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


域扩张

[ L : K ] = [ L : F ] ⋅ [ F : K ] [L:K]=[L:F] \cdot [F:K] [L:K]=[L:F][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 K(a_1,...,a_n)/K K(a1,...,an)/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

g c d ( a 1 , a 2 , . . . , a n ) = 1 gcd(a_1,a_2,...,a_n)=1 gcd(a1,a2,...,an)=1

Ring

( R , + ) (R,+) (R,+) is abelian group

( R , ⋅ ) (R,\cdot) (R,) is semi group

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

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

Z / n Z = { 0 ˉ , 1 ˉ , . . . , n − 1 ˉ } , i ˉ = { i + k n ∣ k ∈ Z } \mathbb{Z}/n\mathbb{Z}=\{ \bar{0}, \bar{1}, ..., \bar{n-1}\}, \bar{i}=\{i+kn|k\in \mathbb{Z}\} Z/nZ={0ˉ,1ˉ,...,n1ˉ},iˉ={i+knkZ}

a ˉ + b ˉ = a + b ˉ = b + a ˉ = b ˉ + a ˉ , a ˉ b ˉ = a b ˉ \bar{a}+\bar{b} = \bar{a+b}=\bar{b+a}=\bar{b}+\bar{a},\bar{a}\bar{b}=\bar{ab} aˉ+bˉ=a+bˉ=b+aˉ=bˉ+aˉ,aˉbˉ=abˉ

polynomial ring

R [ x ] = { a n x n + ⋯ + a 1 x + a 0 ∣ a i ∈ R , i = 0 , 1 , . . . , n , i ∈ Z + } R[x]=\{a_nx^n+ \cdots +a_1x+a_0|a_i \in R,i=0,1,...,n, i \in \mathbb{Z}^+ \} R[x]={anxn++a1x+a0aiR,i=0,1,...,n,iZ+}

f ( α ) ∈ R [ x ] , f ( x ) = ∑ j = 1 n a i x i = 0 f(\alpha) \in R[x], f(x)=\sum_{j=1}^na_ix^i=0 f(α)R[x],f(x)=j=1naixi=0

Z [ x ] → Z / n Z [ x ] \mathbb{Z}[x] \rightarrow \mathbb{Z}/n\mathbb{Z}[x] Z[x]Z/nZ[x] is homomorphism

K e r ϕ = { f ( x ) ∈ R [ x ] ∣ f ( α ) = 0 } = { g ( x ) ∈ R [ x ] ∣ ( x − α ) g ( x ) } Ker \phi=\{f(x) \in R[x]|f(\alpha)=0\}=\{g(x)\in R[x]|(x-\alpha)g(x)\} Kerϕ={f(x)R[x]f(α)=0}={g(x)R[x](xα)g(x)}

polynomial quotient ring

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

f ( x ) = ∑ k = 0 n a k x k , g ( x ) = ∑ k = 0 n b k x k f(x)=\sum_{k=0}^na_kx^k, g(x)=\sum_{k=0}^nb_kx^k f(x)=k=0nakxk,g(x)=k=0nbkxk

f ( x ) + g ( x ) = ∑ k = 0 n a k x k + ∑ 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 ) = ∑ k = 0 n ( a k + b k ) x k f(x)+g(x)=\sum_{k=0}^na_kx^k + \sum_{k=0}^nb_kx^k =(a_0+a_1x+...+a_nx^n)+(b_0+b_1x+...+b_nx^n)= \sum_{k=0}^n(a_k+b_k)x^k f(x)+g(x)=k=0nakxk+k=0nbkxk=(a0+a1x+...+anxn)+(b0+b1x+...+bnxn)=k=0n(ak+bk)xk

f ( x ) g ( x ) = ∑ k = 0 n a k x k ⊗ ∑ 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 ) = ∑ i = 0 n ∑ j = 0 n a i b j x i + j f(x)g(x)=\sum_{k=0}^na_kx^k \otimes \sum_{k=0}^nb_kx^k=(a_0+a_1x+...+a_nx^n) \otimes (b_0+b_1x+...+b_nx^n)=\sum_{i=0}^n\sum_{j=0}^n{a_ib_jx^{i+j}} f(x)g(x)=k=0nakxkk=0nbkxk=(a0+a1x+...+anxn)(b0+b1x+...+bnxn)=i=0nj=0naibjxi+j

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

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

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

Ring R[x] homomorphism

φ : R 1 → R 2 \varphi:R_1 \rightarrow R_2 φ:R1R2

φ ( a i + b j ) = φ ( a i ) + φ ( b j ) \varphi(a_i+b_j)=\varphi(a_i)+\varphi(b_j) φ(ai+bj)=φ(ai)+φ(bj)

φ ( a i b j ) = φ ( a i ) φ ( b j ) \varphi(a_i b_j)=\varphi(a_i)\varphi(b_j) φ(aibj)=φ(ai)φ(bj)

a i , b j ∈ R 1 a_i, b_j \in R_1 ai,bjR1

R_1 and R_2 are homomorphism

ψ ( f ( x ) ) = ∑ i = 0 n φ ( a i ) x i \psi(f(x))=\sum_{i=0}^n\varphi(a_i)x^i ψ(f(x))=i=0nφ(ai)xi

then
ψ : R 1 [ x ] → R 2 [ x ] \psi:R_1[x] \rightarrow R_2[x] ψ:R1[x]R2[x]

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

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

g ( x ) , h ( x ) ∈ R 1 [ x ] g(x), h(x) \in R_1[x] g(x),h(x)R1[x]

detail:
ψ ( g ( x ) + h ( x ) ) = ψ ( ∑ k = 0 n ( a k + b k ) x k ) = ∑ k = 0 n φ ( a k + b k ) x k = ∑ k = 0 n ( φ ( a k ) + φ ( b k ) ) x k = ∑ k = 0 n ( φ ( a k ) ) x k + ∑ k = 0 n ( φ ( b k ) ) x k = ψ ( g ( x ) ) + ψ ( h ( x ) ) \begin{align} \psi(g(x)+h(x))&=\psi \left( \sum_{k=0}^n(a_k+b_k)x^k \right)\\ &=\sum_{k=0}^n\varphi(a_k+b_k)x^k\\ &=\sum_{k=0}^n\left(\varphi(a_k) + \varphi(b_k)\right)x^k \\ &=\sum_{k=0}^n(\varphi(a_k))x^k + \sum_{k=0}^n(\varphi(b_k))x^k\\ &=\psi(g(x)) + \psi(h(x)) \end{align} ψ(g(x)+h(x))=ψ(k=0n(ak+bk)xk)=k=0nφ(ak+bk)xk=k=0n(φ(ak)+φ(bk))xk=k=0n(φ(ak))xk+k=0n(φ(bk))xk=ψ(g(x))+ψ(h(x))

ψ ( g ( x ) h ( x ) ) = ψ ( ∑ i = 0 n ∑ j = 0 n a i b j x i + j ) = ∑ i = 0 n ∑ j = 0 n ψ ( a i b j ) x i + j = ∑ i = 0 n ∑ j = 0 n ψ ( a i ) ψ ( b j ) x i + j = ∑ i = 0 n ψ ( a i ) x i ∑ j = 0 n ψ ( b j ) x j = ψ ( g ( x ) ) ψ ( h ( x ) ) \begin{align} \psi(g(x)h(x))&=\psi \left( \sum_{i=0}^n\sum_{j=0}^n{a_ib_jx^{i+j}} \right)\\ &=\sum_{i=0}^n\sum_{j=0}^n{\psi(a_ib_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{align} ψ(g(x)h(x))=ψ(i=0nj=0naibjxi+j)=i=0nj=0nψ(aibj)xi+j=i=0nj=0nψ(ai)ψ(bj)xi+j=i=0nψ(ai)xij=0nψ(bj)xj=ψ(g(x))ψ(h(x))

quotient ring

( P ( x ) ) = { P ( x ) ∗ h ( x ) ∣ h ( x ) ∈ F ( x ) } (P(x))=\{P(x)*h(x)|h(x)\in F(x)\} (P(x))={P(x)h(x)h(x)F(x)} is ideal

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

0 is zero element in F [ x ] F[x] F[x]

1 is identity in F [ x ] F[x] F[x]

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

( P ( x ) ) + 1 (P(x))+1 (P(x))+1 identity

F [ x ] / ( P ( x ) ) = { ( P ( x ) ) + ∑ i = 1 n − 1 a i x i ∣ a i ∈ F , i = 1 , . . . n } F[x]/(P(x))=\{(P(x))+\sum_{i=1}^{n-1}a_ix^i|a_i \in F, i=1,...n\} F[x]/(P(x))={(P(x))+i=1n1aixiaiF,i=1,...n}

  1. π : R → R / I , ( a → a ˉ ) \pi: R \rightarrow R/I, (a \rightarrow \bar{a}) π:RR/I,(aaˉ) is homomorphism
  2. K e r π = I Ker \pi = I Kerπ=I
    π ( a + b ) = a + b + I = a + b + I + I = a + I + b + I = π ( a ) + π ( b ) \pi(a+b)=a+b+I=a+b+I+I=a+I+b+I=\pi(a)+\pi(b) π(a+b)=a+b+I=a+b+I+I=a+I+b+I=π(a)+π(b)
    π ( a b ) = a b + I = a b + a I + b I + I I = ( a + I ) ( b + I ) = π ( a ) π ( b ) \pi(ab)=ab+I=ab+aI+bI+II=(a+I)(b+I)=\pi(a)\pi(b) π(ab)=ab+I=ab+aI+bI+II=(a+I)(b+I)=π(a)π(b)

[ Z 2 [ x ] , + , ∗ ] , p ( x ) = x 2 + x + 1 [\mathbb{Z}_2[x], +, *], p(x)=x^2+x+1 [Z2[x],+,],p(x)=x2+x+1
Z 2 [ x ] / ( p ( x ) ) = { ( p ( x ) ) , ( p ( x ) ) + 1 , ( p ( x ) ) + x , ( p ( x ) ) + ( x + 1 ) } = { 0 , 1 , x , x + 1 } \mathbb{Z}_2[x]/(p(x)) \\ =\{(p(x)), (p(x))+1, (p(x))+x, (p(x))+(x+1)\} \\ =\{0, 1,x,x+1\} Z2[x]/(p(x))={(p(x)),(p(x))+1,(p(x))+x,(p(x))+(x+1)}={0,1,x,x+1}

+ 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

Z 3 [ x ] / ( x 2 + 1 ) = { a x + b ∣ a , b ∈ Z 3 } \mathbb{Z}_3[x]/(x^2+1)=\{ax+b|a,b\in \mathbb{Z}_3\} Z3[x]/(x2+1)={ax+ba,bZ3}

Example. The polynomial f ( x ) = x 3 + 2 x + 4 ∈ Z 5 [ x ] f(x) = x^3 + 2x+ 4 \in \mathbb{Z}_5[x] f(x)=x3+2x+4Z5[x]of degree 3 is irreducible because it has no roots in Z 5 \mathbb{Z}_5 Z5

Z n [ x ] = { ∑ i = 0 m a i x i ∣ a i ∈ Z n , m ∈ Z + } \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 \} Zn[x]={i=0maixiaiZn,mZ+}
Z n [ x ] / ( p ( x ) ) = { p ( x ) + ∑ i = 0 m a i x i ∣ a i ∈ Z n , m = d e g ( p ( x ) ) − 1 } \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\} Zn[x]/(p(x))={p(x)+i=0maixiaiZn,m=deg(p(x))1}

( p ( x ) , r ( x ) ) = a ∈ F ∗ , s ( x ) , t ( x ) ∈ F ( x ) , p ( x ) s ( x ) + r ( x ) t ( x ) = a (p(x), r(x))=a\in F^*,s(x), t(x) \in F(x), p(x)s(x)+r(x)t(x)=a (p(x),r(x))=aF,s(x),t(x)F(x),p(x)s(x)+r(x)t(x)=a
→ ( p ( x ) ) + a − 1 t ( x ) = ( ( p ( x ) ) + r ( x ) ) − 1 \rightarrow (p(x))+a^{-1}t(x)=((p(x))+r(x))^{-1} (p(x))+a1t(x)=((p(x))+r(x))1

长除法

x 5 + x 4 + x + 1 x 2 + x + 1 = x 3 + x + 1 ⋯ x \frac{x^5+x^4+x+1}{x^2+x+1}=x^3+x+1\cdots x x2+x+1x5+x4+x+1=x3+x+1x

          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 m m阶的不可约多项式 f ( x ) f(x) f(x),如果 f ( x ) f(x) f(x)整除 x n + 1 x^n+1 xn+1的最小正整数 n n n满足 n = 2 m − 1 n =2m−1 n=2m1,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

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