Theorem Related To Rings

Theorem Related To Rings : 

   Theorem 1 : 

                  If  R is a commutative ring with unit element and M is an ideal of R , then M is maximal ideal of R iff R / M is a field .

Proof : 

        Since R is a commutative ring with unity , therefore R / M is also a commutative ring with unity . The zero element of the ring R / M is M and the unit element of the coset M +1 where 1 is the unit element of R .

             Let the ideal M be maximal . Then to prove that R / M is a field .

       Let M + b be any non zero element of    R  / M  . Then M + b ≠ M i.e b doesn't belongs to M . To prove that M + b is inversible .

         If (b) is the principal ideal of R generated by b, then M(b) is also an ideal of R . Since b doesn't belong to M , therefore the ideal M is properly contained M +(b) . But M is a maximal ideal of R . Hence we must has M +(b) = R

      Since 1∈ R , therefore we must obtain 1 on adding an element of M to an element of (b) . Therefore ∃ an element a ∈ M and         β ∈ R such that a + βb = 1    [ Note that (b) =     { βb : β ∈R  } ]
                   ∴   1 = βb = a ∈ M

Consequently

                  M +1 = M + βb = (M + β)(M + b)

         ∴  M + β = (M + b)⁻¹

Thus M+b is conversible .
∴ R / M is a field .

Conversly :

           Let M be an ideal of R such that R / M  is a field . We shall prove that M is a maximal ideal of R .

          Let S ' be an ideal of R properly containing M i.e . M ⊂ S ' and M ≠ S' . Then M will be maximal if S' = R . The elements of R contained in M already belong to S' since  M ⊂ S' . Therefore R will be a subset of S' if every element β of R not contained in M also belongs to S' . If β ∈ R is such that β doesn't belong to M , then M + β ≠ S i.e        M + β  is a non zero element of R / M . Also   S' properly contains M . Therefore ∃ an element γ of S' not contained in M sothat   M + γ is also a non zero elements of R / M .

              Now the non zero elements of R / M form a group with respect to multiplication because R / M is a field . Therefore ∃ a non zero element M + y of R / M such that 

                (M + y)(M + γ) = M + β 

  [ We may take M + y = (M + β)(M + γ)⁻¹]

Now  (M +y)(M + γ) = M + β 

       ⇒ M + yγ = M + β 

       ⇒ yγ - β ∈ M 

       ⇒ yγ - β ∈ S'      [ ∵ M ⊂ S' ]

Now S' is an ideal . Therefore y ∈ R , γ ∈ S' 

 yγ ∈ S' . Again yγ ∈ S´ , yγ - β ∈ S´

⇒ yγ - (yγ - β ) ∈ S´ i.e β ∈ S´

           Thus R ⊂ S´ . Also S´⊂ R as S´ is an ideal of R .

   ∴  S' = R 

Hence the theorem is proved .