Page 200 - DMTH403_ABSTRACT_ALGEBRA
P. 200

Unit 19: The Field of Quotient Euclidean Domains




          (ii)  + is commutative :.For [a,b], [c,d]  F,                                        Notes
               [a,b] + [c,d] = [ad+bc,bd] = [cb+da,db] = [c,d] + [a,b]

          (iii)  [0,1] is the additive identity for F : For [a,b]  F,
               [0,1] + [a,b] = [0.b+l.a, l.b] = [a,b]
          (iv)  The additive inverse of [a,b]  F is [–a,b] :
               [a,b] + [–a,b] = [ab-ab,b ] = [0,b ] = [0,1], since 0.1 = 0.b .
                                 2
                                                          2
                                        2
               We would like you to prove the rest of the requirements for F to be a field.
          So we have put our heads together and proved that F is a field.
          Now, let us define f : R  F : f(.a) = [a,1]. We want to show that f is a monomorphism.
          Firstly, for a, b  R,
          f(a+b) = [a+b,1] = [a,]] + [b,l].
          = f(a) + f(b), and

          Thus, f is a ring homomorphism.
          Next, let a,b  R such that f(a) = f(b). Then [a,1] = [b,l], i.e., a = b. Therefore, f is 1–1.
          Thus, f is a monomorphism.
          So, Im f = (R) is a subring of F which is isomorphic to R.
          As you know, isomorphic structures are algebraically identical.
          So, we can identify R with f(R), and think of R as a subring of F. Now, any element of F is of the
          form [a,b] = [a,1] [l,b] = [a,l] [b,l]  = f(a) f(b) , where b  0. Thus, identifying x  R with f(x)  f(R),
                                            -1
                                   -1
          we can say that any element of F is of the form ab , where a,b  R, b  0.
                                                  -1
          All that we have discussed adds up to the proof of the following theorem.
          Theorem 4: Let R be an integral domain. Then R can be embedded in a field F such that every
          element of F has the form ab  for a, b  R, b  0.
                                 -1
          The field F whose existence we have just proved is called the field of quotients (or the field of
          fraction) of R.
          Thus, Q is the field of quotient of Z. What is the field of quotients of R? The following theorem
          answers this question.
          Theorem 5: Iff : R  K is a monomorphism of an integral domain R into a field K, then there
          exists a monomorphism g : F  K : g([a,1]) = f(a), where F is the field of quotients of R.

          It says that the-field of quotients of an integral domain is the smallest field containing it. Thus,
          the field of quotients of any field is the field itself.
          So, the field of quotients of R is R and of Z  is Z , where p is a prime number.
                                            p
                                                p
          Self Assessment

          1.   An ideal P of a ring R is called a/an .................. ideal of R. If whenever ab  P for a, b  R,
               then either a  P or b  P.
               (a)  prime ideal             (b)  odd ideal
               (c)  even ideal              (d)  integer  ideal






                                           LOVELY PROFESSIONAL UNIVERSITY                                  193
   195   196   197   198   199   200   201   202   203   204   205