Page 162 - DMTH403_ABSTRACT_ALGEBRA

Unit 15: Subrings

Notes
Example: Consider the ring  (X). Show that S = { , X ) is a subring of  (X).
Solution: Note that A A A =  A   (X).  A = – A in  (X).

Now, to apply Theorem 1 we first note that S is non-empty.
Next,    =   S, X  X =   S,
  X = X  S,    =  S, X  X = X  S,   X =   S.
Thus, by Theorem 1, S is a subring of  (X).
For each proper subset of X we get a subring of (X). Thus, a ring can have, several subrings. Let
us consider two subrings of the ring Z .
2

Example: Show that S = ( (n, 0) } | n  Z } is a subring of Z × Z. Also show that
D = ((n,n) | n  Z } is a subring of Z × Z.
Solution: You can recall the ring structure of Z . Both S and D are non-empty. Both of them
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satisfy (a) and (b) of Theorem 1. Thus, S and D are both subrings of Z .
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We would like to make a remark here which is based on the examples of subrings that you have
seen so far.
Remark: (i) If R is a ring with identity, a subring of R may or may not be with identity. For
example, the ring Z has identity 1, but its subring nZ (n  2) is without identity.

(ii) The identity of a subring, if it exists, may not coincide with the identity of the ring. For
example, the identity of the ring Z × Z is (1, 1). But the identity of its subring Z × {0} is (1, 0).

Example: Let R be a ring and a  R. Show that the set aR = ( ax | x  R } is a subring of R.
Solution: Since R  , aR  . Now, for any two elements ax and ay of aR,
ax – ay = a(x – y)  aR and (ax) (ay) = a(xay)  aR.

Thus, by Theorem 1, aR is a subring of R.
Using Example we can immediately say that mZ is a subring of Z  m E Z. This also shows
n
n
us a fact that we have already seen : nZ is a subring of Z  n  Z.
Now let us look at some properties of subrings. From Unit 3 you know that the intersection of
two or more subgroups is a subgroup. The following result says that the same is true for
subrings.
Theorem 2: Let S and S be subrings of a ring R. Then S  S is also a subring of R.
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1
2
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Proof: Since 0 E S and 0 E S , 0 E S S . 2  S S  .
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1
2
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Now, let x, y  S S . Then x, y E S and x, y  S . Thus, by Theorem 1, x – y and xy are in S as
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1
2
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well as in S , i.e,, they lie in S S .
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2
1
Thus, S S is a subring of R.
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On the same lines as the proof above we can prove that the intersection of any family of subrings
of a ring R is a subring of R.
Now let us look at the Cartesian product of subrings.
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