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Let R be a ring. An ideal of R is a subring I of R such that if a element of I, r element of R, then a · r element of I and r · a element of I. The set of integers. If is any nonempty collection of subgroups of the group G, then the intersection H* = is a subgroup of G.

The Arithmetic of Ideals

Let R be a commutative ring with identity. In this section we will begin to study the arithmetic of ideals in R.

Lemma 1: Let collection be a nonempty collections of ideals of R. Then

A = Intersection of idealsI

is an ideal of R.

Proof: From Theorem 3 of the section on subgroups, we know that A is a subgroup of the additive group of R. Let r element of R, a element of A. Then a element of I for all I element of collection, so that r · a element of I for all I element of collection, since I is an ideal of R. Therefore, r · a element of A and A is an ideal of R.

Let S be a subset of R. Then there exists a smallest ideal of R which contains S. Let collection be the set of all ideals of R which contain S. Then

A = Intersection of idealsI

is an idea of R by Lemma 1 and it is clear that A contains S. Furthermore, if A0 is a ideal of R which contains S, then A0 element of collection and A subset of A0. Therefore A is the smallest ideal of R which contains S and A is called the ideal generated by S and is denoted by (S). If I is an ideal and S is a subset of R, then we say that S is a set of generators for I if I = (S).

If S = {a1,...,an}, we will write (a1,...,an) instead of (S). An ideal of the form (a1,...,an) is said to be finitely generated. An ideal of the form (a) is called a principal ideal.

Proposition 2: Let I = (S). Then the set I is the set of sums of the form

(1)
r1s1 + r2s2 + ... + rtst     (ri element of R, si element of S).

Proof: It is clear that every sum of the form (1) belongs to I, since I is an ideal containing S. Therefore, it suffices to show that the set of sums (1) form an ideal containing S, since I is the smallest such ideal. Since the elements of (1) will form a ring. That is an ideal, if there is another element t that is not in the form of (1) it would contradict the fact that (S) is the smallest ideal.

Corollary 3: Let I = (a1, ..., an). Then I consists of all elements of the form

r1a1 + ... + rnan     (ri element of R).

Example 1: Let R be arbitrary. Then (0) is the ideal consisting of only 0, and (1) contains r · 1 for all r element of R, so that (1) = R.

Example 2: Let R = Z, and let I be an ideal. Either I = (0) or I contains a nonzero element a. In this case, I also contains -a, and thus I contains a positive element. Let a be the smallest positive element in I. We assert that I = (a). Let x be a nonzero element of I. By the division algorithm in Z there exist integers q and r such that

x = q · a + r,    0 < r < a.

Since x element of I, a element of I, we have x - q · a = r element of I. Therefore, since a is the smallest positive element in I, and since 0 < r < a, we see that r = 0. Therefore, x = q · a element of (a). Thus, I subset of (a). But it is clear that (a) subset of I, since a element of I. Therefore, I = (a). We have just demonstrated that every ideal of Z is a principal ideal.

Example 3: Let F be a field. Then (0) and (1) are ideals of F. Let Inot equal(0) be an ideal of F. Then I contains an element xnot equal0. But since F is a field and I is an ideal, x-1 · x = 1 element of I. But if y element of F, then y · 1 = y element of I., Thus I = (1), and the only ideals of a field F are (0) and (1).

Example 4: Let F be a field, X an indeterminate over F, R = F[X]. Let Inot equal(0) be an ideal of R. Then I contains a nonzero polynomial f. Let g be the nonzero polynomial of smallest degree in I. We assert that I = (g). Since g element of I, we see that (g) subset of I. Let h element of I. By the division algorithm in F[X], there exist polynomials q,r element of F[X] such that

h = q · g + r,    deg(r) < deg(g).

Since deg(r) < deg(g) and since r = h - q · g element of I, we see that r = 0 by the choice of g. Therefore, h = q · g and h element of (g). Thus, we have proved that I subset of (g), so that I = (g). Thus, every ideal of F[X] is principal.

Example 5: Let us give an example of a ring in which not every ideal is principal. Let F be a field and let X and Y be indeterminates over F. Let us consider the ideal (X,Y). This ideal consists of all polynomials in F[X,Y] having zero constant term. If (X,Y) = (f) for some f element of F[X,Y], then X = k1f, Y = k2f for some k1,k2 element of F[X,Y]. But this impossible, so (X,Y) is not a principal ideal.

A ring R having the property that every ideal is principal is called a principal ideal ring (PIR). If, in addition, R is an integral domain, then R is called a principal ideal domain (PID). For example, Z, a field F, a polynomial ring F[X] over a field F are all examples of principal ideal domains. However, F[X,Y] is not a principal ideal domain.

Let A and B be ideals of R. The sum A + B of A and B is the ideal generated by A union B. By Proposition 2, A + B consists of all sums of the form

r1a1 + ... + rn + s1b1 + ... + smbm,
(ri, sj element of R, ai element of A, bj element of B).

But since A and B are ideals, riai element of A (1 < i < n), sjbj element of B (1 < j < m), and thus r1a1 + ... + rnan element of A, s1b1 + ... + smbm element of B. Therefore, every element of A + B is of the form a + b for a element of A, b element of B. Conversely, every such element belongs to A + B, so that

A + B = {a + b | a element of A, b element of B}.

For example, if R = Z, then

(2) + (4) = {2m + 4n | m,n element of Z}
= {2k | k element of Z}
= (2);
(5) + (7) = {5m + 7n | m,n element of Z}
= (1),

since 1 = 5 · 3 + 7 · (-2) element of (5) + (7). More generally, if a,b element of Z, then

(a) + (b) = {am + bn | m,n element of Z}.

If a = b = 0, then (a) + (b) = (0). Thus, we may assume that a and b are not both 0. Then (a) + (b) contains a = a · 1 + b · 0 and b = a · 0 + b · 1, and is therefore not (0). We have seen above that

(a) + (b) =(c),

where c is the smallest positive integer contained in (a) + (b). However, we have shown that the smallest positive integer of the form am + bn (m,n element of Z) equals (a,b) is the g.c.d. of a and b. Therefore, we have proved the following.

Lemma 4: Let a,b element of Z, a,b not both 0. Then the sum of ideals (a) and (b) is the ideal generated by the greatest common divisor (g.c.d) of a and b:

(a) + (b) = ((a,b)).

By Lemma 4, it is reasonable to regard the sum of two ideals (a), (b) of a general ring R as some sort of greatest common divisor of a and b. This was Kummer's great idea. Define the g.c.d. of two elements a,b element of Z[zeta] to be the ideal (a) + (b) and replace the usual arithmetic properties of the elements Z[zeta] (for example, divisibility, primes, factorization) by corresponding properties of ideals of Z[zeta]. As a first step in this direction, let us define the product of two ideals.

Let A and B be ideals of R. The product A · B of A and B is the ideal generated by all products of the form a · b (a element of A, b element of B). Thus, by proposition 2, A · B consists of all elements of the form

a1 · b1 + ... + an · bn,    ai element of A, bj element of B.

It is easy to see that

(2)
(a) · (b) = (a · b),
(a1,...,am)·(b1,...,bn)=
(3)
(a1 · b1,a1 · b2,...,a1 · bn,a2b1,...,am · bn).

Thus, for example, in F[X,Y], we have

(X,Y) · (X2,Y) = (X3, XY, X2Y, Y2).

The following are properties of sums and products of ideals which can easily be proven. Let A,B and C be ideals of R. Then

(4)
A + B = B + A,
(5)
A + (B + C) = (A + B) + C,
(6)
A · B = B · A,
(7)
A · (B · C) = (A · B) · C,
(8)
A · (B + C) = A · B + A · C.

One surprising property of the product is

(9)
A · B subset of A intersection B.