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数学代写|现代代数代考Modern Algebra代写|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
$$
t \neq 0, \quad \operatorname{deg} s<m-k, \quad \operatorname{deg} t<n-k, \quad \operatorname{deg}(s f+t g)<k .
$$
Proof. ” $\Longrightarrow “$ : Suppose that $k$ does not appear in the degree sequence. Then 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
$$
\begin{aligned}
\operatorname{deg} s & =m-n_{i-1}<m-k, \
0 \leq \operatorname{deg} t & =n-n_{i-1}<n-k .
\end{aligned}
$$
The case $i=\ell+1$ gives $s=g / r_{\ell}$ and $t=-f / r_{\ell}$, where $k<n_{\ell}$ and $r_{\ell+1}=0$.
” $=”:$ Suppose there exist $s, t \in F[x]$ satisfying (11). The Uniqueness Lemma 5.15 implies that there exist $i \in{1, \ldots, \ell+1}$ and $\alpha \in F[x] \backslash{0}$ such that $t=\alpha t_i$ and $r=s f+t g=\alpha r_i$. Then from Lemma 3.15 (b) we find
$$
\begin{aligned}
n-n_{i-1} & \leq \operatorname{deg} \alpha+n-n_{i-1}=\operatorname{deg}\left(\alpha t_i\right)=\operatorname{deg} t<n-k, \
n_i & \leq \operatorname{deg} \alpha+n_i=\operatorname{deg}\left(\alpha r_i\right)=\operatorname{deg} r<k .
\end{aligned}
$$
Together these imply that $n_i<k<n_{i-1}$, so that $k$ is between two consecutive remainder degrees and does not occur in the degree sequence.

数学代写|现代代数代考Modern Algebra代写|Modular Extended Euclidean Algorithms

In this section, we use the subresultants from Section 6.10 to prove a bound on the coefficients of the Extended Euclidean Algorithm 3.14 over $\mathbb{Q}[x]$ and $F(y)[x]$, and to derive modular algorithms for the results of the EEA.
THEOREM 6.52.
Let $f, g \in \mathbb{Z}[x]$ have degrees $n \geq m$ and max-norm $|f|_{\infty},|g|_{\infty}$ at most $A$, and let $\delta=\max \left{n_{i-1}-n_i: 1 \leq i \leq \ell\right}$ be the maximal degree difference of consecutive remainders. The results $r_i, s_i, t_i$ of the Extended Euclidean Algorithm 3.14 for $f$ and $g$ in $\mathbb{Q}[x]$ have numerators and denominators (in lowest terms) absolutely bounded by $B=(n+1)^n A^{n+m}$. The corresponding bound for $q_i$ and $\rho_i$ is $C=(2 B)^{\delta+2}$. The algorithm can be performed with $O\left(n^3 m \delta^2 \log ^2(n A)\right)$ word operations.

PROoF. Let $2 \leq i \leq \ell$ and $n_i=\operatorname{deg} r_i$. In the EEA, $s_i$ and $t_i$ form the unique solution to the system (12) of linear equations, so that $\sigma_{n_i} s_i, \sigma_{n_i} t_i$, and $\sigma_{n_i} r_i=\sigma_{n_i} s_i f+\sigma_{n_i} t_i g$ are in $\mathbb{Z}[x]$, and by Cramer’s rule 25.6 and Hadamard’s inequality 16.6 we have
$$
\begin{aligned}
\left|\sigma_{n_i}\right| & \leq|f|_2^{m-n_i}|g|_2^{n-n_i} \leq(n+1)^{n-n_i} A^{n+m-2 n_i} \leq B, \
\left|\sigma_{n_i} s_i\right|_{\infty} & \leq|f|_2^{m-n_i-1}|g|_2^{n-n_i} \leq(n+1)^{n-n_i-1 / 2} A^{n+m-2 n_i-1} \leq B, \
\left|\sigma_{n_i} t_i\right|_{\infty} & \leq|f|_2^{m-n_i}|g|_2^{n-n_i-1} \leq(n+1)^{n-n_i-1 / 2} A^{n+m-2 n_i-1} \leq B, \
\left|\sigma_{n_i} r_i\right|_{\infty} & =\left|\left(\sigma_{n_i} s_i\right) f+\left(\sigma_{n_i} t_i\right) g\right|_{\infty} \leq\left(n_i+1\right)\left(\left|\sigma_{n_i} s_i\right|_{\infty} \cdot|f|_{\infty}+\left|\sigma_{n_i} t_i\right|_{\infty} \cdot|g|_{\infty}\right) \
& \leq\left(n_i+1\right) \cdot 2(n+1)^{n-n_i-1 / 2} A^{n+m-2 n_i} \leq 2(n+1)^{1 / 2} B .
\end{aligned}
$$

现代代数代写

数学代写|现代代数代考Modern Algebra代写|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
$$
t \neq 0, \quad \operatorname{deg} s<m-k, \quad \operatorname{deg} t<n-k, \quad \operatorname{deg}(s f+t g)<k .
$$
证明。$\Longrightarrow “$:假设$k$不出现在度序列中。然后存在一个$i$和$2 \leq i \leq \ell+1$，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
$$
\begin{aligned}
\operatorname{deg} s & =m-n_{i-1}<m-k, \
0 \leq \operatorname{deg} t & =n-n_{i-1}<n-k .
\end{aligned}
$$
案例$i=\ell+1$给出$s=g / r_{\ell}$和$t=-f / r_{\ell}$，其中$k<n_{\ell}$和$r_{\ell+1}=0$。
“$=”:$假设存在$s, t \in F[x]$满足(11)。唯一性引理5.15暗示存在$i \in{1, \ldots, \ell+1}$和$\alpha \in F[x] \backslash{0}$，使得$t=\alpha t_i$和$r=s f+t g=\alpha r_i$。然后从引理3.15 (b)我们发现
$$
\begin{aligned}
n-n_{i-1} & \leq \operatorname{deg} \alpha+n-n_{i-1}=\operatorname{deg}\left(\alpha t_i\right)=\operatorname{deg} t<n-k, \
n_i & \leq \operatorname{deg} \alpha+n_i=\operatorname{deg}\left(\alpha r_i\right)=\operatorname{deg} r<k .
\end{aligned}
$$
它们合在一起意味着$n_i<k<n_{i-1}$，因此$k$位于两个连续的余数度数之间，不出现在度数序列中。

数学代写|现代代数代考Modern Algebra代写|Modular Extended Euclidean Algorithms

在本节中，我们使用第6.10节的子结果来证明扩展欧几里得算法3.14在$\mathbb{Q}[x]$和$F(y)[x]$上的系数的一个界，并推导出EEA结果的模块化算法。
定理6.52。
设$f, g \in \mathbb{Z}[x]$有度数$n \geq m$，最大范数$|f|{\infty},|g|{\infty}$最多为$A$，设$\delta=\max \left{n_{i-1}-n_i: 1 \leq i \leq \ell\right}$为连续余数的最大度数差。对于$\mathbb{Q}[x]$中的$f$和$g$，扩展欧几里得算法3.14的结果$r_i, s_i, t_i$的分子和分母(最低项)绝对以$B=(n+1)^n A^{n+m}$为界。$q_i$和$\rho_i$对应的绑定为$C=(2 B)^{\delta+2}$。该算法可以通过$O\left(n^3 m \delta^2 \log ^2(n A)\right)$字操作来执行。

证明。让$2 \leq i \leq \ell$和$n_i=\operatorname{deg} r_i$。在EEA中，$s_i$和$t_i$构成线性方程组(12)的唯一解，因此$\sigma_{n_i} s_i, \sigma_{n_i} t_i$和$\sigma_{n_i} r_i=\sigma_{n_i} s_i f+\sigma_{n_i} t_i g$在$\mathbb{Z}[x]$中，根据Cramer规则25.6和Hadamard不等式16.6，我们有
$$
\begin{aligned}
\left|\sigma_{n_i}\right| & \leq|f|2^{m-n_i}|g|_2^{n-n_i} \leq(n+1)^{n-n_i} A^{n+m-2 n_i} \leq B, \ \left|\sigma{n_i} s_i\right|{\infty} & \leq|f|_2^{m-n_i-1}|g|_2^{n-n_i} \leq(n+1)^{n-n_i-1 / 2} A^{n+m-2 n_i-1} \leq B, \ \left|\sigma{n_i} t_i\right|{\infty} & \leq|f|_2^{m-n_i}|g|_2^{n-n_i-1} \leq(n+1)^{n-n_i-1 / 2} A^{n+m-2 n_i-1} \leq B, \ \left|\sigma{n_i} r_i\right|{\infty} & =\left|\left(\sigma{n_i} s_i\right) f+\left(\sigma_{n_i} t_i\right) g\right|{\infty} \leq\left(n_i+1\right)\left(\left|\sigma{n_i} s_i\right|{\infty} \cdot|f|{\infty}+\left|\sigma_{n_i} t_i\right|{\infty} \cdot|g|{\infty}\right) \
& \leq\left(n_i+1\right) \cdot 2(n+1)^{n-n_i-1 / 2} A^{n+m-2 n_i} \leq 2(n+1)^{1 / 2} B .
\end{aligned}
$$

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