In this section, we introduce an important class of the subrings known as ideals, whose role in ring theory is similar to that of the normal subgroups in the study of groups. As shall be seen, ideals lead to the construction of quotient rings which are the appropriate analogs of quotient groups.
De?nition 1.1 (Ideal). A subring (I,+,·) of the ring (R,+,·) is an ideal of (R,+,·) i? for all r ? R and a ? I such that both r·a ? I and a·r ? I.
Remark 1.2 1. If (I,+,·) is a subring of a ring (R,+,·), I is already closed under multiplication. 2. For (I,+,·) to be an ideal, I is closed under multiplication by an arbitrary element of R. Based on De?nition of subring, which gives a minimum set of conditions on I for (I,+,·) to be a subring, so the de?nition of an ideal can be given as follows.
De?nition 1.3 Let (R,+,·) be a ring and I a nonempty subset of R. Then (I,+,·) is an ideal of (R,+,·) i?
1. If a,b ? I then a?b ? I; 2. If r ? R and a ? I then both r·a ? I and a·r ? I.
Remark 1.4 In the case of a commutative ring , we need only require r·a ? I.
Now, the concept of an ideal can be explained by the following examples:
Example 1.5 In any ring (R,+,·), the trivial subrings (R,+,·) and ({0},+,·) are both ideals. A ring which contains no ideals except these two ideals is said to be simple. Any ideal di?erent from (R,+,·) is called proper.
Example 1.6 The subring ({0,3,6,9},+12,·12) is an ideal of (Z,+12,·12), the ring of integers modulo 12. Example 1.7 For a ?xed integer a ?Z, let (a) denotes to the set of all integral multiples of a, that is, (a) = {na : n ?Z}. Show that ((a),+,·) is an ideal ? Solution: The following relations shows the triple ((a),+,·) is an ideal of the ring (Z,+,·): 1. na?ma = (n?m)a ? (a); 2. m(na) = (mn)a ? (a) with m,n ?Z.
1-1
1-2 Lecture 5: Ring Theory
In particular, since (2) = Ze, the ring of even integers (Ze,+,·) is an ideal of (Z,+,·). Example 1.8 Suppose (R,+,·) is the commutative ring of functions f : R ? R such that (f + g)(a) = f(a) + g(b), (f ·g)(a) = f(a)·g(b), a ? R. De?ne I = {f ? R : f(1) = 0}. For functions f,g ? I and h ? R, we have (f ?g)(1) = f(1)?g(1) = 0?0 = 0, and also (h·f)(1) = h(1)·f(1) = h(1)·0 = 0. Since both f ?g and h·f ? I. Then (I,+,·) is an ideal of (R,+,·). Theorem 1.9 If (I,+,·) is a proper ideal of a ring with identity, then no element of I has a multiplicative inverse, that is I ?R? = ?. Proof: Suppose to the contrary that there is some member a ?= 0 of I such that a?1 exists. Since I is closed under multiplication by arbitrary elements of R, a?1·a = 1 ? I. It follows that from same reason I contains r·1 = r, for all r ? R. This means that R ? I. On the other hand, (I,+,·) is a proper ideal of a ring (R,+,·), so I ? R, so I = R. This contradicts the hypothesis that I is a proper subset of R.
Theorem 1.10 If (Ii,+,·) is an arbitrary indexed collection of ideals of the ring (R,+,·), then so also is (? Ii,+,·). Proof: The intersection ? Ii ?= ?, since for all set Ii, 0 ? Ii, why? Suppose the elements a,b ?? Ii and r ? R. Then a,b ? I (where i ranges over the index set). Since the triple (Ii,+,·) is an ideal of (R,+,·), it follows from de?nition of ideal that a?b ? Ii and r·a and a·r ? Ii. But this is true for all i, a?b,r·a,a·r ?? Ii, so (? Ii,+,·) is an ideal of (R,+,·).