Subrings, Ideals, Quotient Rings & Euclidean Rings Subrings
In analogy with the concept of a subgroup of a group, we now introduce the concept of a subring. If (R, +, •) is a ring then a non-empty subset of R with the induced operations +, * as in R can be a ring. Such a ring is called a subring of the ring R.
Definition. (Subring). Let (R,+,•) be a ring and S be a non-empty subset of R. If (S, +, •) is also a ring with respect to the two operations +, • in R then (S, +, •) is a subring of R.
The binary operations in S thus defined are the induced operations in S from R.
Definition. Let (F,+, • ) be a field and (S, +, •) be a subring ofF. If (S, +, •) is a field then we say that S is a subfield of F. If (S, +, •) is an integral domain then we say that S is a subdomain of F.
Note. 1. If (S’, +, •) is a subring of the ring (R, +, •) then (S, + ) is a subgroup of the (R, +) group. Hence zero element in R is also zero element in S.
If (S, +, •) is a subfield of the field (F. +. •) then (i) (S, + ) is a subgroup of (F, + ) group and is a subgroup of (F-{0}) group.
Examples of Subrings, Ideals, Quotient Rings, And Euclidean Rings Subrings
Example1. The set of even integers is a subring of (Z, +,”•) ring or integral domain.
Example 2. (Z, +, •), (Q, +, •) are subrings of the field of real numbers (R, +, •).
Example3. Let (Q, +,•) be the ring of rational numbers. Then S = {sr2|a Zj is a nonempty subset of Q and (S, +) is a subgroup of the group (Q, +).
But for 1 / 2 ∈S we have (1 / 2). (1 / 2). = (1 / 4) S and hence is not a binary operation in S. Thus (S, +, •) is not a subring of (Q, +,•)
Example4. Let (R, +, •) be a ring and 0 e R be the zero elements of R. Then S = {0} is a non-empty subset of R so that (S, +, •) is itself a ring. Therefore (S, +, •) is a subring of R.
{0} is called trivial subring and (R, +,•) is called improper subring of R.
Example5. For each positive integer n, the set nz = (0,± n,± 2n,± 3n,…….. } is a subring of Z. ‘
Example6. The set of Gaussian integers Z, i] = {a+b ia,b∈Z,i2 =-1} is a subring of the complex number field C.
Theorem 1. (Subring Test). Let S be a non-empty subset of a ring R. Then S is a subring of R if and only if a-b in S and a/b in S for all a/b in S.
Proof. Let S be a subring of R.
We now prove that a-b e S and ab e S V a,b e S . Since S’ is a subring of R, S is a ring with respect to the addition and multiplication operations in R
a,b ∈ S=>a,-b ∈ S => a + (-b) = a-b <=S and a,b ∈ S => ab ∈ S
Let a-b ∈ S and ab ∈ S ∀a,b ∈ S.
We now prove that S is a ring.
Since S is a nonempty subset of the commutative group (R, +) with the condition a-b ∈ S and ab ∈ S; by group theory (S, +) is a commutative subgroup of (R, +).
Since ab ∈ S∀a, b ∈ S, multiplication (•) is a binary operation in S.
Also, a,b,c ∈ S => a,b,c ∈ R => a (bc) = (ab) c
Further a, b,c ∈ S=>a,b,c ∈ R => a (b + c) = ab + ac and (b + c) a = ba + ca
(S, +, •) is a ring and hence (S, +, •) is a subring of R.
Note. Every subring contains at least zero elements of the ring.
Theorem 2. (Subfield Test). Let K be a non-empty subset of a field F. Then K is a Subfield of F if and only if a,b ∈ K => a-b ∈ K and a ∈ K,b ≠ 0 ∈ K => ab-1 ∈ K
Example 1. nZ is a subdomain of Z.
we know that (Z,+, •) where Z = the set of all integers is an integral domain.
For a fixed n∈ Z we have nZ ={ nx /x ∈Z }
0∈Z is Zero element and n0 =0=> 0 ∈nZ.
∴ nZ ≠Φ and n Z⊂ Z
Let x,y ∈ Z. Then nx, ny ∈ nZ.
nx-ny =n(x-y) ∈ nZ ; ( x-y ∈ Z)
Also (nx) (ny) =n(x n y) ∈ nZ; ( x. n y ∈ Z ) n Z is a subdomain of Z.
Example 2. Z is not a subfield of Q. For 2,3 ∈ Z and 3 ≠ 0 => 3-1 = (1 /3)∈ Q.
But 2.3-1 = (2/3)∈ Z
Example 3. The Unity element of a ring need not be the same as the unity element of a subring.
Consider Z6 = {0, 1, 2, 3, 4, 5} the ring with unity element 1.
For the subring s = { 0,2,4 } ; we have 0. 4 = 4. 0 =0; 2. 4 = 4. 2 = 2; 4. 4 = 4. 4 = 4
=> 4 is the unity element of S. Hence the unity of Z6 ≠ unity of S.
Theorem 3. The intersection of two subrings of an R is a subring of R.
Prood: let \(S_1, S_2\) be two subrings of R. Let \(0 \in R\) be the zero element.
Since every subring contains at least zero elements of the ring, \(0 \in S_1\) and \(0 \in S_2.\)
∴ \(0 \in S_1 \cap S_2\) and hence \(S_1 \cap S_2 \neq \phi\) and \(S_1 \cap S_2\) subset R.
Let a, b \(\in S_1 \cap S_2\). Then a, b \(\in S_1\) and \(a, b \in S_2\).
a, b \(\in S_1\) and \(S_1\) is a subring of R \(\Rightarrow a-b \in S_1\) and \(a b \in S_1\)
a, b \(\in S_2\) and \(S_2\) is a subring of R
⇒ \(a-b \in S_2\) and \(a, b \in S_2\)
From (1) and (2) we have a, b\( \in S_1 \cap S_2 \Rightarrow a-b \in S_1 \cap S_2\) and \(a b \in S_1 \cap S_2\)
∴ \(S_1 \cap S_2\) is a subring of R.