环
环的定义与基本性质
下面我们定义一种新的代数结构。如果一个集合R R R + + + ⋅ \cdot ⋅ ( R , + ) (R,+) ( R , + ) ( R , ⋅ ) (R,\cdot) ( R , ⋅ ) ∀ a , b , c ∈ R , a ⋅ ( b + c ) = a ⋅ b + a ⋅ c , ( a + b ) ⋅ c = a ⋅ c + b ⋅ c \forall a,b,c\in R,a\cdot (b+c)=a\cdot b+a\cdot c,(a+b)\cdot c=a\cdot c+b\cdot c ∀ a , b , c ∈ R , a ⋅ ( b + c ) = a ⋅ b + a ⋅ c , ( a + b ) ⋅ c = a ⋅ c + b ⋅ c ( R , + , ⋅ ) (R,+,\cdot) ( R , + , ⋅ )
注意到，在环的定义中并没有要求乘法单位元存在，存在乘法单位元的环称为有幺环。其次，环中的加法始终满足交换律，但乘法交换律是没有保证的，所以当我们说“交换环”时特指环中的乘法运算也满足交换律。在环中，加法单位元记为0 0 0 1 1 1 R = { 0 } R=\{0\} R = { 0 }
基于环的定义，可以证明以下运算法则成立：① a ⋅ 0 = 0 ⋅ a = 0 a\cdot 0=0\cdot a=0 a ⋅ 0 = 0 ⋅ a = 0 a ⋅ ( 0 + 0 ) = a ⋅ 0 + a ⋅ 0 a\cdot(0+0)=a\cdot0+a\cdot 0 a ⋅ ( 0 + 0 ) = a ⋅ 0 + a ⋅ 0 0 + 0 = 0 0+0=0 0 + 0 = 0 a ⋅ ( 0 + 0 ) = a ⋅ 0 a\cdot (0+0)=a\cdot 0 a ⋅ ( 0 + 0 ) = a ⋅ 0 a ⋅ 0 = 0 a\cdot 0=0 a ⋅ 0 = 0 ( − a ) ⋅ b = a ⋅ ( − b ) = − ( a ⋅ b ) (-a)\cdot b=a\cdot (-b)=-(a\cdot b) ( − a ) ⋅ b = a ⋅ ( − b ) = − ( a ⋅ b ) − a -a − a a a a ( a + ( − a ) ) ⋅ b = 0 ⋅ b = 0 (a+(-a))\cdot b=0\cdot b=0 ( a + ( − a )) ⋅ b = 0 ⋅ b = 0 ( − a ) ⋅ b = − ( a ⋅ b ) (-a)\cdot b=-(a\cdot b) ( − a ) ⋅ b = − ( a ⋅ b ) ( − a ) ⋅ ( − b ) = a ⋅ b (-a)\cdot (-b)= a\cdot b ( − a ) ⋅ ( − b ) = a ⋅ b ( − a ) ⋅ ( − b ) = − ( a ⋅ ( − b ) ) = − ( − ( a ⋅ b ) ) = a ⋅ b (-a)\cdot (-b)=-(a\cdot (-b))=-(-(a\cdot b))=a\cdot b ( − a ) ⋅ ( − b ) = − ( a ⋅ ( − b )) = − ( − ( a ⋅ b )) = a ⋅ b 1 ≠ 0 1\neq 0 1 = 0 1 = 0 1=0 1 = 0 ∀ a ∈ R , a = 1 ⋅ a = 0 ⋅ a = 0 \forall a\in R,a=1\cdot a=0\cdot a=0 ∀ a ∈ R , a = 1 ⋅ a = 0 ⋅ a = 0 1 ≠ 1 ′ 1\neq 1' 1 = 1 ′ 1 ⋅ 1 ′ = 1 = 1 ′ 1\cdot 1'=1=1' 1 ⋅ 1 ′ = 1 = 1 ′
如果a ≠ 0 , b ≠ 0 a\neq 0,b\neq 0 a = 0 , b = 0 a ⋅ b = 0 a\cdot b=0 a ⋅ b = 0 a a a b b b R R R a ∈ R a\in R a ∈ R a a a ( Z , + , ⋅ ) (\Z,+,\cdot) ( Z , + , ⋅ ) ± 1 \pm 1 ± 1 ( Z n , + , ⋅ ) (\Z_n,+,\cdot) ( Z n , + , ⋅ ) n n n n = 4 n=4 n = 4 2 2 2 1 1 1 3 3 3 w ∈ R w\in R w ∈ R u ≠ 0 u\neq 0 u = 0 u w = 0 uw=0 u w = 0 v v v w v = 1 wv=1 w v = 1 u w v = u uwv=u u w v = u u w v = ( u w ) v = 0 uwv=(uw)v=0 u w v = ( u w ) v = 0 u = 0 u=0 u = 0
把有幺环R R R a , b , c , d ∈ R a,b,c,d\in R a , b , c , d ∈ R a c = 1 , b d = 1 ac=1,bd=1 a c = 1 , b d = 1 ( a b ) ( d c ) = a ( b d ) c = 1 (ab)(dc)=a(bd)c=1 ( ab ) ( d c ) = a ( b d ) c = 1 a b ab ab ( Z n , + , ⋅ ) (\Z_n,+,\cdot) ( Z n , + , ⋅ ) n n n ( Z n ∗ , ⋅ ) (\Z_n^*,\cdot) ( Z n ∗ , ⋅ )
环中乘法消去律成立（左消去律与右消去律，并且我们特别要求只能消去非零元）当且仅当环中无零因子。左推右，如果存在零因子，那么∃ a , b ≠ 0 \exists a,b\neq 0 ∃ a , b = 0 a ⋅ b = 0 a\cdot b = 0 a ⋅ b = 0 a ⋅ b = a ⋅ 0 a\cdot b = a\cdot 0 a ⋅ b = a ⋅ 0 b = 0 b=0 b = 0 a b = a c ab=ac ab = a c a ( b − c ) = 0 a(b-c)=0 a ( b − c ) = 0 a = 0 a=0 a = 0 b − c = 0 b-c=0 b − c = 0 a ≠ 0 ⟹ b = c a\neq 0\implies b=c a = 0 ⟹ b = c b a = c a ba=ca ba = c a a ≠ 0 ⟹ b = c a\neq 0\implies b=c a = 0 ⟹ b = c
整环、除环、域
如果一个环有乘法单位元、乘法满足交换律、无零因子，就称它为整环(integral domain)。整环是无零因子的有幺交换环，或满足乘法消去律的有幺交换环。( Z , + , ⋅ ) (\Z,+,\cdot) ( Z , + , ⋅ )
如果一个环中的所有非0元素都有乘法逆元，就称它为除环(division ring)。可见，除环中的所有非零 元素不仅关于加法构成了群，关于乘法也构成了乘法群。进一步，如果这个乘法群是阿贝尔群，就称它为域(field)。域是满足乘法交换律的除环。有理数、实数、复数关于加法和乘法都构成域，分别称为有理数域、实数域、复数域。域满足加法交换律、加法结合律、乘法交换律、乘法结合律、分配律，是一个拥有非常完好性质的代数结构。
Rmk. 除环无零因子：如果a b = 0 ab=0 ab = 0 b = a − 1 ⋅ 0 = 0 b=a^{-1}\cdot 0=0 b = a − 1 ⋅ 0 = 0 a = 0 ⋅ b − 1 = 0 a=0\cdot b^{-1}=0 a = 0 ⋅ b − 1 = 0
Rmk. 域是除环，因此无零因子，同时满足交换律，因此是整环。然而整环不一定是域，比如( Z , + , ⋅ ) (\Z,+,\cdot) ( Z , + , ⋅ )
有限整环一定是域。由于整环已经满足交换律和有单位元，只需证明有限整环中所有非零元素都有乘法逆元。为此，对于环R R R ∀ a ∈ R ∖ { 0 } \forall a\in R\setminus\{ 0\} ∀ a ∈ R ∖ { 0 } R → R R\to R R → R f ( r ) = a r f(r)=ar f ( r ) = a r f ( r ) f(r) f ( r ) R R R f f f f f f 0 0 0 0 0 0 R ∖ { 0 } R\setminus \{0\} R ∖ { 0 } ∀ a ∈ R ∖ { 0 } \forall a\in R\setminus \{0\} ∀ a ∈ R ∖ { 0 } r ≠ 0 r\neq 0 r = 0 f ( r ) = a r = 1 f(r)=ar=1 f ( r ) = a r = 1
子环
和群类似，如果环R R R S S S S S S R R R R R R S S S S ⊆ R ′ S\subseteq R' S ⊆ R ′ R ′ R' R ′ S S S ⟨ S ⟩ \lang S\rang ⟨ S ⟩ S S S R ′ R' R ′ S S S S S S
要判定子环，一般先用判定子群的方法验证是否是子群，再考虑乘法是否满足封闭性即可（结合律和分配律是继承的）。例如，考虑整数环Z \Z Z Z \Z Z d Z d\Z d Z d Z d\Z d Z Z \Z Z d Z d\Z d Z { a + b i ∣ a , b ∈ Z } \{a+bi\mid a,b\in \Z\} { a + bi ∣ a , b ∈ Z } i = − 1 i=\sqrt{-1} i = − 1 \C \C \C Z [ i ] \Z[i] Z [ i ] { a + b θ ∣ a , b ∈ Z } , θ = − d , d ∈ N + \{a+b\theta\mid a,b\in \Z\},\theta=\sqrt{-d},d\in \N^+ { a + b θ ∣ a , b ∈ Z } , θ = − d , d ∈ N + Z [ θ ] \Z[\theta] Z [ θ ]
环同态
我们已经多次看到，一个代数结构的“结构”完全体现在它定义的运算上。当两个代数结构之间存在一个保运算的映射时，我们就说它们有相同的结构。当元素个数相同时，称之为“同构”；当元素个数不同时，称之为“同态”。前者是以一一对应的方式保持结构，后者是按照一个确定的颗粒度以多对一的方式保持结构。因此我们这样定义环同态(Ring Homomorphism)：称f f f R R R S S S ∀ a , b ∈ R \forall a,b\in R ∀ a , b ∈ R f ( a + b ) = f ( a ) + f ( b ) f(a+b)=f(a)+f(b) f ( a + b ) = f ( a ) + f ( b ) f ( a b ) = f ( a ) f ( b ) f(ab)=f(a)f(b) f ( ab ) = f ( a ) f ( b ) f ( 1 R ) = 1 S f(1_R)=1_S f ( 1 R ) = 1 S f f f f f f R → S R\to S R → S
定义环同态f : R → S f:R\to S f : R → S 0 0 0 ker ( f ) = { r ∈ R ∣ f ( r ) = 0 } \ker (f)=\{r\in R\mid f(r)=0\} ker ( f ) = { r ∈ R ∣ f ( r ) = 0 } f f f ker ( f ) = { 0 } \ker(f)=\{0\} ker ( f ) = { 0 } ( ker ( f ) , + ) (\ker(f),+) ( ker ( f ) , + ) ( R , + ) (R,+) ( R , + ) ker ( f ) \ker(f) ker ( f ) R R R ker ( f ) \ker(f) ker ( f ) ∀ a , b ∈ ker ( f ) \forall a,b\in \ker(f) ∀ a , b ∈ ker ( f ) f ( a b ) = f ( a ) f ( b ) = 0 ⋅ 0 = 0 f(ab)=f(a)f(b)=0\cdot 0=0 f ( ab ) = f ( a ) f ( b ) = 0 ⋅ 0 = 0 a b ∈ ker ( f ) ab\in \ker(f) ab ∈ ker ( f )
理想(Ideals)
我们发现ker ( f ) \ker(f) ker ( f ) ∀ r ∈ R \forall r\in R ∀ r ∈ R r ker ( f ) ⊆ ker ( f ) r\ker(f)\subseteq \ker (f) r ker ( f ) ⊆ ker ( f ) ∀ a ∈ ker ( f ) \forall a\in \ker(f) ∀ a ∈ ker ( f ) f ( r a ) = f ( r ) f ( a ) = f ( r ) ⋅ 0 = 0 f(ra)=f(r)f(a)=f(r)\cdot 0 = 0 f ( r a ) = f ( r ) f ( a ) = f ( r ) ⋅ 0 = 0 r a ∈ ker ( f ) ra\in \ker(f) r a ∈ ker ( f ) ∀ r ∈ R , ker ( f ) r ⊆ ker ( f ) \forall r\in R,\ker(f)r\subseteq \ker(f) ∀ r ∈ R , ker ( f ) r ⊆ ker ( f ) R R R I I I I I I I I I R R R ∀ r ∈ R , r I ⊆ I \forall r\in R,rI\subseteq I ∀ r ∈ R , r I ⊆ I ∀ r ∈ R , I r ⊆ I \forall r\in R,Ir\subseteq I ∀ r ∈ R , I r ⊆ I ∀ i 1 , i 2 ∈ I , i 1 ⋅ i 2 ∈ i 1 I ⊆ I \forall i_1,i_2\in I,i_1\cdot i_2\in i_1 I\subseteq I ∀ i 1 , i 2 ∈ I , i 1 ⋅ i 2 ∈ i 1 I ⊆ I ker ( f ) \ker (f) ker ( f ) R R R
设I I I R R R I I I R R R R / I = { r + I ∣ r ∈ R } R/I=\{r+I\mid r\in R\} R / I = { r + I ∣ r ∈ R } r + I r+I r + I ( r + I ) + ( s + I ) = ( r + s ) + I (r+I)+(s+I)=(r+s)+I ( r + I ) + ( s + I ) = ( r + s ) + I ( r + I ) ( s + I ) = r s + I (r+I)(s+I)=rs+I ( r + I ) ( s + I ) = r s + I R / I × R / I → R / I R/I\times R/I\to R/I R / I × R / I → R / I ∀ r ′ ∈ r + I , s ′ ∈ s + I \forall r'\in r+I,s'\in s+I ∀ r ′ ∈ r + I , s ′ ∈ s + I r s + I = r ′ s ′ + I rs+I=r's'+I r s + I = r ′ s ′ + I r − r ′ ∈ I , s − s ′ ∈ I r-r'\in I,s-s'\in I r − r ′ ∈ I , s − s ′ ∈ I i 1 = r − r ′ , i 2 = s − s ′ i_1=r-r',i_2=s-s' i 1 = r − r ′ , i 2 = s − s ′ r s − r ′ s ′ = ( i 1 + r ′ ) ( i 2 + s ′ ) − r ′ s ′ = i 1 i 2 + i 1 s ′ + r ′ i 2 rs-r's'=(i_1+r')(i_2+s')-r's'=i_1i_2+i_1s'+r'i_2 r s − r ′ s ′ = ( i 1 + r ′ ) ( i 2 + s ′ ) − r ′ s ′ = i 1 i 2 + i 1 s ′ + r ′ i 2 i 1 i 2 ∈ I , i 2 s ′ ∈ I s ′ ⊆ I , r ′ i 2 ∈ r ′ I ⊆ I i_1i_2\in I,i_2s'\in Is'\subseteq I,r'i_2\in r'I\subseteq I i 1 i 2 ∈ I , i 2 s ′ ∈ I s ′ ⊆ I , r ′ i 2 ∈ r ′ I ⊆ I r s − r ′ s ′ ∈ I rs-r's'\in I r s − r ′ s ′ ∈ I r s + I = r ′ s ′ + I rs+I=r's'+I r s + I = r ′ s ′ + I R R R
注意，商环中的乘法并不是集合的乘法关系。考虑( Z , + , ⋅ ) (\Z,+,\cdot) ( Z , + , ⋅ ) 2 Z 2\Z 2 Z ( 0 + 2 Z ) ( 0 + 2 Z ) = ( 2 Z ) ( 2 Z ) = 4 Z (0+2\Z)(0+2\Z)=(2\Z)(2\Z)=4\Z ( 0 + 2 Z ) ( 0 + 2 Z ) = ( 2 Z ) ( 2 Z ) = 4 Z ( 0 + 2 Z ) ( 0 + 2 Z ) = 0 + 2 Z = 2 Z (0+2\Z)(0+2\Z)=0+2\Z=2\Z ( 0 + 2 Z ) ( 0 + 2 Z ) = 0 + 2 Z = 2 Z
{ 0 } \{0\} { 0 } R R R R R R R R R { 0 } ⊊ I \{0\}\subsetneq I { 0 } ⊊ I a ∈ I a\in I a ∈ I a − 1 a^{-1} a − 1 a − 1 I ⊆ I a^{-1}I\subseteq I a − 1 I ⊆ I a − 1 a = 1 ∈ I a^{-1}a=1\in I a − 1 a = 1 ∈ I ∀ r ∈ R , r ⋅ 1 = r ∈ r I ⊆ I \forall r\in R,r\cdot 1=r\in rI\subseteq I ∀ r ∈ R , r ⋅ 1 = r ∈ r I ⊆ I R ⊆ I R\subseteq I R ⊆ I I = R I=R I = R R R R { 0 } \{0\} { 0 } 1 R 1_R 1 R 0 0 0 { 0 } \{0\} { 0 }
对于环R R R X X X ( X ) (X) ( X ) X X X X X X R R R ( X ) = { ∑ x ∈ X r 1 x r 2 ∣ r 1 , r 2 ∈ R } (X)=\{\sum\limits_{x\in X}r_1 xr_2\mid r_1,r_2\in R\} ( X ) = { x ∈ X ∑ r 1 x r 2 ∣ r 1 , r 2 ∈ R } ( X ) (X) ( X ) ( X ) (X) ( X ) ( X ) = { ∑ x ∈ X r x ∣ r ∈ R } (X)=\{\sum\limits_{x\in X}r x\mid r \in R\} ( X ) = { x ∈ X ∑ r x ∣ r ∈ R } a a a ( a ) (a) ( a ) ( a ) = { r a ∣ r ∈ R } = R a = a R (a)=\{ra\mid r\in R\}=Ra=aR ( a ) = { r a ∣ r ∈ R } = R a = a R
两个理想的加法定义为I + J : = { x + y ∣ x ∈ I , y ∈ J } I+J:=\{x+y\mid x\in I,y\in J\} I + J := { x + y ∣ x ∈ I , y ∈ J } ∀ r ∈ R \forall r\in R ∀ r ∈ R r ( I + J ) = { r x + r y ∣ x ∈ I , y ∈ J } ⊆ r I + r J ⊆ I + J r(I+J)=\{rx+ry\mid x\in I,y\in J\}\subseteq rI+rJ\subseteq I+J r ( I + J ) = { r x + r y ∣ x ∈ I , y ∈ J } ⊆ r I + r J ⊆ I + J ∀ r ∈ R , ( I + J ) r ⊆ I + J \forall r\in R,(I+J)r\subseteq I+J ∀ r ∈ R , ( I + J ) r ⊆ I + J ∀ r ∈ R \forall r\in R ∀ r ∈ R r ( I ∩ J ) = { r x ∣ x ∈ I ∧ x ∈ J } ⊆ { r x ∣ x ∈ I } = r I ⊆ I r(I\cap J)=\{rx\mid x\in I\land x\in J\}\subseteq \{rx\mid x\in I\}=rI\subseteq I r ( I ∩ J ) = { r x ∣ x ∈ I ∧ x ∈ J } ⊆ { r x ∣ x ∈ I } = r I ⊆ I r ( I ∩ J ) ⊆ J r(I\cap J)\subseteq J r ( I ∩ J ) ⊆ J r ( I ∩ J ) ⊆ I ∩ J r(I\cap J)\subseteq I\cap J r ( I ∩ J ) ⊆ I ∩ J
环的同构定理
对于任意的理想I I I π ( r ) = r + I \pi(r)=r+I π ( r ) = r + I π ( r s ) = r s + I = ( r + I ) ( s + I ) = π ( r ) π ( s ) \pi(rs)=rs+I=(r+I)(s+I)=\pi(r)\pi(s) π ( r s ) = r s + I = ( r + I ) ( s + I ) = π ( r ) π ( s ) π ( 1 R ) = 1 R + I = 1 R / I \pi(1_R)=1_R+I=1_{R/I} π ( 1 R ) = 1 R + I = 1 R / I
在讨论群同构中，分解定理告诉我们任何群同态都可以分解为自然同态与另一个群同态的复合：对于群同态f : R → S f:R\to S f : R → S I ⊆ ker ( f ) I\subseteq \ker(f) I ⊆ ker ( f ) R / I R/I R / I S S S f ˉ ( a + I ) = f ( a ) \bar f(a+I)=f(a) f ˉ ( a + I ) = f ( a ) f = f ˉ ∘ π f=\bar f\circ \pi f = f ˉ ∘ π I I I ( R , + ) (R,+) ( R , + ) f f f π \pi π f ˉ \bar f f ˉ f ˉ \bar f f ˉ f ˉ ( ( a + I ) ( b + I ) ) = f ˉ ( a b + I ) = f ( a b ) \bar f((a+I)(b+I))=\bar f(ab+I)=f(ab) f ˉ (( a + I ) ( b + I )) = f ˉ ( ab + I ) = f ( ab ) = f ( a ) f ( b ) = f ˉ ( a + I ) f ˉ ( b + I ) =f(a)f(b)=\bar f(a+I)\bar f(b+I) = f ( a ) f ( b ) = f ˉ ( a + I ) f ˉ ( b + I ) f ˉ ( 1 R + I ) = f ( 1 R ) = 1 S \bar f(1_R+I)=f(1_R)=1_S f ˉ ( 1 R + I ) = f ( 1 R ) = 1 S
接下来只需要再次运用讨论群同态时的论证就能得到，当ker ( f ) = I \ker(f)=I ker ( f ) = I f f f R / ker ( f ) ≅ f ( R ) R/\ker (f)\cong f(R) R / ker ( f ) ≅ f ( R ) π \pi π f ˉ \bar f f ˉ π \pi π S S S ( S + I ) / I ≅ S / ( S ∩ I ) (S+I)/I\cong S/(S\cap I) ( S + I ) / I ≅ S / ( S ∩ I ) I , J I,J I , J R R R I ⊆ J I\subseteq J I ⊆ J R / J ≅ ( R / I ) / ( J / I ) R/J\cong (R/I)/(J/I) R / J ≅ ( R / I ) / ( J / I )
同样，我们需要补充论证环同态下“左子环右子环”的系列性质。由于环同态一定意味着群同态，而群同态下左子群右子群的性质已经成立，我们只需验证乘法封闭性即可（结合律始终成立）。设f : R → R ′ f:R\to R' f : R → R ′ S S S R R R ⟹ f ( S ) \implies f(S) ⟹ f ( S ) R ′ R' R ′ ∀ s 1 , s 2 ∈ S \forall s_1,s_2\in S ∀ s 1 , s 2 ∈ S f ( s 1 ) f ( s 2 ) = f ( s 1 s 2 ) ∈ f ( S ) f(s_1)f(s_2)=f(s_1s_2)\in f(S) f ( s 1 ) f ( s 2 ) = f ( s 1 s 2 ) ∈ f ( S ) f ( R ) f(R) f ( R ) R ′ R' R ′ S ′ S' S ′ R ′ R' R ′ ⟹ f − 1 ( S ′ ) \implies f^{-1}(S') ⟹ f − 1 ( S ′ ) R R R ∀ s 1 , s 2 ∈ f − 1 ( S ′ ) \forall s_1,s_2\in f^{-1}(S') ∀ s 1 , s 2 ∈ f − 1 ( S ′ ) f ( s 1 ) , f ( s 2 ) ∈ S ′ f(s_1),f(s_2)\in S' f ( s 1 ) , f ( s 2 ) ∈ S ′ f ( s 1 ) f ( s 2 ) = f ( s 1 s 2 ) ∈ S ′ f(s_1)f(s_2)=f(s_1s_2)\in S' f ( s 1 ) f ( s 2 ) = f ( s 1 s 2 ) ∈ S ′ s 1 s 2 ∈ f − 1 ( S ′ ) s_1s_2\in f^{-1}(S') s 1 s 2 ∈ f − 1 ( S ′ ) I I I R R R ⟹ f ( I ) \implies f(I) ⟹ f ( I ) f ( R ) f(R) f ( R ) ∀ f ( r ) ∈ f ( R ) \forall f(r)\in f(R) ∀ f ( r ) ∈ f ( R ) f ( r ) f ( I ) = f ( r I ) ⊆ f ( I ) f(r)f(I)=f(rI)\subseteq f(I) f ( r ) f ( I ) = f ( r I ) ⊆ f ( I ) I ′ I' I ′ R ′ R' R ′ ⟹ f − 1 ( I ′ ) \implies f^{-1}(I') ⟹ f − 1 ( I ′ ) R R R ∀ r ∈ R \forall r\in R ∀ r ∈ R ∀ i ∈ f − 1 ( I ′ ) \forall i\in f^{-1}(I') ∀ i ∈ f − 1 ( I ′ ) f ( i ) ∈ I ′ f(i)\in I' f ( i ) ∈ I ′ f ( r ) f ( i ) = f ( r i ) ∈ I ′ f(r)f(i)=f(ri)\in I' f ( r ) f ( i ) = f ( r i ) ∈ I ′ r i ∈ f − 1 ( I ′ ) ri\in f^{-1}(I') r i ∈ f − 1 ( I ′ )
环与商环间的一一对应
同样，和群与商群间的一一对应相同，有环与商环间的一一对应关系：对R R R I I I S 1 = { S ∣ S is a subring of R ∧ I ⊆ S ⊆ R } S_1=\{S\mid S \text{ is a subring of }R \land I\subseteq S\subseteq R\} S 1 = { S ∣ S is a subring of R ∧ I ⊆ S ⊆ R } S 2 = { K ∣ K is a subring of R / I } S_2=\{K\mid K \text{ is a subring of }R/I\} S 2 = { K ∣ K is a subring of R / I } S 1 S_1 S 1 S 2 S_2 S 2 π ( S ) = S / I \pi (S)=S/I π ( S ) = S / I
环的外直积
同样，可以定义环的外直积：R 1 × ⋯ × R n = { ( a 1 , ⋯ , a n ) ∣ a i ∈ R i } R_1\times \cdots \times R_n=\{(a_1,\cdots,a_n)\mid a_i\in R_i\} R 1 × ⋯ × R n = {( a 1 , ⋯ , a n ) ∣ a i ∈ R i }
中国剩余定理(Chinese Remainder Theorem, CRT)
数论中的中国剩余定理是求解同余方程组的算法。设n 1 , ⋯ , n m n_1,\cdots,n_m n 1 , ⋯ , n m x ≡ a 1 ( m o d n 1 ) , ⋯ , x ≡ a m ( m o d n m ) x\equiv a_1\pmod {n_1},\cdots,x\equiv a_m\pmod {n_m} x ≡ a 1 ( mod n 1 ) , ⋯ , x ≡ a m ( mod n m ) u 1 ≡ 1 ( m o d n 1 ) , u 1 ≡ 0 ( m o d n i , i ≠ 1 ) u_1\equiv 1\pmod{n_1},u_1\equiv 0\pmod{n_i,i\neq 1} u 1 ≡ 1 ( mod n 1 ) , u 1 ≡ 0 ( mod n i , i = 1 ) u 1 = k n 2 n 3 ⋯ n m , u 1 − 1 = s n 1 u_1=kn_2n_3\cdots n_m,u_1-1=sn_1 u 1 = k n 2 n 3 ⋯ n m , u 1 − 1 = s n 1 k ( n 2 ⋯ n m ) − s n 1 = 1 k(n_2\cdots n_m)-sn_1=1 k ( n 2 ⋯ n m ) − s n 1 = 1 n i n_i n i gcd ( n 2 ⋯ n m , n 1 ) = 1 \gcd(n_2\cdots n_m,n_1)=1 g cd( n 2 ⋯ n m , n 1 ) = 1 k , s k,s k , s u 1 u_1 u 1 u 2 ≡ 1 ( m o d n 2 ) , u 2 ≡ 0 ( m o d n i , i ≠ 2 ) u_2\equiv 1\pmod{n_2},u_2\equiv 0\pmod{n_i,i\neq 2} u 2 ≡ 1 ( mod n 2 ) , u 2 ≡ 0 ( mod n i , i = 2 ) u 1 , ⋯ , u m u_1,\cdots,u_m u 1 , ⋯ , u m u i u_i u i x x x x ≡ a 1 u 1 + a 2 u 2 + ⋯ + a m u m ( m o d n 1 n 2 ⋯ n m ) x\equiv a_1u_1+a_2u_2+\cdots+a_mu_m\pmod{n_1n_2\cdots n_m} x ≡ a 1 u 1 + a 2 u 2 + ⋯ + a m u m ( mod n 1 n 2 ⋯ n m )
中国剩余定理本质上指出了这样一件事：我们知道显然可以构造映射x → ( x m o d n 1 , ⋯ , x m o d n m ) x\to (x\mod n_1,\cdots,x\mod n_m) x → ( x mod n 1 , ⋯ , x mod n m ) ( x m o d n 1 , ⋯ , x m o d n m ) → x (x\mod n_1,\cdots,x\mod n_m)\to x ( x mod n 1 , ⋯ , x mod n m ) → x x x x ( x m o d n 1 , ⋯ , x m o d n m ) (x\mod n_1,\cdots,x\mod n_m) ( x mod n 1 , ⋯ , x mod n m ) Z n ≅ Z n 1 × ⋯ × Z n m \Z_{n}\cong \Z_{n_1}\times \cdots \times \Z_{n_m} Z n ≅ Z n 1 × ⋯ × Z n m
既然数论的中国剩余定理是基于整数环( Z , + , ⋅ ) (\Z,+,\cdot) ( Z , + , ⋅ ) R R R n i n_i n i ( n i ) = n i Z (n_i)=n_i\Z ( n i ) = n i Z n i n_i n i n i n_i n i a i a_i a i a i + ( n i ) a_i+(n_i) a i + ( n i ) x ≡ a i ( m o d n i ) ⟺ x ∈ a i + ( n i ) x\equiv a_i \pmod{n_i}\iff x\in a_i+(n_i) x ≡ a i ( mod n i ) ⟺ x ∈ a i + ( n i ) ⟺ x − a i ∈ ( n i ) \iff x-a_i\in (n_i) ⟺ x − a i ∈ ( n i ) I I I a − b ∈ I a-b\in I a − b ∈ I a ≡ b ( m o d I ) a\equiv b\pmod{I} a ≡ b ( mod I ) gcd ( n i , n j ) = 1 ⟺ 1 ∈ ( n i ) + ( n j ) \gcd(n_i,n_j)=1\iff 1\in (n_i)+(n_j) g cd( n i , n j ) = 1 ⟺ 1 ∈ ( n i ) + ( n j ) ⟺ ( n i ) + ( n j ) = Z \iff (n_i)+(n_j)=\Z ⟺ ( n i ) + ( n j ) = Z I , J I,J I , J I + J = R I+J=R I + J = R Z / ⋂ i = 1 m ( n i ) ≅ Z / ( n 1 ) × ⋯ × Z / ( n m ) \Z/\bigcap\limits_{i=1}^{m}(n_i)\cong \Z/(n_1)\times \cdots \times \Z/(n_m) Z / i = 1 ⋂ m ( n i ) ≅ Z / ( n 1 ) × ⋯ × Z / ( n m ) R R R m m m I 1 , ⋯ , I m I_1,\cdots,I_m I 1 , ⋯ , I m R / ⋂ i = 1 m I i ≅ R / I 1 × ⋯ × R / I m R/\bigcap\limits_{i=1}^{m}I_i\cong R/I_1\times \cdots \times R/I_m R / i = 1 ⋂ m I i ≅ R / I 1 × ⋯ × R / I m
我们在一般形式下再次证明中国剩余定理：构造映射f : R → R / I 1 × ⋯ × R / I m f:R\to R/I_1\times \cdots \times R/I_m f : R → R / I 1 × ⋯ × R / I m ∀ r ∈ R , f ( r ) = ( r + I 1 , ⋯ , r + I m ) \forall r\in R,f(r)=(r+I_1,\cdots,r+I_m) ∀ r ∈ R , f ( r ) = ( r + I 1 , ⋯ , r + I m ) f f f R / ker ( f ) ≅ f ( R ) R/\ker(f)\cong f(R) R / ker ( f ) ≅ f ( R ) ∀ x ∈ ker ( f ) \forall x\in \ker(f) ∀ x ∈ ker ( f ) x + I i = I i , ∀ i ∈ [ m ] x+I_i=I_i,\forall i\in [m] x + I i = I i , ∀ i ∈ [ m ] ∀ i ∈ [ m ] , x ∈ I i \forall i\in [m],x\in I_i ∀ i ∈ [ m ] , x ∈ I i ker ( f ) ⊆ ⋂ i = 1 m I i \ker(f)\subseteq \bigcap\limits_{i=1}^{m}I_i ker ( f ) ⊆ i = 1 ⋂ m I i ⋂ i = 1 m I i ⊆ ker ( f ) \bigcap\limits_{i=1}^{m}I_i\subseteq \ker(f) i = 1 ⋂ m I i ⊆ ker ( f ) ker ( f ) = ⋂ i = 1 m I i \ker(f)=\bigcap\limits_{i=1}^{m}I_i ker ( f ) = i = 1 ⋂ m I i f f f ∀ r 1 + I 1 ∈ R / I 1 , ⋯ , r m + I m ∈ R / I m \forall r_1+I_1\in R/I_1,\cdots, r_m+I_m\in R/I_m ∀ r 1 + I 1 ∈ R / I 1 , ⋯ , r m + I m ∈ R / I m r ∈ R r\in R r ∈ R r + I 1 = r 1 + I 1 , ⋯ , r + I m = r m + I m r+I_1=r_1+I_1,\cdots,r+I_m=r_m+I_m r + I 1 = r 1 + I 1 , ⋯ , r + I m = r m + I m ∀ i ∈ [ m ] , r − r i ∈ I i \forall i\in [m],r-r_i\in I_i ∀ i ∈ [ m ] , r − r i ∈ I i r ≡ r i ( m o d I i ) r\equiv r_i\pmod{I_i} r ≡ r i ( mod I i ) m m m u j ≡ 1 ( m o d I j ) , u j ≡ 0 ( m o d I i , i ≠ j ) u_j\equiv 1\pmod{I_j},u_j\equiv 0\pmod{I_i,i\neq j} u j ≡ 1 ( mod I j ) , u j ≡ 0 ( mod I i , i = j ) r = r 1 u 1 + r 2 u 2 + ⋯ + r m u m r=r_1u_1+r_2u_2+\cdots+r_mu_m r = r 1 u 1 + r 2 u 2 + ⋯ + r m u m r − r i = r 1 u 1 + ⋯ + r i ( u i − 1 ) + ⋯ + r m u m r-r_i=r_1u_1+\cdots+r_i(u_i-1)+\cdots+r_mu_m r − r i = r 1 u 1 + ⋯ + r i ( u i − 1 ) + ⋯ + r m u m ∀ j ≠ i , u j ∈ I i \forall j\neq i,u_j\in I_i ∀ j = i , u j ∈ I i r j u j ∈ I i r_ju_j\in I_i r j u j ∈ I i u i − 1 ∈ I i u_i-1\in I_i u i − 1 ∈ I i r i ( u i − 1 ) ∈ I i r_i(u_i-1)\in I_i r i ( u i − 1 ) ∈ I i r − r i ∈ I i r-r_i\in I_i r − r i ∈ I i u j u_j u j u 1 u_1 u 1 ∀ i > 1 , 1 ∈ I 1 + I i \forall i>1,1\in I_1+I_i ∀ i > 1 , 1 ∈ I 1 + I i b i ∈ I 1 , c i ∈ I i b_i\in I_1,c_i\in I_i b i ∈ I 1 , c i ∈ I i b i + c i = 1 b_i+c_i=1 b i + c i = 1 ∏ i = 2 m ( b i + c i ) = 1 \prod\limits_{i=2}^{m}(b_i+c_i)=1 i = 2 ∏ m ( b i + c i ) = 1 ( ∑ ⋯ ) + ∏ i = 2 m c i = 1 (\sum \cdots)+\prod\limits_{i=2}^{m}c_i=1 ( ∑ ⋯ ) + i = 2 ∏ m c i = 1 ( ∑ ⋯ ) (\sum\cdots) ( ∑ ⋯ ) b j ∈ I 1 b_j\in I_1 b j ∈ I 1 ∈ I 1 \in I_1 ∈ I 1 C = ∏ i = 2 m c i C=\prod\limits_{i=2}^{m}c_i C = i = 2 ∏ m c i C ∈ I 2 ∧ C ∈ I 3 ∧ ⋯ ∧ C ∈ I m C\in I_2\land C\in I_3\land \cdots\land C\in I_m C ∈ I 2 ∧ C ∈ I 3 ∧ ⋯ ∧ C ∈ I m C ≡ 0 ( m o d I i ) , ∀ i > 1 C\equiv 0\pmod{I_i},\forall i>1 C ≡ 0 ( mod I i ) , ∀ i > 1 1 − C ∈ I 1 1-C\in I_1 1 − C ∈ I 1 C ≡ 1 ( m o d I 1 ) C\equiv 1\pmod{I_1} C ≡ 1 ( mod I 1 ) C C C u 1 u_1 u 1 u j u_j u j f ( R ) = R / I 1 × ⋯ × R / I m f(R)= R/I_1\times \cdots \times R/I_m f ( R ) = R / I 1 × ⋯ × R / I m R / ⋂ i = 1 m I i ≅ R / I 1 × ⋯ × R / I m R/\bigcap\limits_{i=1}^{m}I_i\cong R/I_1\times \cdots \times R/I_m R / i = 1 ⋂ m I i ≅ R / I 1 × ⋯ × R / I m
极大理想与素理想
极大理想
如果环R R R I I I R R R I I I I I I R R R R R R R R R R R R R R R S = { I ∣ I is a proper ideal of R } S=\{I\mid I\text{ is a proper ideal of }R\} S = { I ∣ I is a proper ideal of R } S S S S S S S S S I 1 ⊆ I 2 ⊆ ⋯ ⊆ ⋯ I_1\subseteq I_2\subseteq \cdots \subseteq \cdots I 1 ⊆ I 2 ⊆ ⋯ ⊆ ⋯ ⋃ i = 1 ∞ I i \bigcup\limits_{i=1}^{\infty}I_i i = 1 ⋃ ∞ I i ∀ i , I i ⊊ R \forall i,I_i\subsetneq R ∀ i , I i ⊊ R I i I_i I i 1 1 1 1 ∉ ⋃ i = 1 ∞ I i 1\not\in \bigcup\limits_{i=1}^{\infty}I_i 1 ∈ i = 1 ⋃ ∞ I i ⋃ i = 1 ∞ I i \bigcup\limits_{i=1}^{\infty}I_i i = 1 ⋃ ∞ I i ∀ a , b ∈ ⋃ i = 1 ∞ I i \forall a,b\in \bigcup\limits_{i=1}^{\infty}I_i ∀ a , b ∈ i = 1 ⋃ ∞ I i I k ⊆ I m I_k\subseteq I_m I k ⊆ I m a ∈ I k , b ∈ I m , a\in I_k,b\in I_m, a ∈ I k , b ∈ I m , a ∈ I m a\in I_m a ∈ I m a − b ∈ I m , a b ∈ I m a-b\in I_m,ab\in I_m a − b ∈ I m , ab ∈ I m ⋃ i = 1 ∞ I i \bigcup\limits_{i=1}^{\infty}I_i i = 1 ⋃ ∞ I i ∀ r ∈ R , i ∈ ⋃ i = 1 ∞ I i \forall r\in R,i\in \bigcup\limits_{i=1}^{\infty}I_i ∀ r ∈ R , i ∈ i = 1 ⋃ ∞ I i i ∈ I m i\in I_m i ∈ I m r i ∈ I m ∈ ⋃ i = 1 ∞ I i ri\in I_m\in \bigcup\limits_{i=1}^{\infty}I_i r i ∈ I m ∈ i = 1 ⋃ ∞ I i ⋃ i = 1 ∞ I i \bigcup\limits_{i=1}^{\infty}I_i i = 1 ⋃ ∞ I i R R R
交换 环R R R M M M R / M R/M R / M R R R R / M R/M R / M R R R R / M R/M R / M R / M R/M R / M R / M R/M R / M M M M ⟺ \iff ⟺ ( ∀ a ∈ R , a + M ≠ M ⟹ ∃ b ∈ R s.t. ( a + M ) ( b + M ) = 1 + M ) (\forall a\in R,a+M\neq M\implies\exists b\in R\text{ s.t. }(a+M)(b+M)=1+M) ( ∀ a ∈ R , a + M = M ⟹ ∃ b ∈ R s.t. ( a + M ) ( b + M ) = 1 + M ) a ∉ M a\not\in M a ∈ M M ⊊ ( a ) + M M\subsetneq (a)+M M ⊊ ( a ) + M ( a ) + M (a)+M ( a ) + M M M M ( a ) + M = R (a)+M=R ( a ) + M = R ( a ) = a R (a)=aR ( a ) = a R a R + M = R aR+M=R a R + M = R ∀ r ′ ∈ R , ∃ r ∈ R , m ∈ M \forall r'\in R,\exists r\in R,m\in M ∀ r ′ ∈ R , ∃ r ∈ R , m ∈ M a r + m = r ′ ar+m=r' a r + m = r ′ r ′ = 1 r'=1 r ′ = 1 a r + m = 1 ar+m=1 a r + m = 1 a r − 1 ∈ M ar-1\in M a r − 1 ∈ M a r + M = 1 + M ar+M=1+M a r + M = 1 + M ( a + M ) ( r + M ) = 1 + M (a+M)(r+M)=1+M ( a + M ) ( r + M ) = 1 + M b = r b=r b = r I I I M ⊊ I M\subsetneq I M ⊊ I I = R I=R I = R ∀ a ∈ I \forall a\in I ∀ a ∈ I a ∉ M a\not\in M a ∈ M ∃ b ∈ R \exists b\in R ∃ b ∈ R ( a + M ) ( b + M ) = 1 + M (a+M)(b+M)=1+M ( a + M ) ( b + M ) = 1 + M ∃ m ∈ M \exists m\in M ∃ m ∈ M a b − 1 = m ab-1=m ab − 1 = m a b − m = 1 ab-m=1 ab − m = 1 a ∈ I a\in I a ∈ I a b ∈ I ab\in I ab ∈ I m ∈ M ⊊ I m\in M\subsetneq I m ∈ M ⊊ I m ∈ I m\in I m ∈ I a b − m = 1 ∈ I ab-m=1\in I ab − m = 1 ∈ I I = R I=R I = R
域是具有非常良好性质的代数结构。以上定理表明，在交换环中只要能找到极大理想，我们就能用这个极大理想的商环构造一个域。
素理想
整数（环）具有这样的性质：设n = a b n=ab n = ab p p p p ∣ n p\mid n p ∣ n p ∣ a p\mid a p ∣ a p ∣ b p\mid b p ∣ b p p p a b ∈ ( p ) ab\in (p) ab ∈ ( p ) a ∈ ( p ) a\in (p) a ∈ ( p ) b ∈ ( p ) b\in (p) b ∈ ( p ) p p p 50 = 2 × 25 50=2\times 25 50 = 2 × 25 10 ∣ 50 10\mid 50 10 ∣ 50 10 ∣ 2 10\mid 2 10 ∣ 2 10 ∣ 25 10\mid 25 10 ∣ 25
我们把以上整数环中的素数性质抽象出来，定义理想的“素数”性质：称交换 环R R R P P P ∀ a , b ∈ R , a b ∈ P ⟹ \forall a,b\in R,ab\in P\implies ∀ a , b ∈ R , ab ∈ P ⟹ a ∈ P \or b ∈ P a\in P\or b\in P a ∈ P \or b ∈ P
交换 环R R R P P P R / P R/P R / P R R R R / P R/P R / P R R R R / P R/P R / P R / P R/P R / P R / P R/P R / P P P P ⟺ \iff ⟺ ( ∀ a , b ∈ R ， ( a + P ) ( b + P ) = P ⟹ a + P = P \or b + P = P ) (\forall a,b\in R， (a+P)(b+P)= P\implies a+P= P\or b+P= P) ( ∀ a , b ∈ R ， ( a + P ) ( b + P ) = P ⟹ a + P = P \or b + P = P ) ∀ a , b ∈ R , a b ∈ P ⟹ a ∈ P \or b ∈ P \forall a,b\in R,ab\in P\implies a\in P\or b\in P ∀ a , b ∈ R , ab ∈ P ⟹ a ∈ P \or b ∈ P
极大理想与素理想的关系
我们看到，极大理想与素理想作为理想的特殊性会在其商环上表现出来，这种关系是当且仅当的，我们可以通过理想的商环是否是域或整环反推出理想是否是极大理想或素理想。环同态下的同构定理就给出了kernel作为极大理想和素理想的例子：设f f f R R R R / ker ( f ) ≅ f ( R ) R/\ker(f)\cong f(R) R / ker ( f ) ≅ f ( R ) f ( R ) f(R) f ( R ) R / ker ( f ) R/\ker(f) R / ker ( f ) ker ( f ) \ker(f) ker ( f ) R R R f ( R ) f(R) f ( R ) R / ker ( f ) R/\ker(f) R / ker ( f ) ker ( f ) \ker(f) ker ( f ) R R R
我们发现，在交换环中 极大理想一定是素理想！Pf：若M M M R / M R/M R / M R / M R/M R / M M M M Z [ x ] \Z[x] Z [ x ] f : Z [ x ] → Z f:\Z[x]\to \Z f : Z [ x ] → Z f ( a 0 + a 1 x 1 + ⋯ + a n x n + ⋯ ) = a 0 f(a_0+a_1x_1+\cdots+a_nx^n+\cdots)=a_0 f ( a 0 + a 1 x 1 + ⋯ + a n x n + ⋯ ) = a 0 f f f Z [ x ] / ker ( f ) ≅ Z \Z[x]/\ker(f)\cong \Z Z [ x ] / ker ( f ) ≅ Z Z \Z Z ker ( f ) \ker(f) ker ( f ) ker ( f ) \ker(f) ker ( f ) x Z [ x ] x\Z[x] x Z [ x ] x x x ( x ) (x) ( x ) ( x ) ⊊ ( 2 , x ) ⊊ Z [ x ] (x)\subsetneq (2,x)\subsetneq \Z[x] ( x ) ⊊ ( 2 , x ) ⊊ Z [ x ] ( x ) (x) ( x )