Ring Theory Oct. 10th, 2016 Lecture 4: Integral Domain and Subrings Dr. Ruma Kareem K. Ajeena Math. Dept., Class: 3

Definition 1.1 (Integral Domain). An integral domain is a commutative ring with identity which does not have divisors of zero.

Remark 1.2 The cancellation law for multiplication is satisfied in any integral domain.

1.0.1 Subrings

Definition 1.3 (Subring). Let (R, +, ·) be a ring and S ⊆ R be a nonempty subset of R. If the triple (S, +, ·) is itself a ring, then (S, +, ·) is said to be a subring of (R, +, ·).

Here, one needs to apply the definition of a ring to show that the triple (S, +, ·) is a ring. But both the distributive and associative laws hold automatically in S as a consequence of their validity in R. Since these laws are inherited from R, there is no need to satisfy them in the definition of a subring. So, we can give alternative definition of a subring as follows:

1. S is a nonempty subset of R; 2. (S, +) is a subgroup of (R, +); 3. S is closed under multiplication.

Furthermore, the definition of a subring can be improved as follows:

Definition 1.4 Let (R, +, ·) be a ring and Φ ≠ S ⊆ R. Then the triple (S, +, ·) is a subring of (R, +, ·) iff

1. a − b ∈ S, where a, b ∈ S (closed under differences), 2. a · b ∈ S, where a, b ∈ S (closed under multiplication).

Example 1.5 Every ring (R, +, ·) has two trivial subrings, if 0 is a zero element in a ring (R, +, ·), then both ({0}, +, ·) and (R, +, ·) are subrings of (R, +, ·).

Example 1.6 In the ring of integers (Z, +, ·), the triple (Ze, +, ·) is a subring, while (Z0, +, ·) is not. In particular, we can deduce that in a ring with identity, a subring does not need to contain the identity element.

Example 1.7 Let (Z6, +6, ·6), the ring of integers modulo 6. If S = {0, 2, 4}, then (S6, +6, ·6) which has operation tables are given as follows: (S6, +6, ·6) is a subring of (Z6, +6, ·6). √ Example 1.8 Let S = {a + b 3 : a, b ∈ Z}. Then (S, +, ·) is a subring of (R, +, ·). For a, b, c, d ∈ Z then √ √ √ (a + b 3) − (c + d 3) = (a − c) + (b − d) 3 ∈ S, √ √ √ (a + b 3) · (c + d 3) = (ac + 3bd) + (bc − ad) 3 ∈ S. This means that the set S is closed under both differences and products

1-1 1-2 Lecture 2: Ring Theory

Table 1.1:

+6 0 2 4 0 0 2 4 2 2 4 0 4 4 0 2

Table 1.2:

·6 0 2 4 0 0 0 0 2 0 4 2 4 0 2 4

Remark 1.9 One can point out to the following situations:

1. When a ring has an identity, this need not be true of its subrings. 2. Some subring has multiplicative identity, but the entire ring does not. 3. Both the ring and one of its subrings have identity elements which are distinct.

The identity for the subring must be a divisor of zero in the parent ring in each of these cases. To justify that, let 1′ is the identity element of the subring (S, +, ·), we assume 1′ is not identity for the entire ring (R, +, ·). So, there exists an element a ∈ R for which a · 1′ ≠ a. It is clear that

(a · 1′) · 1′ = a · (1′ · 1′) = a · 1′ or (a · 1′ − a) · 1′ = 0. Since neither a · 1′ − a nor 1′ is zero, the ring (r, +, ·) has zero divisors, so, 1′ is a zero divisor of (R, +, ·).

Example 1.10 The possibility (3) can be illustrated as follows. Remember the system (R × R, +, ·) as given in Example (??), this system is a ring. The triple (R × 0, +, ·) forms a subring with identity element (1,0), in this case, the element (1,0) differs from the identity for the parent ring which is (1,1).

Definition 1.11 Let (R, +, ·) be an arbitrary ring. If there exists a positive integer n such that na for all a ∈ R, then the least positive integer with this property is called the characteristic of the ring. If no such positive integer exists (that is, na = 0 for all a ∈ R implies n = 0), then we say (R, +, ·) has characteristic zero.

Example 1.12 The rings of integers, rational numbers and real numbers are examples of systems having characteristic zero.

Theorem 1.13 Let (R, +, ·) be a ring with identity. Then (R, +, ·) has characteristic n > 0 iff n is the least positive integer for which n1 = 0. Lecture 2: Ring Theory 1-3

Proof ⇒ If the ring (R, +, ·) is of characteristic n > 0, it follows trivially that (from Definition (1.11)) n1 = 0. If m1 = 0, where 0 < m < n, then

ma = m(1 · a) = (m1) · a = 0 · a = 0, ∀a ∈ R.

This means that the characteristic of (R, +, ·) is less than n (that is , m). This contradicts our hypothesis, so n is least positive integer for which n1 = 0. The converse (⇐) can be established in much the same way. (Prove that as a H.W.).

Corollary 1.14 In an integral domain, all the nonzero elements have the same additive order, which is the characteristic of the domain.

Proof. Suppose the integral domain (R, +, ·) has positive characteristic n. From definition of , any a ∈ R, a ≠ 0 will then possess a finite additive order m, with m ≤ n. But 0 = ma = (m1) · a, this means that m1 = 0, since (R, +, ·) is free of zero divisors. We can conclude n ≤ m (from Thm (1.13) n is least positive integer). Hence m = n and every nonzero element of R has additive order n.