Page 161 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Definition: 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.
Definition: 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  S whenever a, b  S,
(ii) 0  S, and

(iii) for each a  S, - a  S.
Even this definition can be improved upon. For this, recall from Unit 3 that (S, +)  (R, +) if a – b
 S whenever a, b  S. This observation allows us to give a set of conditions for a subset to be a
subring, which are easy to verify.
Theorem 1: Let S be a non-empty subset of (R, +, .). Then S is a subring of R if and only if

(a) x – y  S  x, y  S; and
(b) xy  S  x, y  S.
Proof: We need to show that S is a subring of R according to our definition iff S satisfies (a) and
(b). Now, S is a subring of R iff (S, f )  (R, f ) and S is closed under multiplication, i.e., iff (a)
and (b) hold.

So, we have proved the theorem.
This theorem allows us a neat way of showing that a subset is a subring.
Let us look at some examples.
We have already noted that Z is a subring of Q. In fact, you can use Theorem 1 to check that Z is
subring of R, C and Z + iZ too. You can also verify that Q is a subring of R, C and

Q  2Q  {a  2 | ,  Q}.

Example: Consider Z , the ring of integers modulo 6. Show, that 3Z = (3.0, 3.1, ....., 3.5) is
6
6
a subring of Z .
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Solution: Firstly, do you agree that 3Z = (0, 3)? Remember that 6 = 0, 9 = 3, and so on. Also,
6
0 – 5 = – 3 = 5. Thus, x - y  3Z  x, y  3Z . You can also verify that xy  3Z  x, y  3Z . Thus,
6
by Theorem 1, 3Z is a subring of Z . 6 6 6 6
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