
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 be a nonempty collections of ideals of R. Then
A = I
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 R, a A. Then a I for all I , so that r · a I for all I , since I is an ideal of R. Therefore, r · a 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 be the set of all ideals of R which contain S. Then
A = I
is an idea of R by Lemma 1 and it is clear that A contains S. Furthermore, if A_{0} is a ideal of R which contains S, then A_{0} and A A_{0}. 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 = {a_{1},...,a_{n}}, we will write (a_{1},...,a_{n}) instead of (S). An ideal of the form (a_{1},...,a_{n}) 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)
r _{1}s _{1} + r _{2}s _{2} + ... + r _{t}s _{t} (r _{i} R, s _{i} 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 = (a_{1}, ..., a_{n}). Then I consists of all elements of the form
r _{1}a _{1} + ... + r _{n}a _{n} (r _{i} R).
Example 1: Let R be arbitrary. Then (0) is the ideal consisting of only 0, and (1) contains r · 1 for all r 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 I, a I, we have x  q · a = r 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 (a). Thus, I (a). But it is clear that (a) I, since a 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 I(0) be an ideal of F. Then I contains an element x0. But since F is a field and I is an ideal, x^{1} · x = 1 I. But if y F, then y · 1 = y 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 I(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 I, we see that (g) I. Let h I. By the division algorithm in F[X], there exist polynomials q,r F[X] such that
h = q · g + r, deg(r) < deg(g).
Since deg(r) < deg(g) and since r = h  q · g I, we see that r = 0 by the choice of g. Therefore, h = q · g and h (g). Thus, we have proved that I (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 F[X,Y], then X = k_{1}f, Y = k_{2}f for some k_{1},k_{2} 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 B. By Proposition 2, A + B consists of all sums of the form
r_{1}a_{1} + ... + r_{n} + s_{1}b_{1} + ... + s_{m}b_{m},
(r _{i}, s _{j} R, a _{i} A, b _{j} B).
But since A and B are ideals, r_{i}a_{i} A (1 < i < n), s_{j}b_{j} B (1 < j < m), and thus r_{1}a_{1} + ... + r_{n}a_{n} A, s_{1}b_{1} + ... + s_{m}b_{m} B. Therefore, every element of A + B is of the form a + b for a A, b B. Conversely, every such element belongs to A + B, so that
A + B = {a + b  a A, b B}.
For example, if R = Z, then
(2) + (4) = {2m + 4n  m,n Z}
= {2k  k Z}
= (2);
(5) + (7) = {5m + 7n  m,n Z}
= (1),
since 1 = 5 · 3 + 7 · (2) (5) + (7). More generally, if a,b Z, then
(a) + (b) = {am + bn  m,n 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 Z) equals (a,b) is the g.c.d. of a and b. Therefore, we have proved the following.
Lemma 4: Let a,b 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 Z[] to be the ideal (a) + (b) and replace the usual arithmetic properties of the elements Z[] (for example, divisibility, primes, factorization) by corresponding properties of ideals of Z[]. 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 A, b B). Thus, by proposition 2, A · B consists of all elements of the form
a _{1} · b _{1} + ... + a _{n} · b _{n}, a _{i} A, b _{j} B.
It is easy to see that
(2)
(a) · (b) = (a · b),
(a_{1},...,a_{m})·(b_{1},...,b_{n})=
(3)
(a_{1} · b_{1},a_{1} · b_{2},...,a_{1} · b_{n},a_{2}b_{1},...,a_{m} · b_{n}).
Thus, for example, in F[X,Y], we have
(X,Y) · (X^{2},Y) = (X^{3}, XY, X^{2}Y, Y^{2}).
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 A B.
