Page 163 - DMTH403_ABSTRACT_ALGEBRA
P. 163
Abstract Algebra
Notes Theorem 3: Let S and S be subrings of the rings R and R , respectively. Then S × S is a subring
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of R × R .
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Proof: Since S and S are subrings of R and R , S and S . S × S .
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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,
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a. a S and b b, b b S . 2
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(We are using + and . for both R and R here, for convenience.) Hence,
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(a, b) (a, b) = (a a, b b) S × Sz, and
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(a, b) . (a, b) = (aa, bb) S × S .
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Thus, by Theorem 1, S × S is a subring of R × R .
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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 dont 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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