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P. 211

Abstract Algebra




                    Notes          Proof: Since p p , ...p, = q q  ... ,.q , p  | q q . ... q ,.
                                              1  2     1 2   m  1   1 2   m
                                   Thus, p  | q for some j = 1. .... ..,m. By changing the order of the q , if necessary, we can assume
                                            j
                                                                                       i
                                         1
                                   that j = 1, i.e., p | q. Let q  = p u . Since q  is irreducible, u  must be a unit in R. So p  and q  are
                                               1
                                                           l
                                                            l
                                                                   1
                                                                                 1
                                                                                                           1
                                                       1
                                                                                                      1
                                   associates. Now we have
                                   p p  = P  (p u )q ....q .
                                    1
                                     2
                                         n
                                             1
                                                2
                                                   m
                                            1
                                   Cancelling p  from both sides, we get
                                             1
                                   p p ...p  = u q ...,q .
                                     3
                                            1 2
                                        n
                                                 m
                                    2
                                   Now, if m > n, we can apply the same process to p , p , and so on.
                                                                           2
                                                                              3
                                   Then we will get
                                   1 = u u  .... u  q  .... q .
                                        2
                                       1
                                             n
                                                    m
                                               n+1
                                   This shows that q  is a unit. But this contradicts the fact that q  is irreducible.
                                                                                     n+1
                                                 n+1
                                   Thus, m  n.
                                   Interchanging the roles of the ps and qs and by using a similar argument, we get n  m.
                                   Thus, n = m.
                                   During the proof we have also shown that each pi is an associate of some q, and vice versa.
                                                                                               j
                                   What Theorem 12 says is that any two prime factorisations of an element in a PID are identical,
                                   apart from the order in which the factors appear and apart from replacement of the factors by
                                   their associates.
                                   Thus, Theorems 11 and 12 say that every non-zero element in a PID R, which is not a unit, can be
                                   expressed uniquely (up to associates) as a product of a finite number of prime elements.
                                   For example, x  – 1  R[x] can be written as (x-1)(x+1) or (x-1) (x-1) or [2(x-tl)] [2(x-1)] in R[x].
                                              2
                                   The property that we have shown for a PID in Theorems 11 and 12 is true for several other
                                   domains also. Let us discuss such rings now.
                                   Self Assessment
                                   1.  An integral domains R a .................. if every ideal in R is a principle ideal.
                                       (a)  principle ideal domain   (b)  unique ideal domain
                                       (c)  special ideal domain     (d)  range ideal domain
                                   2.  g.c.d. represent ..................
                                       (a)  greatest common  divisor   (b)    greatest common dividend
                                       (c)  greatest common  domain    (d)    greatest commutated domain
                                   3.  Let R be .................. and a, b  R, if a g.c.d. of a and b exists, then it is unique up to unit.
                                       (a)  domain and range         (b)  integral  domain
                                       (c)  UID                      (d)  SID
                                   4.  PID stands for ..................

                                       (a)  principal integral  domain  (b)  pair ideal domain
                                       (c)  principle ideal domain   (d)  principle ideal divisor





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