \leq m \leq n$。则当且仅当存在$s, t \in F[x]$满足时，$k$不出现在度序列中
$$
t \neq 0, \quad \operatorname{deg} s<m-k, \quad \operatorname{deg} t<n-k, \quad \operatorname{deg}(s f+t g)<k .
$$
存在一个与$2 \leq i \leq \ell+1$对应的$i$，使得$n_i<k<n_{i-1}$。我们声称$s=s_i$和$t=t_i$可以完成这项工作。我们有$s f+t g=r_i$和$\operatorname{deg} r_i=n_i<k$。更进一步，由引理3.15 (b)我们得到
$$
\begin{aligned}
\operatorname{deg} s & =m-n_{i-1}<m-k, \
0 \leq \operatorname{deg} t & =n-n_{i-1}<n-k .
\end{aligned}
$$

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写