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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 â‰ 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 âˆˆ I with m âˆ‰ (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 âˆˆ I and mt âˆˆ I, this means 1 âˆˆ 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 âˆˆ 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 âˆˆ I, a âˆ‰ M. Then a + M is not the zero element of R/M, and so (a + M)(b + M) = 1 + M, for some b âˆˆ R. Then ab - 1 âˆˆ M; let m = ab - 1. Now 1 = ab - m and so 1 âˆˆ I since a âˆˆ I and m âˆˆ I. It follows that I = R and so M is a maximal ideal of R.

(â‡’): 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 âˆˆ2 R/M with (a + M)(b + M) = 1 + M. This means ab + M = 1 + M, or ab - 1 âˆˆ M.

So we need to show that there exists b âˆˆ R for which ab - 1 âˆˆ M. Let M' denote the set of elements of R of the form

or + s, for some r âˆˆ R and s âˆˆ M.

Then M' is an ideal of R (check), and M' properly contains M since a âˆˆ M' and a âˆ‰ M. Then M' = R since M is a maximal ideal of R. In particular then 1 âˆˆ M' and 1 = ab + m for some b âˆˆ R and m âˆˆ M. Then ab - 1 âˆˆ 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 â‰ R and whenever ab âˆˆ I for elements a and b of R, either a âˆˆ I or b âˆˆ 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 âˆˆ 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) âˆˆ F[x].

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

1. ( ) Assume p(x) is irreducible and let I1 be an ideal of F[x] containing I. Then I_{1} = (f(x)) for some f(x) âˆˆ F[x]. Since p(x) âˆˆ I_{1} we have p(x) = f(x)q(x) for some q(x) âˆˆ 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 I_{1} = F[x]. If q(x) has degree zero then p(x) = af(x) for some nonzero a âˆˆ F, and f(x) = a^{-1} p(x); then f(x) âˆˆ I and I_{1} = I. Thus either I_{1} = I or I_{1} = F[x], so I is a maximal ideal of F[x].

(â‡’): Suppose I = (p(x)) is a maximal ideal of F[x]. Then p(x) â‰ 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.

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