Ideal generated by a subset

In the following discussion R will be a commutative ring with unit.

1.6.1 Definition. Let S = {a₁, a₂, ..., a_n} a non-empty subset of R. The intersection of all ideals of R containing S, is said ideal generated by S.  □

This definition makes sense since the intersection of a family of ideals of a ring R, is an ideal. The elements a₁, a₂, ..., a_n are said generators of the ideal, and the ideal generated from them is indicated as follows

1.6.2 Proposition. The ideal (a₁, a₂, ..., a_n) is the smallest ideal containing the elements a₁, a₂, ..., a_n and we have

Proof. The first part is obvious. Now every ideal containing a₁, a₂, ..., a_n must contain also ∑^m_i=1r_ia_i with r_i ∈ R. Moreover must contain as well ∑^m_i=1z_ia_i with z_i ∈ ℤ, and the set

is an ideal containing a₁, a₂, ..., a_n. Thus we proved that

Remark. The reason why the set T contains also the integer sums z_ia_i is given to the fact that the set

is an ideal, but it does not contain necessarily the a_i are required. For example if R = 2ℤ, S = {4}, the set T' = {4r | r ∈ 2ℤ} is the ideal of 2ℤ made by multiples of 8, hence not containing 4. If the ring R has the unit then the set T can be reduced to T'

If the set S contains infinite elements and R is a unitary ring, then

thus (S) consists of finite sums (of variable length) of elements r_is_i.

1.6.3 Examples.

1. R = ℤ, S = (5). Then

   (5) = {5z | z ∈ ℤ} = 5ℤ

2. Let R a ring with unit, S = {1}. Then

   (1) = R

3. Let R = 2ℤ, S = {6}. Then

   (6) = {6r + 6z| r ∈ 2ℤ, z ∈ ℤ} = {0, ± 6, ± 12,...} = 6ℤ

   Remark. The set T' = {6r | r ∈ 2ℤ} without integer multiples of 6, is the set containing multiples of 12, hence an ideal, but not the smallest ideal since it does not contain 6.

4. Let R = 𝕂[x,y], S = {x,y}. Then

   (x,y) = {f{x,y) ⋅ x + g{x,y) ⋅ y | f(x,y), g(x,y) ∈ 𝕂[x,y]} = {polynomials with null constant term}.

Principal ideals

If an ideal I of R contains a, then by the absorption property it must contain all elements divisible by a. In other words, Ra ⊆ I, where

It can be easily verified that Ra is itself a right-ideal, hence the smallest ideal containing a. Similarly Ra is a left-ideal. In the commutative case, these two subsets coincide and are denoted (a).

An ideal can be generated by different elements or by a single element a; In this latter case the ideal is said principal ideal generated by the element a. Let R be a commutative ring with identity; Then aR = (a) = {ar | r ∈ R} is a principal ideal. In this case aR = Ra since commutations holds.

Notation. aR stands for the set of all multiples of a by elements in R. (a) = aR is usually only true when R is commutative and with identity.

1.6.4 Definition. A principal ring is a ring with unit such that every ideal is principal.  □

1.6.5 Examples.

1. In ℤ the ideal generated by S = {6,15} is

   (6,15) = {z₁6 + z₂ 15| z_i ∈ ℤ }

   among the elements that can be written as linear combination of elements in ℤ, there is the gcd(6,15). Thus

   (6,15) = (3) = {3z | z ∈ ℤ}

   6 in (3) and 15 in (3) so (6,15) is a subset of (3); 3 in (6,15) so (3) is a subset of (6,15)... so they're equal.

2. In the ring R = ℚ[x,y] the set I of all polynomials with two variable x,y with null constant term, is an ideal generated by S = {x,y} but it's not a principal ideal (see exercise xx). Thus ℚ[x,y] is not a principal ring.

1.6.6 Definition. An ideal I ≠ R of R is said prime if for every a,b ∈ R

1.6.7 Examples. Consider in ℤ the ideal (5) = 5ℤ. If ab ∈ (5), then ab is a multiple of 5, hence 5 divides ab; But 5 is a prime number we must have a multiple of 5 (a ∈ (5)) or b multiple of 5 (b ∈ (5)). On the other hand (10) is not a prime ideal 4 ⋅ 5 = 20 ∈ (10) but neither 4 or 5 are in (10).  ■

We can notice easily that in ℤ, if n ≠ 0, (n) is a prime ideal if and only if the integer n is prime.

1.6.7 Definition. An ideal I of a ring R, I ≠ R is said maximal if

In other words, I is a maximal ideal if it is not properly contained in any proper ideal of R i.e. there are no other ideals contained between I and R.  □

1.6.8 Examples. Suppose first that n = p, where p is an irreducible element, and let I be an ideal of ℤ such that (p) ⊆ I ⊆ ℤ. Then there exists an integer k in I such that k ∉ (p). That is, k is not a multiple of p. Since p is a prime, this implies that k and p are relatively prime and there exist integers s and t such that

Now sk ∈ I, since k ∈ I. We also have tp ∈ I, since p ∈ I. Therefore, sk + tp = 1 is in I, since I is an ideal. But 1 ∈ I implies immediately that I = ℤ, and this proves that (p) is a maximal ideal if p is irreducible.

1. In ℤ, (5) is a maximal ideal. For suppose that (5) ⫋ (m) in ℤ, m must be a divisor of 5 distinct by 5. But then it must be m = 1, hence (m) = (1) = ℤ. The prime ideals of ℤ are (0),(2),(3),(5) ,...; these are all maximal except (0).

2. Example 1 shows that the ideal (4) is not maximal in ℤ. However, (4) is a maximal ideal of the ring E of all even integers. To see that this is true, let I be an ideal of E such that (4) ⊆ I ⊆ E. Let x be any element of I that is not in (4). Then x has the form

   x = 4k + 2 = 2(2k + 1)

   where k ∈ ℤ. Since I is an ideal

   x ∈ I and 4k ∈ I   ⇒   x − 4k = 2 ∈ I

   But 2 ∈ I ⇒ I = E. Thus (4) is a maximal ideal of E

3. In ℤ the ideal {0} is prime, as the integers are an integral domain, but not maximal as it is contained in any other ideal.

1.6.9 Lemma. A commutative ring R, with unit is a field if and only if it contains only the trivial ideals {0} and R

Proof. Let R a field and let I and ideal of R different from {0}. Then there exists in I an element a ≠ 0 such that the smallest ideal of R that contains a is Ra = {ra∣r ∈ R}. Then by definition of an ideal, I contains as well aa⁻¹ = 1, which implies that I contains every r = r1. Thus I = R. Conversely, let R a field, to show that R is a field, it's sufficient to prove that every non null element is invertible. Let a ≠ 0 an element of R. Consider the set I = {ar | r ∈ R}. This is an ideal different from {0}, because I contains a (R has the unit), hence I = R, and we would have 1 = ar̅ for some r̅ ∈ R. Thus a is invertible. □

The importance of maximal ideals is evident from the result of the following theorem.

1.6.10 Theorem. Let R a commutative ring with unit. An ideal I is maximal if and only if R/I is a field.

Proof. The The correspondence theorem for rings shows that I is a maximal ideal if and only if R/I has no ideals other than {0} and R/I itself, otherwise there would exist an additional ideal J/R in R/I which would correspond to a proper ideal J of R properly containing I, contradicting the maximality of I in R; The previous Lemma shows that a commutative ring with unit is simple if and only if R/I is a field. □

Proof 2. You can also prove it quite directly: Say I is maximal. Now pick any y outside of I, so the element y + I of R/I is nonzero in this quotient ring. But I maximal means the ideal (I,y) = {i + ry : i ∈ I, r ∈ R} = all of R. So some particular i in I and r in R exist with i + ry = 1, so ry + I = 1 − i + I = 1 + I = the multliplicative identity of R/I. But ry + I = (r + I)(y + I), so r + I is the inverse of y + I. We've shown that every nonzero element of R/I has an inverse, so R/I is a field.

1.6.11 Theorem. Let R a commutative ring (not necessarily with unit) and let I one of its ideal. The ideal I is prime if and only if R/I is an integral domain.

Proof. Assume that R/I is an integral domain. Thus in R/I we have a product of cosets equal to the zero coset:

Remember that the zero element of R/I is the coset consisting of all elements of I.

If both a + I and b + I, differ from the identity 0 + I, we would have zero divisors. The relation above is equal to

that is ab ∈ I and either a ∈ I or b ∈ I, and so I must be a prime ideal.

For the other direction, assume that I is a prime ideal. If a,b ∈ R with (a + I) (b + I) = 0 + I in R/I, then we have ab ∈ I, and so by assumption either a ∈ I or b ∈ I. This shows that either a + I = 0 + I or b + I = 0 + I, and so R/I is an integral domain.  □

The two theorems just proved state that in a commutative ring with unit, a maximal ideal is prime (remember that every field is also an integral domain).

We give now an example of a prime ideal which is not maximal.

1.6.12 Example. Let R = 𝕂[x,y]. The ideal (x) is a prime ideal which is not maximal. Consider the mapping

an epimorphism, with kernel (x). Owing to the fundamental theorem on ring homomorphism, we have

since the quotient is isomorphic to an integral domain, (x) is a prime ideal but since the quotient is not a field (x) is not a maximal ideal. Remember that 𝕂[y] is not a field.

For example (x) is a prime ideal in ℤ[x], as ℤ[x]/(x) ~ ℤ is an integral domain.  ■

Remark. Pay attention to the fact that in a ring without unit it does not hold true that a maximal ideal is prime. The following is an example.

1.6.13 Example. Let R = 2ℤ and let I = (4) {0, ±4, ± 8,..}. I is maximal: indeed let U an ideal of of R such U ⊃ I and U ≠ I. We must have U = 2ℤ, because since U ⊋ I, U must contain an element of the form 2k with k odd. But then k + 1 is even and 2 = 2(k + 1) −2k ∈ U hence U = 2ℤ. However I is not prime, e.g. 4 = 2 ⋅ 2 ∈ (4), but 2 ≠ (4). We could have also verify that I is not prime by checking the quotient ring 2ℤ/(4) ={(4), 2 + (4)}. The product of two whatsoever elements of such ring is not an integral domain as it should be in the case of the ideal being prime.  ■