多项式 商环 域(群论笔记)

多项式 商环 域

标签（空格分隔）： 多项式 商环

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域扩张

$[L:K]=[L:F]\cdot [F:K]$

有限扩张是代数扩
每个有限扩张都是代数扩张
任何一个有限代数扩张 $K(a_1,...,a_n)/K$ 都是有限生成的$

多项式商环在扩张域

多项式在域上可约

不能写成两个次数较低的多项式之乘积的多项式

最小多项式

$gcd(a_1,a_2,...,a_n)=1$

Ring

$(R,+)$ is abelian group

$(R,\cdot )$ is semi group

$\forall a,b,c\in R,a(b+c)=ab+ac,(b+c)a=ba+ca$

then $(R,+,\cdot )$ is a ring

$\mathbb{Z}/n\mathbb{Z}=\{\bar{0},\bar{1},...,\overline{n-1}\},\bar{i}=\{i+kn|k\in \mathbb{Z}\}$

$\bar{a}+\bar{b}=\overline{a+b}=\overline{b+a}=\bar{b}+\bar{a},\bar{a}\bar{b}=\overline{ab}$

polynomial ring

$R[x]=\{a_n{x}^n+\cdots +a_{1}x+a_0|a_i\in R,i=0,1,...,n,i\in {\mathbb{Z}}^{+}\}$

$f(\alpha )\in R[x],f(x)={\sum }_{j=1}^n{a}_i{x}^i=0$

$\mathbb{Z}[x]\to \mathbb{Z}/n\mathbb{Z}[x]$ is homomorphism

$Ker\varphi =\{f(x)\in R[x]|f(\alpha )=0\}=\{g(x)\in R[x]|(x-\alpha )g(x)\}$

polynomial quotient ring

$f(x),g(x)\in R[x]$

$$f(x)=\sum _{k=0}^n{a}_k{x}^k,g(x)=\sum _{k=0}^n{b}_k{x}^k$$

$$f(x)+g(x)=\sum _{k=0}^n{a}_k{x}^k+\sum _{k=0}^n{b}_k{x}^k=(a_0+a_{1}x+...+a_n{x}^n)+(b_0+b_{1}x+...+b_n{x}^n)=\sum _{k=0}^n(a_k+b_k)x^k$$

$$f(x)g(x)=\sum _{k=0}^n{a}_k{x}^k\otimes \sum _{k=0}^n{b}_k{x}^k=(a_0+a_{1}x+...+a_n{x}^n)\otimes (b_0+b_{1}x+...+b_n{x}^n)=\sum _{i=0}^n\sum _{j=0}^n{a}_i{b}_j{x}^{i+j}$$

$\forall f(x)\in I[x],\forall g(x)\in R[x]$

$f(x)g(x)\in I[x],g(x)f(x)\in I[x]$

$I[x]$ is ideal in $R[x]$

Ring R[x] homomorphism

$\phi :R_1\to R_2$

$\phi (a_i+b_j)=\phi (a_i)+\phi (b_j)$

$\phi (a_i{b}_j)=\phi (a_i)\phi (b_j)$

$a_i,b_j\in R_1$

R_1 and R_2 are homomorphism

$\psi (f(x))={\sum }_{i=0}^n\phi (a_i)x^i$

then
$\psi :R_1[x]\to R_2[x]$

$\psi (g(x)+h(x))=\psi (g(x))+\psi (h(x))$

$\psi (g(x)h(x))=\psi (g(x))\psi (h(x))$

$g(x),h(x)\in R_1[x]$

detail:
$$\begin{array}{rl}\psi (g(x)+h(x))& =\psi \left(\sum _{k=0}^n(a_k+b_k)x^k\right)\\ & =\sum _{k=0}^n\phi (a_k+b_k)x^k\\ & =\sum _{k=0}^n\left(\phi (a_k)+\phi (b_k)\right)x^k\\ & =\sum _{k=0}^n(\phi (a_k))x^k+\sum _{k=0}^n(\phi (b_k))x^k\\ & =\psi (g(x))+\psi (h(x))\end{array}$$

$$\begin{array}{rl}\psi (g(x)h(x))& =\psi \left(\sum _{i=0}^n\sum _{j=0}^n{a}_i{b}_j{x}^{i+j}\right)\\ & =\sum _{i=0}^n\sum _{j=0}^n\psi (a_i{b}_j)x^{i+j}\\ & =\sum _{i=0}^n\sum _{j=0}^n\psi (a_i)\psi (b_j)x^{i+j}\\ & =\sum _{i=0}^n\psi (a_i)x^i\sum _{j=0}^n\psi (b_j)x^j\\ & =\psi (g(x))\psi (h(x))\end{array}$$

quotient ring

$(P(x))=\{P(x)\ast h(x)|h(x)\in F(x)\}$ is ideal

$[F[x]/(P(x)),\oplus ,\otimes ]$ is Ring

0 is zero element in $F[x]$

1 is identity in $F[x]$

$(P(x))+0$ zero element

$(P(x))+1$ identity

$F[x]/(P(x))=\{(P(x))+{\sum }_{i=1}^{n-1}{a}_i{x}^i|a_i\in F,i=1,...n\}$

1. $\pi :R\to R/I,(a\to \bar{a})$ is homomorphism
2. $Ker\pi =I$
   $\pi (a+b)=a+b+I=a+b+I+I=a+I+b+I=\pi (a)+\pi (b)$
   $\pi (ab)=ab+I=ab+aI+bI+II=(a+I)(b+I)=\pi (a)\pi (b)$

$[{\mathbb{Z}}_2[x],+,\ast ],p(x)=x^2+x+1$
$$\begin{gathered} {\mathbb{Z}}_2[x]/(p(x)) \\ =\{(p(x)),(p(x))+1,(p(x))+x,(p(x))+(x+1)\} \\ =\{0,1,x,x+1\} \end{gathered}$$

+	0	1	x	x+1
0	0	1	x	x+1
1	1	0	x+1	x
x	x	x+1	0	1
x+1	x+1	x	1	0

*	0	1	x	x+1
0	0	0	0	0
1	0	1	x	x+1
x	0	x	x+1	1
x+1	0	x+1	1	x

${\mathbb{Z}}_3[x]/(x^2+1)=\{ax+b|a,b\in {\mathbb{Z}}_3\}$

Example. The polynomial $f(x)=x^3+2x+4\in {\mathbb{Z}}_5[x]$of degree 3 is irreducible because it has no roots in ${\mathbb{Z}}_5$

$${\mathbb{Z}}_n[x]=\left\{\sum _{i=0}^m{a}_i{x}^i\mid a_i\in {\mathbb{Z}}_n,m\in {\mathbb{Z}}^{+}\right\}$$
$${\mathbb{Z}}_n[x]/(p(x))=\left\{p(x)+\sum _{i=0}^m{a}_i{x}^i\mid a_i\in {\mathbb{Z}}_n,m=deg(p(x))-1\right\}$$

$(p(x),r(x))=a\in F^{\ast },s(x),t(x)\in F(x),p(x)s(x)+r(x)t(x)=a$
$\to (p(x))+a^{-1}t(x)=((p(x))+r(x){)}^{-1}$

长除法

$$\frac{x^5+x^4+x+1}{x^2+x+1}=x^3+x+1\cdots x$$

          x^3   +x        +1
        ------------------------
x^2+x+1| x^5+x^4        +x+1
         x^5+x^4+x^3
        ------------------------
                 x^3    +x+1
                 x^3+x^2+x
        ------------------------
                	 x^2  +1
                     x^2+x+1
        ------------------------
                         x

Field

Prime-Polynomial
一个$m$阶的不可约多项式$f(x)$，如果$f(x)$整除$x^n+1$的最小正整数$n$满足$n=2m-1$，则该多项式是本原的。

References:

1. 环R[x]的性质, 2012届学士学位论文

2. 离散数学教学课件 (12)

3. math.byu.edu bakker M371Lec19

4. 求解本原多项式的快速算法 1000—3428(2008)15—0146—02