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【商环定义】（最低要求）
设 R ≠ { 0 } R \neq \left\{ 0 \right\} R={0}为交换幺环，设子集 S ⊆ R S \subseteq R S⊆R满足乘法运算封闭且含单位元 1 1 1。在 R × S R \times S R×S上定义如下的等价关系 ∼ \sim ∼：
( ∀ ⟨ r 1 , s 1 ⟩ , ⟨ r 2 , s 2 ⟩ ∈ R × S ) [ ⟨ r 1 , s 1 ⟩ ∼ ⟨ r 2 , s 2 ⟩ ⇔ ( ∃ s 3 ∈ S ) [ s 3 ( s 2 r 1 − r 2 s 1 ) = 0 ] ] \left( \forall\left\langle r_{1},s_{1} \right\rangle,\left\langle r_{2},s_{2} \right\rangle \in R \times S\ \right)\left\lbrack \left\langle r_{1},s_{1} \right\rangle\sim\left\langle r_{2},s_{2} \right\rangle \Leftrightarrow \left( \exists s_{3} \in S \right)\left\lbrack s_{3}\left( s_{2}r_{1} - r_{2}s_{1} \right) = 0 \right\rbrack \right\rbrack (∀⟨r1,s1⟩,⟨r2,s2⟩∈R×S )[⟨r1,s1⟩∼⟨r2,s2⟩⇔(∃s3∈S)[s3(s2r1−r2s1)=0]]
设 R S = R × S / ∼ R_{S} = R \times S/\sim RS=R×S/∼为集合 R × S R \times S R×S上的等价类。同时定义等价类 R S R_{S} RS上的加法和乘法为：
r 1 s 1 + r 2 s 2 = s 2 r 1 + r 2 s 1 s 1 s 2 r 1 s 1 × r 2 s 2 = r 1 r 2 s 1 s 2 \frac{r_{1}}{s_{1}} + \frac{r_{2}}{s_{2}} = \frac{s_{2}r_{1} + r_{2}s_{1}}{s_{1}s_{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{r_{1}}{s_{1}} \times \frac{r_{2}}{s_{2}} = \frac{r_{1}r_{2}}{s_{1}s_{2}} s1r1+s2r2=s1s2s2r1+r2s1 s1r1×s2r2=s1s2r1r2
可以看出元素 0 1 \frac{0}{1} 10为 R S R_{S} RS的零元，元素 1 1 \frac{1}{1} 11为 R S R_{S} RS的单位元，并且 R S R_{S} RS为交换环。环 R S R_{S} RS被称为 R R R的以 S S S为分母的商环或者分数环。
【有效性验证】
1. 等价关系 ∼ \sim ∼具有自反性、传递性、对称性
自反性和对称性比较容易验证，下面证明传递性：
设 ⟨ r 1 , s 1 ⟩ ∼ ⟨ r 2 , s 2 ⟩ 、 ⟨ r 2 , s 2 ⟩ ∼ ⟨ r 3 , s 3 ⟩ \left\langle r_{1},s_{1} \right\rangle\sim\left\langle r_{2},s_{2} \right\rangle 、\left\langle r_{2},s_{2} \right\rangle\sim\left\langle r_{3},s_{3} \right\rangle ⟨r1,s1⟩∼⟨r2,s2⟩、⟨r2,s2⟩∼⟨r3,s3⟩，根据定义有
s 12 ( s 2 r 1 − r 2 s 1 ) = 0 s_{12}\left( s_{2}r_{1} - r_{2}s_{1} \right) = 0 s12(s2r1−r2s1)=0
s 23 ( s 3 r 2 − r 3 s 2 ) = 0 s_{23}\left( s_{3}r_{2} - r_{3}s_{2} \right) = 0 s23(s3r2−r3s2)=0
那么
s 12 s 2 r 1 = s 12 r 2 s 1 s_{12}s_{2}r_{1} = s_{12}r_{2}s_{1} s12s2r1=s12r2s1
s 23 s 3 r 2 = s 23 r 3 s 2 s_{23}s_{3}r_{2} = s_{23}r_{3}s_{2} s23s3r2=s23r3s2
消去 r 2 r_{2} r2可得 s 12 s 2 s 23 s 3 r 1 = s 12 s 1 s 23 s 2 r 3 s_{12}s_{2}s_{23}s_{3}r_{1} = s_{12}s_{1}s_{23}s_{2}r_{3} s12s2s23s3r1=s12s1s23s2r3，即 s 12 s 23 s 2 ( s 3 r 1 − s 1 r 3 ) = 0 s_{12}s_{23}s_{2}\left( s_{3}r_{1} - s_{1}r_{3} \right) = 0 s12s23s2(s3r1−s1r3)=0，那么 ⟨ r 1 , s 1 ⟩ ∼ ⟨ r 3 , s 3 ⟩ \left\langle r_{1},s_{1} \right\rangle\sim\left\langle r_{3},s_{3} \right\rangle ⟨r1,s1⟩∼⟨r3,s3⟩，也就是满足传递性。
2. 加法是有效的
设 ⟨ r 1 , s 1 ⟩ ∼ ⟨ r 3 , s 3 ⟩ 、 ⟨ r 2 , s 2 ⟩ ∼ ⟨ r 4 , s 4 ⟩ \left\langle r_{1},s_{1} \right\rangle\sim\left\langle r_{3},s_{3} \right\rangle 、\left\langle r_{2},s_{2} \right\rangle\sim\left\langle r_{4},s_{4} \right\rangle ⟨r1,s1⟩∼⟨r3,s3⟩、⟨r2,s2⟩∼⟨r4,s4⟩，接下来验证：
r 1 s 1 + r 2 s 2 = r 3 s 3 + r 4 s 4 \frac{r_{1}}{s_{1}} + \frac{r_{2}}{s_{2}} = \frac{r_{3}}{s_{3}} + \frac{r_{4}}{s_{4}} s1r1+s2r2=s3r3+s4r4
根据定义，需要验证
s 2 r 1 + r 2 s 1 s 1 s 2 = s 4 r 3 + r 4 s 3 s 3 s 4 \frac{s_{2}r_{1} + r_{2}s_{1}}{s_{1}s_{2}} = \frac{s_{4}r_{3} + r_{4}s_{3}}{s_{3}s_{4}} s1s2s2r1+r2s1=s3s4s4r3+r4s3
因为 ⟨ r 1 , s 1 ⟩ ∼ ⟨ r 3 , s 3 ⟩ 、 ⟨ r 2 , s 2 ⟩ ∼ ⟨ r 4 , s 4 ⟩ \left\langle r_{1},s_{1} \right\rangle\sim\left\langle r_{3},s_{3} \right\rangle 、\left\langle r_{2},s_{2} \right\rangle\sim\left\langle r_{4},s_{4} \right\rangle ⟨r1,s1⟩∼⟨r3,s3⟩、⟨r2,s2⟩∼⟨r4,s4⟩，所以
s 13 ( s 3 r 1 − r 3 s 1 ) = 0 s_{13}\left( s_{3}r_{1} - r_{3}s_{1} \right) = 0 s13(s3r1−r3s1)=0
s 24 ( s 4 r 2 − r 4 s 2 ) = 0 s_{24}\left( s_{4}r_{2} - r_{4}s_{2} \right) = 0 s24(s4r2−r4s2)=0
从而
s 13 s 24 ( ( s 2 r 1 + r 2 s 1 ) ( s 3 s 4 ) − ( s 4 r 3 + r 4 s 3 ) ( s 1 s 2 ) ) s_{13}s_{24}\left( \left( s_{2}r_{1} + r_{2}s_{1} \right)\left( s_{3}s_{4} \right) - \left( s_{4}r_{3} + r_{4}s_{3} \right)\left( s_{1}s_{2} \right) \right) s13s24((s2r1+r2s1)(s3s4)−(s4r3+r4s3)(s1s2))
= s 13 s 24 s 4 s 2 ( s 3 r 1 − r 3 s 1 ) + s 13 s 24 s 1 s 3 ( s 4 r 2 − r 4 s 2 ) = 0 = s_{13}s_{24}s_{4}s_{2}\left( s_{3}r_{1} - r_{3}s_{1} \right) + s_{13}s_{24}s_{1}s_{3}\left( s_{4}r_{2} - r_{4}s_{2} \right) = 0 =s13s24s4s2(s3r1−r3s1)+s13s24s1s3(s4r2−r4s2)=0
即
s 2 r 1 + r 2 s 1 s 1 s 2 = s 4 r 3 + r 4 s 3 s 3 s 4 \frac{s_{2}r_{1} + r_{2}s_{1}}{s_{1}s_{2}} = \frac{s_{4}r_{3} + r_{4}s_{3}}{s_{3}s_{4}} s1s2s2r1+r2s1=s3s4s4r3+r4s3
r 1 s 1 + r 2 s 2 = r 3 s 3 + r 4 s 4 \frac{r_{1}}{s_{1}} + \frac{r_{2}}{s_{2}} = \frac{r_{3}}{s_{3}} + \frac{r_{4}}{s_{4}} s1r1+s2r2=s3r3+s4r4
3. 乘法是有效的
设 ⟨ r 1 , s 1 ⟩ ∼ ⟨ r 3 , s 3 ⟩ 、 ⟨ r 2 , s 2 ⟩ ∼ ⟨ r 4 , s 4 ⟩ \left\langle r_{1},s_{1} \right\rangle\sim\left\langle r_{3},s_{3} \right\rangle 、\left\langle r_{2},s_{2} \right\rangle\sim\left\langle r_{4},s_{4} \right\rangle ⟨r1,s1⟩∼⟨r3,s3⟩、⟨r2,s2⟩∼⟨r4,s4⟩，接下来验证：
r 1 s 1 × r 2 s 2 = r 3 s 3 × r 4 s 4 \frac{r_{1}}{s_{1}} \times \frac{r_{2}}{s_{2}} = \frac{r_{3}}{s_{3}} \times \frac{r_{4}}{s_{4}} s1r1×s2r2=s3r3×s4r4
根据定义，需要验证
r 1 r 2 s 1 s 2 = r 3 r 4 s 3 s 4 \frac{r_{1}r_{2}}{s_{1}s_{2}} = \frac{r_{3}r_{4}}{s_{3}s_{4}} s1s2r1r2=s3s4r3r4
因为 ⟨ r 1 , s 1 ⟩ ∼ ⟨ r 3 , s 3 ⟩ 、 ⟨ r 2 , s 2 ⟩ ∼ ⟨ r 4 , s 4 ⟩ \left\langle r_{1},s_{1} \right\rangle\sim\left\langle r_{3},s_{3} \right\rangle 、\left\langle r_{2},s_{2} \right\rangle\sim\left\langle r_{4},s_{4} \right\rangle ⟨r1,s1⟩∼⟨r3,s3⟩、⟨r2,s2⟩∼⟨r4,s4⟩，所以
s 13 ( s 3 r 1 − r 3 s 1 ) = 0 s_{13}\left( s_{3}r_{1} - r_{3}s_{1} \right) = 0 s13(s3r1−r3s1)=0
s 24 ( s 4 r 2 − r 4 s 2 ) = 0 s_{24}\left( s_{4}r_{2} - r_{4}s_{2} \right) = 0 s24(s4r2−r4s2)=0
s 13 s 3 r 1 = s 13 r 3 s 1 s_{13}s_{3}r_{1} = s_{13}r_{3}s_{1} s13s3r1=s13r3s1
s 24 s 4 r 2 = s 24 r 4 s 2 s_{24}s_{4}r_{2} = s_{24}r_{4}s_{2} s24s4r2=s24r4s2
从而
s 13 s 24 ( s 3 s 4 r 1 r 2 − s 1 s 2 r 3 r 4 ) = s 24 s 13 r 3 s 1 s 4 r 2 − s 13 s 24 r 3 s 1 s 4 r 2 = 0 s_{13}s_{24}\left( s_{3}s_{4}r_{1}r_{2} - s_{1}s_{2}r_{3}r_{4} \right) = s_{24}s_{13}r_{3}s_{1}s_{4}r_{2} - s_{13}s_{24}r_{3}s_{1}s_{4}r_{2} = 0 s13s24(s3s4r1r2−s1s2r3r4)=s24s13r3s1s4r2−s13s24r3s1s4r2=0
即
r 1 r 2 s 1 s 2 = r 3 r 4 s 3 s 4 \frac{r_{1}r_{2}}{s_{1}s_{2}} = \frac{r_{3}r_{4}}{s_{3}s_{4}} s1s2r1r2=s3s4r3r4
r 1 s 1 × r 2 s 2 = r 3 s 3 × r 4 s 4 \frac{r_{1}}{s_{1}} \times \frac{r_{2}}{s_{2}} = \frac{r_{3}}{s_{3}} \times \frac{r_{4}}{s_{4}} s1r1×s2r2=s3r3×s4r4
4. 加法乘法都满足交换律结合律，还满足分配律
通过字母运算容易验证，不再赘述。
【备注】
当 S S S选择元素时排除了所有 R R R的乘法零因子后，等价关系可以化简为
( ∀ ⟨ r 1 , s 1 ⟩ , ⟨ r 2 , s 2 ⟩ ∈ R × S ) [ ⟨ r 1 , s 1 ⟩ ∼ ⟨ r 2 , s 2 ⟩ ⇔ s 2 r 1 − r 2 s 1 = 0 ] \left( \forall\left\langle r_{1},s_{1} \right\rangle,\left\langle r_{2},s_{2} \right\rangle \in R \times S\ \right)\left\lbrack \left\langle r_{1},s_{1} \right\rangle\sim\left\langle r_{2},s_{2} \right\rangle \Leftrightarrow s_{2}r_{1} - r_{2}s_{1} = 0 \right\rbrack (∀⟨r1,s1⟩,⟨r2,s2⟩∈R×S )[⟨r1,s1⟩∼⟨r2,s2⟩⇔s2r1−r2s1=0]