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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
有理数函数重构的整数类比是，给定整数$m>g \geq 0$和$k \in{1, \ldots, m}$，用$r, t \in \mathbb{Z}$计算有理数$r / t \in \mathbb{Q}$，使得
$$
\operatorname{gcd}(t, m)=1 \text { and } r t^{-1} \equiv g \bmod m, \quad|r|<k, \quad 0 \leq t \leq \frac{m}{k},
$$
其中$t^{-1}$是$t$模$m$的倒数。在多项式的情况下，我们会看到相关的问题
$$
r \equiv t g \bmod m, \quad|r|<k, \quad 0 \leq t \leq \frac{m}{k},
$$
是永远可解的，而(24)不必有解。然而，唯一性语句比多项式的情况弱一些。下面的引理是唯一性引理5.15的整数类比。
引理5.25。让$f, g \in \mathbb{N}$和$r, s, t \in \mathbb{Z}$等于$r=s f+t g$，假设
$$
|r|<k \text { and } 0<t \leq \frac{f}{k} \text { for some } k \in{1, \ldots, f}
$$
我们设$r_i, s_i, t_i \in \mathbb{Z}$为$0 \leq i \leq \ell+1$的传统扩展欧几里得算法对$f, g$的结果，$r_i \geq 0$为所有$i$。此外，我们通过定义$j \in$${1, \ldots, \ell+1}$
$$
r_j<k \leq r_{j-1}
$$
如果$j \leq \ell$，我们选择$q \in \mathbb{N}{\geq 1}$使得$$ r{j-1}-q r_j<k \leq r_{j-1}-(q-1) r_j
$$
让$q=0$ if $j=\ell+1$。那么存在一个非零$\alpha \in \mathbb{Z}$，使得
$$
\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), $$，其中$r_j^=r_{j-1}-q r_j, s_j^=s_{j-1}-q s_j$和$t_j^=t_{j-1}-q t_j$。
数学代写|现代代数代写Modern Algebra代考|Partial fraction decomposition
设$F$为一个域，$f_1, \ldots, f_r \in F[x]$为非常一元多项式和对素数多项式，$e_1, \ldots, e_r \in \mathbb{N}$为正整数，$f=f_1^{e_1} \cdots f_r^{e_r}$为正整数。(我们将在第三部分看到如何将有限域和$\mathbb{Q}$上的多项式分解为不可约因子，但在这里我们不假设$f_i$不可约。)对于另一个次小于$n=\operatorname{deg} f$的多项式$g \in F[x]$，有理函数$g / f \in F(x)$相对于给定的分母$f$的因式分解的部分分式分解为
$$
\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}},
$$
对于所有$i, j$, $g_{i j} \in F[x]$的度数小于$f_i$。如果所有的$f_i$都是线性多项式，那么$g_{i j}$就是常数。
例5.28。设$F=\mathbb{Q}, f=x^4-x^2$和$g=x^3+4 x^2-x-2$。$g / f$对于$f$的因式分解$f=x^2(x-1)(x+1)$的部分分式分解为线性多项式为
$$
\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} .
$$
下面的问题提出了:(31)中的分解是否总是唯一存在，我们如何计算它?下一个引理是找到答案的第一步。
引理5.29。对于所有$i$存在唯一的多项式$c_i \in F[x]$和$\operatorname{deg} c_i<e_i \operatorname{deg} f_i$，使得
$$
\frac{g}{f}=\frac{c_1}{f_1^{e_1}}+\cdots+\frac{c_r}{f_r^{e_r}} .
$$
证明。我们在(33)两边乘以$f$得到线性方程
$$
g=c_1 \prod_{j \neq 1} f_j^{e_j}+\cdots+c_r \prod_{j \neq r} f_j^{e_j}
$$
与“未知”$c_1, \ldots, c_r$。(我们已经在第4.5节看到了如何找到这些方程的多项式解。)对于任何$i \leq r$，除$i$外，每个求和都可以被$f_i^{e_i}$整除，因此$g \equiv c_i \prod_{j \neq i} f_j^{e_j} \bmod f_i^{e_i}$。现在每个$f_j$都是$f_i$的互素数因此对$f_i^{e_i}$取可逆模，我们得到
$$
c_i \equiv g \prod_{j \neq i} f_j^{-e_j} \bmod f_i^{e_i},
$$
它和$\operatorname{deg} c_i<\operatorname{deg} f_i^{e_i}$一起决定了$c_i$。
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