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Abstract Algebra




                    Notes          Theorem 3: Let S  and S  be subrings of the rings R  and R , respectively. Then S  × S  is a subring
                                                                                                 1
                                                1
                                                                          1
                                                                                2
                                                     2
                                                                                                     2
                                   of R × R .
                                      1   2
                                   Proof: Since S  and S  are subrings of R  and R , S    and S   . S  × S   .
                                                   2
                                              1
                                                                       2
                                                                                           2
                                                                                       1
                                                                          1
                                                                                  2
                                                                  1
                                   Now, let (a, b) and (a’, b’)  S  × S . Then a, a’ E S  and b, b’ E S . As S  and S  are subrings, a – a’,
                                                                                    2
                                                                                               2
                                                                         1
                                                                                         1
                                                          1
                                                             2
                                   a. a’  S  and b – b’, b b’  S . 2
                                         1
                                   (We are using + and . for both R  and R  here, for convenience.) Hence,
                                                            1
                                                                  2
                                   (a, b) – (a’, b’) = (a – a’, b – b’)  S  × Sz, and
                                                             1
                                   (a, b) . (a’, b’) = (aa’, bb’)  S  × S .
                                                         1
                                                            2
                                   Thus, by Theorem 1, S  × S  is a subring of R  × R .
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                                                        2
                                                                      1
                                                                          2
                                   Self Assessment
                                   1.  Let S be a subset of a ring (R, +,.). S is called ................ of R of O  S
                                       (a)  ring                     (b)  subring
                                       (c)  polynomial  ring         (d)  ideals
                                   2.  S be a ................ subset of (R, +,.). Then S is a subring of R. If and only if x – y  SY X, y  S,
                                       (a)  empty                    (b)  non-empty
                                       (c)  null set                 (d)  real set
                                   3.  Let R be a ring and a  R then the set aR = (ax | x  R) is a ................ of R.
                                       (a)  subring                  (b)  ring
                                       (c)  ideal                    (d)  polynomial
                                   15.2 Summary
                                       Let (R, +, .) be a ring and S be a subset of R. We say that S is a subring of R, if (S, +, .) is itself
                                   
                                       a ring, i.e., S is a ring with respect to the operations on R.
                                       For example, we can say that 2Z, the set of even integers, is a subring of Z.
                                       Before giving more examples, let us analyse  the definition of a subring. The definition
                                       says that  a subring of a ring R is a ring with respect to the operations on R. Now, the
                                       distributive, commutative and associative laws hold good in R. Therefore, they hold good
                                       in any subset of R also. So, to prove that a subset S of R is a ring we don’t need to check all
                                       the 6 axioms R1-R6 for S. It is enough to check that
                                       (i)  S is closed under both + and . ,
                                       (ii)  0  S, and

                                       (iii)  for each a  S, – a  S.
                                       If S satisfies these three conditions, then S is a subring of R. So we have an alternative
                                       definition for a subring.

                                       Let S be a subset of a ring (R, +, .). S is called a subring of R if
                                   
                                       (i)  S is closed under + and . , i.e., a + b, a. b E S whenever a, b E S,







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