Page 190 - DMTH403_ABSTRACT_ALGEBRA

Unit 18: Integral Domains

Theorem 3: Let m be a positive integer and R be an integral domain. Then the following Notes
conditions are equivalent.
(a) m = 0.
1
(b) ma = 0 for all a  R.
(c) ma = 0 for some a  0 in R.

Proof: We will prove (a)  (b)  (c)  (a).
(a)  (b) : We know that m 1 = 0.
Thus, for any a  R, ma = m (la) = (ml) (a) = 0a = 0, i.e., (b) holds.

(b) (c) : If ma = 0  a  R, then if is certainly true for some a  0 in R.
(c)  (a) : Let ma = 0 for some a  0 in R. Then 0 = ma = m (la) = (ml) a. As a  0 and R is without
zero divisors, we get m = 0.
1
What Theorem 3 tells us is that to find the characteristic of a domain we only need to look at the
set in {n.1 | n  N}.
Let us look at some examples.

(i) char Q = 0, since n.1  0 for any n  N.
(ii) Similarly, char R = 0 and char C = 0.
(iii) You have already seen that chat Z, = n. Thus, for any positive integer n, there exists a ring
with characteristic n.
Now let us look at a peculiarity of the characteristic of a domain.
Theorem 4: The characteristic of an integral domain is either zero or a prime number.?
Proof: Let R be a domain. We will prove that if the characteristic of K is not zero, then it is a
prime number. So suppose char R = m, where m  0. So m is the least positive integer such that
m.1 = 0. We will show that m is a prime number by supposing that it is not, and then proving that
our supposition is wrong.
So suppose m = st, where s, t  N, 1 < s < m and 1 < t < m. Then m.1 = 0  (st).l = 0 * (s.1) (t.1) = 0.
As R is without zero divisors, we get s.1 = 0 or t.1 = 0. But, s and t are less than m. So, we reach a
contradiction to the fact that m = char R. Therefore, our assumption that m = st, where 1 < s < m,
1 < t < m is wrong. Thus, the only factors of m are 1 and itself. That is, m is a prime number.
We will now see what algebraic structure we get after we impose certain restrictions on the
multiplication of a domain.

18.2 Field

Let (R, +, .) be a ring. We know that (R, +) is an abelian group. We also know that the operation
is commutative and associative. But (R,.) is not an abelian group. Actually, even if R has identity,
(R,.) will never be a group since there is no element a  R such that a.0 = 1. But can (R\{0}) be a
group? It can, in some cases. For example, from Unit 2 you know that Q* and R* are groups with
respect to multiplication. This allows us to say that Q and R are fields, a term we will now define.
Definition: A ring (R, +,.) is called a field if (R\{0},.) is an abelian group.

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