Prime and Maximal Ideals | Algebra - Mathematics PDF Download

The goal of this section is to characterize those ideals of commutative rings with identity which correspond to factor rings that are either integral domains or fields.

Definition 3.4.1 Let R be a ring. A two-sided ideal I of R is called maximal if I ≠ R and no proper ideal of R properly contains I.

EXAMPLES

1. In Z, the ideal (6) = 6Z is not maximal since (3) is a proper ideal of Z properly containing h6i (by a proper ideal we mean one which is not equal to the whole ring).

2. In Z, the ideal (5) is maximal. For suppose that I is an ideal of Z properly containing (5). Then there exists some m ∈ I with m ∉ (5), i.e. 5 does not divide m. Then gcd(5, m) = 1 since 5 is prime, and we can write 1 = 5s + mt for integers s and t. Since 5s ∈ I and mt ∈ I, this means 1 ∈ I. Then I = Z, and (5) is a maximal ideal in Z.

3. The maximal ideals in Z are precisely the ideals of the form (p), where p is prime.

The following is a generalization of the statement that Z/nZ is a field precisely when n is prime.

Theorem 3.4.2 Let R be a commutative ring with identity, and let M be an ideal of R. Then the factor ring R/M is a field if and only if M is a maximal ideal of R.

COMMENT ON PROOF: There are two things to be shown here. We must show that if R/M is a field (i.e. if every non-zero element of R/M is a unit), then M is a maximal ideal of R. A useful strategy for doing this is to suppose that I is an ideal of R properly containing M, and try to show that I must be equal to R.
We must also show that if M is a maximal ideal of R, then every non-zero element of R/M is a unit. A strategy for doing this is as follows : if a ∈ R does not belong to M (so a+ M is not the zero element in R/M), then the fact that M is maximal as an ideal of R means that the only ideal of R that contains both M and the element a is R itself. In particular the only ideal of R that contains both M and the element a contains the identity element of R.

Proof of Theorem 4.2.6: (  ) Suppose that R/M is a field and let I be an ideal of R properly containing M. Let a ∈ I, a ∉ M. Then a + M is not the zero element of R/M, and so (a + M)(b + M) = 1 + M, for some b ∈ R. Then ab - 1 ∈ M; let m = ab - 1. Now 1 = ab - m and so 1 ∈ I since a ∈ I and m ∈ I. It follows that I = R and so M is a maximal ideal of R.

(⇒): Suppose that M is a maximal ideal of R and let a+ M be a non-zero element of R/M. We need to show the existence of b + m ∈2 R/M with (a + M)(b + M) = 1 + M. This means ab + M = 1 + M, or ab - 1 ∈ M.
So we need to show that there exists b ∈ R for which ab - 1 ∈ M. Let M' denote the set of elements of R of the form

or + s, for some r ∈ R and s ∈ M.

Then M' is an ideal of R (check), and M' properly contains M since a ∈ M' and a ∉ M. Then M' = R since M is a maximal ideal of R. In particular then 1 ∈ M' and 1 = ab + m for some b ∈ R and m ∈ M. Then ab - 1 ∈ M and (a + M)(b + M) = 1 + M in R/M.

So a + M has an inverse in R/M as required.

We will now characterize those ideals I of R for which R=I is an integral domain.

Definition 3.4.3 Let R be a commutative ring. An ideal I of R is called prime if I ≠ R and whenever ab ∈ I for elements a and b of R, either a ∈ I or b ∈ I.

EXAMPLE: The ideal (6) is not a prime ideal in Z, since 2x3 2 (6) although neither 2 nor 3 belongs to (6). However the ideal (5) is prime in Z, since the product of two integers is a multiple of 5 only if at least one of the two is a multiple of 5.
The prime ideals of Z are precisely the maximal ideals; they have the form hpi for a prime p.

Theorem 3.4.4 Let R be a commutative ring with identity, and let I be an ideal of R. Then the factor ring R/I is an integral domain if and only if I is a prime ideal of R.

Proof: R/I is certainly a commutative ring with identity, so we need to show that R/I contains zero-divisors if and only if I is not a prime ideal of R. So let a+ I, b+ I be non-zero elements of R/I. This means neither a nor b belongs to I. We have (a + I)(b + I) = 0 + I in R/I if and only if ab ∈ I. This happens for some pair a and b if and only if I is not prime.

Corollary 3.4.5 Let R be a commutative ring with identity. Then every maximal ideal of R is prime.

Proof: Let M be a maximal ideal of R. Then R=M is a eld so in particular it is an integral domain. Thus M is a prime ideal of R.

QUESTION FOR THE SEMINAR: Try to prove Corollary 3.4.5 using only the denitions of prime and maximal ideals.

It is not true that every prime ideal of a commutative ring with identity is maximal. For example

1. We have already seen that the zero ideal of Z is prime but not maximal.

2. In Z[x], let I denote the ideal consisting of all elements whose constant term is 0 (I is the principal ideal generated by x). The I is a prime ideal of Z[x] but it is not maximal, since it is contained for example in the ideal of Z[x] consisting of all those polynomials whose constant term is even.

Theorem 3.4.6 Let F be a field and let I be an ideal of the polynomial ring F[x]. Then

1. I is maximal if and only if I = (p(x)) for some irreducible polynomial p(x) in F[x].

2. I is prime if and only if I = {0} or I = (p(x)) for an irreducible p(x) ∈ F[x].

Proof: By Lemma 3.2.3 I is principal, I = (p(x)) for some p(x) ∈ F[x].

1. (  ) Assume p(x) is irreducible and let I1 be an ideal of F[x] containing I. Then I₁ = (f(x)) for some f(x) ∈ F[x]. Since p(x) ∈ I₁ we have p(x) = f(x)q(x) for some q(x) ∈ F[x]. Since p(x) is irreducible this means that either f(x) has degree zero (i.e. is a non-zero element of F) or q(x) has degree zero.

If f(x) has degree zero then f(x) is a unit in F[x] and I₁ = F[x]. If q(x) has degree zero then p(x) = af(x) for some nonzero a ∈ F, and f(x) = a^-1 p(x); then f(x) ∈ I and I₁ = I. Thus either I₁ = I or I₁ = F[x], so I is a maximal ideal of F[x].

(⇒): Suppose I = (p(x)) is a maximal ideal of F[x]. Then p(x) ≠ 0. If p(x) = g(x)h(x) is a proper factorization of p(x) then g(x) and h(x) both have degree at least 1 and (g(x)) and hh(x)i are proper ideals of F[x] properly containing I. This contradicts the maximality of I, so we conclude that p(x) is irreducible. This proves 1.

2. Certainly the zero ideal of F[x] and the principal ideals generated by irreducible polynomials are prime. Every other ideal has the form (f(x)) for a reducible f(x). If I = (f(x)) and f(x) = g(x)h(x) where g(x) and h(x) both have degree less than that of f(x) then neither g(x) nor h(x) belongs to I but their product does. Thus I is not prime.