分式环
在本节中 R R R a a a a b = a c {ab} = {ac} ab = a c R R R b = c b = c b = c R R R a a a R R R a a a a b = a c {ab} = {ac} ab = a c a a a b = c b = c b = c c = 0 c = 0 c = 0 R R R Q Q Q Q Q Q R R R Q Q Q Q Q Q R R R Q Q Q Z \mathbb{Z} Z
为了看到从整环 Z \mathbb{Z} Z Q \mathbb{Q} Q 1 2 = 2 4 = 3 6 = … \frac{1}{2} = \frac{2}{4} = \frac{3}{6} = \ldots 2 1 = 4 2 = 6 3 = …
a b = c d if and only if a d = b c . \frac{a}{b} = \frac{c}{d}\;\text{ if and only if }\;{ad} = {bc}. b a = d c if and only if a d = b c . 更精确地说，分数 a b \frac{a}{b} b a b ≠ 0 b \neq 0 b = 0 ( a , b ) ∼ ( c , d ) \left( {a,b}\right) \sim \left( {c,d}\right) ( a , b ) ∼ ( c , d ) a d = b c {ad} = {bc} a d = b c
a b + c d = a d + b c b d and a b × c d = a c b d . \frac{a}{b} + \frac{c}{d} = \frac{{ad} + {bc}}{bd}\;\text{ and }\;\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}. b a + d c = b d a d + b c and b a × d c = b d a c . 这些定义良好（独立于等价类代表的选择）并使得分数集合成为一个交换环（实际上是一个域），Q。整数 Z \mathbb{Z} Z Q \mathbb{Q} Q { a 1 ∣ a ∈ Z } \left\{ {\frac{a}{1} \mid a \in \mathbb{Z}}\right\} { 1 a ∣ a ∈ Z } a a a Q \mathbb{Q} Q 1 a \frac{1}{a} a 1
对于任意的交换环 R R R b b b R R R b d = 0 {bd} = 0 b d = 0 b b b
d = d 1 = b d b = 0 b = 0 d = \frac{d}{1} = \frac{bd}{b} = \frac{0}{b} = 0 d = 1 d = b b d = b 0 = 0 在“分式环”中（为了方便，我们假设 R R R R R R b b b d d d b d {bd} b d R R R R R R Z \mathbb{Z} Z Q \mathbb{Q} Q
定理15
设 R R R D D D R R R a , b ∈ D a,b \in D a , b ∈ D a b ∈ D {ab} \in D ab ∈ D Q Q Q Q Q Q R R R D D D Q Q Q Q Q Q
（1）Q Q Q r d − 1 r{d}^{-1} r d − 1 r ∈ R r \in R r ∈ R d ∈ D d \in D d ∈ D D = R − { 0 } D = R - \{ 0\} D = R − { 0 } Q Q Q
（2）（Q Q Q Q Q Q R R R D D D S S S φ : R → S \varphi : R \rightarrow S φ : R → S d ∈ D d \in D d ∈ D φ ( d ) \varphi \left( d\right) φ ( d ) S S S Φ : Q → S \Phi : Q \rightarrow S Φ : Q → S Φ ∣ R = φ {\left. \Phi \right| }_{R} = \varphi Φ ∣ R = φ R R R D D D Q Q Q
注：在第15.4节中给出了一个更一般的构造。一般结果的证明更为技术性，但依赖于与定理15证明相同的基本原理和步骤。希望获得更大一般性的读者可以一起阅读下面的证明和第15.4节的开头部分。
证明：设 F = { ( r , d ) ∣ r ∈ R , d ∈ D } \mathcal{F} = \{ \left( {r,d}\right) \mid r \in R,d \in D\} F = { ( r , d ) ∣ r ∈ R , d ∈ D } F \mathcal{F} F ∼ \sim ∼
( r , d ) ∼ ( s , e ) if and only if r e = s d . \left( {r,d}\right) \sim \left( {s,e}\right) \;\text{ if and only if }\;{re} = {sd}. ( r , d ) ∼ ( s , e ) if and only if re = s d . 立即可得这个关系是自反的和对称的。假设 ( r , d ) ∼ ( s , e ) \left( {r,d}\right) \sim \left( {s,e}\right) ( r , d ) ∼ ( s , e ) ( s , e ) ∼ ( t , f ) \left( {s,e}\right) \sim \left( {t,f}\right) ( s , e ) ∼ ( t , f ) r e − s d = 0 {re} - {sd} = 0 re − s d = 0 s f − t e = 0 {sf} - {te} = 0 s f − t e = 0 f f f d d d ( r f − t d ) e = 0 \left( {{rf} - {td}}\right) e = 0 ( r f − t d ) e = 0 e ∈ D e \in D e ∈ D r f − t d = 0 {rf} - {td} = 0 r f − t d = 0 ( r , d ) ∼ ( t , f ) \left( {r,d}\right) \sim \left( {t,f}\right) ( r , d ) ∼ ( t , f ) ∼ \sim ∼ r d \frac{r}{d} d r
r d = { ( a , b ) ∣ a ∈ R , b ∈ D and r b = a d } . \frac{r}{d} = \{ \left( {a,b}\right) \mid a \in R,b \in D\text{ and }{rb} = {ad}\} . d r = { ( a , b ) ∣ a ∈ R , b ∈ D and r b = a d } . 设 Q Q Q ∼ \sim ∼ r d = r e d e \frac{r}{d} = \frac{re}{de} d r = d e re Q Q Q e ∈ D e \in D e ∈ D D D D
我们现在在 Q Q Q
a b + c d = a d + b c b d and a b × c d = a c b d . \frac{a}{b} + \frac{c}{d} = \frac{{ad} + {bc}}{bd}\;\text{ and }\;\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}. b a + d c = b d a d + b c and b a × d c = b d a c . 为了证明 Q Q Q
(1) 这些操作是良定义的（即不依赖于等价类代表元的选择），
(2) Q Q Q 0 d \frac{0}{d} d 0 d ∈ D d \in D d ∈ D a d \frac{a}{d} d a − a d \frac{-a}{d} d − a
(3) 乘法是结合的、分配的和交换的，以及
(4) Q Q Q d ∈ D ) \left. {d \in D}\right) d ∈ D ) ( = d d \left( { = \frac{d}{d}}\right. ( = d d
这些都是仅涉及 R R R ∼ \sim ∼ D D D
例如，为了检查加法是否定义良好，假设 a b = a ′ b ′ \frac{a}{b} = \frac{{a}^{\prime }}{{b}^{\prime }} b a = b ′ a ′ a b ′ = a ′ b a{b}^{\prime } = {a}^{\prime }b a b ′ = a ′ b c d = c ′ d ′ \frac{c}{d} = \frac{{c}^{\prime }}{{d}^{\prime }} d c = d ′ c ′ c d ′ = c ′ d c{d}^{\prime } = {c}^{\prime }d c d ′ = c ′ d a d + b c b d = a ′ d ′ + b ′ c ′ b ′ d ′ \frac{{ad} + {bc}}{bd} = \frac{{a}^{\prime }{d}^{\prime } + {b}^{\prime }{c}^{\prime }}{{b}^{\prime }{d}^{\prime }} b d a d + b c = b ′ d ′ a ′ d ′ + b ′ c ′
( a d + b c ) ( b ′ d ′ ) = ( a ′ d ′ + b ′ c ′ ) ( b d ) . \left( {{ad} + {bc}}\right) \left( {{b}^{\prime }{d}^{\prime }}\right) = \left( {{a}^{\prime }{d}^{\prime } + {b}^{\prime }{c}^{\prime }}\right) \left( {bd}\right) . ( a d + b c ) ( b ′ d ′ ) = ( a ′ d ′ + b ′ c ′ ) ( b d ) . 这个方程的左边是 a b ′ d d ′ + c d ′ b b ′ a{b}^{\prime }d{d}^{\prime } + c{d}^{\prime }b{b}^{\prime } a b ′ d d ′ + c d ′ b b ′ a ′ b {a}^{\prime }b a ′ b a b ′ a{b}^{\prime } a b ′ c ′ d {c}^{\prime }d c ′ d c d ′ c{d}^{\prime } c d ′ a ′ b d d ′ + c ′ d b b ′ {a}^{\prime }{bd}{d}^{\prime } + {c}^{\prime }{db}{b}^{\prime } a ′ b d d ′ + c ′ d b b ′
接下来，我们通过定义将 R R R Q Q Q
ι : R → Q by ι : r ↦ r d d where d is any element of D . \iota : R \rightarrow Q\;\text{ by }\;\iota : r \mapsto \frac{rd}{d}\text{ where }d\text{ is any element of }D\text{. } ι : R → Q by ι : r ↦ d r d where d is any element of D . 由于 r d d = r e e \frac{rd}{d} = \frac{re}{e} d r d = e re d , e ∈ D , ι ( r ) d,e \in D,\iota \left( r\right) d , e ∈ D , ι ( r ) d ∈ D d \in D d ∈ D D D D ι \iota ι
此外，ι \iota ι
ι ( r ) = 0 ⇔ r d d = 0 d ⇔ r d 2 = 0 ⇔ r = 0 \iota \left( r\right) = 0 \Leftrightarrow \frac{rd}{d} = \frac{0}{d} \Leftrightarrow r{d}^{2} = 0 \Leftrightarrow r = 0 ι ( r ) = 0 ⇔ d r d = d 0 ⇔ r d 2 = 0 ⇔ r = 0 因为 d d d d 2 {d}^{2} d 2 Q Q Q ι ( R ) \iota \left( R\right) ι ( R ) R R R r ∈ R r \in R r ∈ R ι ( r ) \iota \left( r\right) ι ( r ) R R R Q Q Q
接下来注意，每个 d ∈ D d \in D d ∈ D Q Q Q d d d d e e \frac{de}{e} e d e e d e \frac{e}{de} d e e Q Q Q r ⋅ d − 1 r \cdot {d}^{-1} r ⋅ d − 1 r ∈ R r \in R r ∈ R d ∈ D d \in D d ∈ D D = R − { 0 } , \;D = R - \{ 0\} , D = R − { 0 } , Q \;Q\; Q Q Q Q
仍需确定 Q Q Q φ : R → S \varphi : R \rightarrow S φ : R → S d ∈ D d \in D d ∈ D φ ( d ) \varphi \left( d\right) φ ( d ) S S S φ \varphi φ Φ : Q → S \Phi : Q \rightarrow S Φ : Q → S Φ ( r d − 1 ) = φ ( r ) φ ( d ) − 1 \Phi \left( {r{d}^{-1}}\right) = \varphi \left( r\right) \varphi {\left( d\right) }^{-1} Φ ( r d − 1 ) = φ ( r ) φ ( d ) − 1 r ∈ R , d ∈ D r \in R,d \in D r ∈ R , d ∈ D r d − 1 = s e − 1 r{d}^{-1} = s{e}^{-1} r d − 1 = s e − 1 r e = s d {re} = {sd} re = s d φ ( r ) φ ( e ) = φ ( s ) φ ( d ) \varphi \left( r\right) \varphi \left( e\right) = \varphi \left( s\right) \varphi \left( d\right) φ ( r ) φ ( e ) = φ ( s ) φ ( d )
Φ ( r d − 1 ) = φ ( r ) φ ( d ) − 1 = φ ( s ) φ ( e ) − 1 = Φ ( s e − 1 ) . \Phi \left( {r{d}^{-1}}\right) = \varphi \left( r\right) \varphi {\left( d\right) }^{-1} = \varphi \left( s\right) \varphi {\left( e\right) }^{-1} = \Phi \left( {s{e}^{-1}}\right) . Φ ( r d − 1 ) = φ ( r ) φ ( d ) − 1 = φ ( s ) φ ( e ) − 1 = Φ ( s e − 1 ) . 很容易验证 Φ \Phi Φ Φ \Phi Φ r d − 1 ∈ ker Φ r{d}^{-1} \in \ker \Phi r d − 1 ∈ ker Φ r ∈ ker Φ ∩ R = ker φ r \in \ker \Phi \cap R = \ker \varphi r ∈ ker Φ ∩ R = ker φ φ \varphi φ r r r r d − 1 r{d}^{-1} r d − 1
定义
令 R , D R,D R , D Q Q Q
（1）环 Q Q Q R R R D D D D − 1 R {D}^{-1}R D − 1 R
（2）如果 R R R D = R − { 0 } , Q D = R - \{ 0\} ,Q D = R − { 0 } , Q R R R
如果 A A A F F F A A A F F F A A A F F F F F F A A A A A A F F F A A A F F F A A A
下一个推论表明，包含一个整环 R R R
推论 16
设 R R R Q Q Q R R R F F F R R R R ′ {R}^{\prime } R ′ F F F R ′ {R}^{\prime } R ′ Q Q Q
证明：设 φ : R ≅ R ′ ⊆ F \varphi : R \cong {R}^{\prime } \subseteq F φ : R ≅ R ′ ⊆ F R R R R ′ {R}^{\prime } R ′ φ : R → F \varphi : R \rightarrow F φ : R → F R R R F F F Φ : Q → F \Phi : Q \rightarrow F Φ : Q → F φ \varphi φ Q Q Q Φ \Phi Φ Φ ( Q ) \Phi \left( Q\right) Φ ( Q ) φ ( R ) = R ′ \varphi \left( R\right) = {R}^{\prime } φ ( R ) = R ′ F F F Q Q Q R ′ = φ ( R ) {R}^{\prime } = \varphi \left( R\right) R ′ = φ ( R ) F F F r 1 , r 2 ∈ R {r}_{1},{r}_{2} \in R r 1 , r 2 ∈ R φ ( r 1 ) φ ( r 2 ) − 1 = φ ( r 1 r 2 − 1 ) \varphi \left( {r}_{1}\right) \varphi {\left( {r}_{2}\right) }^{-1} = \varphi \left( {{r}_{1}{r}_{2}^{-1}}\right) φ ( r 1 ) φ ( r 2 ) − 1 = φ ( r 1 r 2 − 1 ) Q Q Q r 1 , r 2 ∈ R {r}_{1},{r}_{2} \in R r 1 , r 2 ∈ R r 1 r 2 − 1 {r}_{1}{r}_{2}^{-1} r 1 r 2 − 1 R ′ {R}^{\prime } R ′ F F F Φ ( Q ) \Phi \left( Q\right) Φ ( Q ) Φ ( Q ) \Phi \left( Q\right) Φ ( Q ) R ′ {R}^{\prime } R ′ F F F
示例
(1)如果 R R R R R R
(2)整数Z \mathbb{Z} Z Q \mathbb{Q} Q O \mathcal{O} O Q ( D ) \mathbb{Q}\left(\sqrt{D}\right) Q ( D )
(3)Z \mathbb{Z} Z 2 Z 2\mathbb{Z} 2 Z Q \mathbb{Q} Q
(4)如果R R R R [ x ] R[x] R [ x ] x x x R R R p ( x ) q ( x ) \frac{p(x)}{q(x)} q ( x ) p ( x ) p ( x ) p(x) p ( x ) q ( x ) q(x) q ( x ) R R R q ( x ) q(x) q ( x ) p ( x ) p(x) p ( x ) q ( x ) q(x) q ( x ) R R R a b \frac{a}{b} b a a , b ∈ R a,b\in R a , b ∈ R b ≠ 0 b\neq0 b = 0 F F F F ( x ) F(x) F ( x ) F F F R R R R R R F F F F ( x ) F(x) F ( x )
例如，假设 R = Z R = \mathbb{Z} R = Z F = Q F = \mathbb{Q} F = Q p ( x ) , q ( x ) p\left( x\right) ,q\left( x\right) p ( x ) , q ( x ) Q [ x ] \mathbb{Q}\left\lbrack x\right\rbrack Q [ x ] N , N p ( x ) , N q ( x ) N,{Np}\left( x\right) ,{Nq}\left( x\right) N , Np ( x ) , Nq ( x ) N N N p ( x ) p\left( x\right) p ( x ) q ( x ) q\left( x\right) q ( x ) p ( x ) q ( x ) = N p ( x ) N q ( x ) \frac{p\left( x\right) }{q\left( x\right) } = \frac{{Np}\left( x\right) }{{Nq}\left( x\right) } q ( x ) p ( x ) = Nq ( x ) Np ( x ) Q [ x ] \mathbb{Q}\left\lbrack x\right\rbrack Q [ x ] Z [ x ] \mathbb{Z}\left\lbrack x\right\rbrack Z [ x ]
(5) 如果 R R R d d d R R R D = { 1 , d , d 2 , d 3 , … } D = \{ 1,d,{d}^{2},{d}^{3},\ldots \} D = { 1 , d , d 2 , d 3 , … } R [ 1 / d ] R\left\lbrack {1/d}\right\rbrack R [ 1/ d ] D − 1 R {D}^{-1}R D − 1 R R [ 1 / d ] R\left\lbrack {1/d}\right\rbrack R [ 1/ d ] R R R r 1 \frac{r}{1} 1 r R R R R R R R [ 1 / d ] R\left\lbrack {1/d}\right\rbrack R [ 1/ d ] 1 / d 1/d 1/ d R R R
