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Unit 14: Rings




          14.2 Elementary Properties                                                            Notes

          In this section we will prove some simple but important properties of rings which are immediate
          consequences of the definition of a ring. As we go along you must not forget that for any ring R,
          (R, +) is an abelian group. Hence, the results obtained for groups in the earlier units are applicable
          to the abelian group (R, +). In particular,
          (i)  the zero element, 0, and the additive inverse of any element is unique.

          (ii)  the cancellation law holds for addition; i.e.,    a, b, c  R , a + c = b + c  a = b.
          As we have mentioned earlier, we will write a – b for a + (–b) and ab for a. b, where a, b  R.
          So let us state some properties which follow from the axiom R6, mainly.
          Theorem 1: Let R be a ring. Then, for any a, b, c  R,

          (i)  a0 = 0 = 0a,
          (ii)  a(–b) = (–a)b = –(ab),
          (iii)  (– a) (– b) = ab,

          (iv)  a(b – c) = ab – ac, and
          Proof:
          (i)  Now, 0 + 0 = 0
                a(0 + 0) = a0
                a0 + a0 = a0, applying the distributive law.

                             = a0 + 0, since 0 is the additive identity.
                a0 = 0, by the cancellation law for (R, +).
               Using the other distributive law, we can similarly show that 0a = 0.
               Thus, a0 = 0 = 0a for all a  R.

          (ii)  From the definition of additive inverse, we know that b + (– b) = 0.
                Now, 0 = a0, from (i) above.
                      = a(b + (– b)), as 0 = b + (– b).
                      = ab + a(– b), by distributivity.

               Now, ab + [– (ab)] = 0 and ab + a(– b) = 0. But you know that the additive inverse of an
               element is unique.
               Hence, we get – (ab) = a(– b).

               In the same manner, using the fact that a + (–a) = 0, we get – (ab) = (– a)b.
               Thus, a(– b) = (– a)b = – (ab) for all a, b  R.
          (iii)  For a, b  R,
               (– a) (– b) = – (a(– b)), from (ii) above.

                      = a(– (– b)), from (ii) above.
                      = ab, since b is the additive inverse of (– b).






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