Page 47 - DMTH502_LINEAR_ALGEBRA

Unit 1: Vector Space over Fields

Also S is closed with respect to multiplication, Notes
a S, b S ab S.
Now to prove that the conditions are sufficient suppose S is a non-empty subset of R for which
the conditions (i) and (ii) are satisfied.
From condition (i)
a S a – a S 0 S.
Hence additive identity is in S.

Now 0 S, a S – a S
i.e., each element of S possesses additive inverse.
Let a, b S then – b S and then from condition (i)
a S, – b S a – (–b) S (a + b) S

Thus S is closed under addition. S being subset of R, associative and commutative laws hold in
S. Therefore, (S, +) is an abelian group.
From condition (ii) S is closed under multiplication.

Since S is a subset of R, the associative law for multiplication and distributive laws of
multiplication over addition hold in S. Thus S is a subring of R.

Intersection of Subrings

Theorem 13: The intersection of two subrings is a subring.
Proof: Let S and S be two subrings of ring R.
1 2
Since 0 S and 0 S at least 0 S S . Therefore S S is non-empty.
1 2 1 2 1 2
Let a, b S S , then
1 2
a S S a S and a S
1 2 1 2
and b S S b S and b S .
1 2 1 2
But S and S are subrings of R, therefore
1 2
a, b S a – b S and a b S .
1 1 1
and a, b S a – b S and a b S .
2 2 2
Consequently, a, b S S a – b S S and a b S S .
1 2 1 2 1 2
Hence S S is a subring of R.
1 2
Illustrative Examples

Example 17: If R is a ring with additive identity 0, then for all a, b R, prove that

a (b – c) = ab – ac
and (b – c) a = ba – ca.
Solution: We have, a (b – c) = a [b + (–c)]
= ab + a (–c) [left distributive law]

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