    # Prime and Maximal ideals

“There was a street evangelist who was trying to get the attention of passersby, he was urging them to flee from the wrath to come -I warn you, there will be weeping, wailing, and gnashing of teeth! An old woman in the crowd replied, -Sir, I do not have teeth! And the evangelist retorted, -Lady, teeth will be provided!” Apocalypse, Anawim, #justtothepoint.

# Recall

💍 A ring R is a non-empty set with two binary operations, addition (a + b) and multiplication (ab), such that ∀ a, b, c ∈ R:

1. Both operations are closed: a + b ∈ R, a·b ∈ R.
2. Commutative under addition: a + b = b + a.
3. Associative under addition: (a + b) + c = a + (b + c).
4. There is an additive identity 0 ∈ R such that a + 0 = a, ∀ a ∈ R.
5. There are inverse elements for addition, ∃-a ∈ R: a + (-a) = (-a) + a = 0.
6. Associative under product: a(bc) = (ab)c.
7. Multiplication is distributive over addition: a(bc) = ab + ac, (b + c)a = ba + ca.

A subring A of a ring R is called a (two-sided) ideal of R if it absorbs multiplication from the left and right from R, that is, ∀r ∈ R, a ∈ A, ra ∈ A and ar ∈ A. In other words, ∀r ∈ R, rA = {ra | a ∈ A} ⊆ A and Ar = {ar | a ∈ A} ⊆ A.

Let A ⊂ R, A ≠ R, R be a ring, A is an ideal of R if the following conditions are satisfied:

1. a - b ∈ A, ∀a, b ∈ A.
2. ∀r ∈ R, ∀a ∈ A, ra ∈ A and ar ∈ A.

Let R be a ring and let I be an ideal of R. In particular, R is an Abelian group under addition, and I is a normal subgroup of R, I ◁ R, therefore we could form the factor group R/I = {r + I | r ∈ R}.

Theorem. Let R be a ring and let I be a subring of R. The set of cosets R/I = {r + I | r ∈ R} is a ring under the operations (s + I) + (t + I) = (s + t) + I and (s + I)(t + I) = st + I iff I is an ideal of R.

# More examples

• Consider the factor ring of the Gaussian integers R = ℤ[i]/⟨2 -i⟩.

The Gaussian integers are the set ℤ[i] = {x + iy : x, y ∈ ℤ} of complex numbers whose real and imaginary parts are both integers.

ℤ[i] is a ring (it is indeed a subring of ℂ) since it is closed under addition and multiplication: (x + iy) + (p + iq) = (x + p) + i(y + q), (x + iy)(p + iq) =[i2 = -1] (xp − yq) + i(xq + yp).

2 -i + ⟨2 -i⟩ =[aH = H ↭ a ∈ H = 0 + H] 0 + ⟨2 -i⟩, so when dealing with coset representatives 2 -i ≡ 0 or 2 = i, e.g., 5 + 6i + ⟨2 -i⟩ = 5 + 6·2 + ⟨2 -i⟩ = 17 + ⟨2 -i⟩. In other words, ∀r ∈ R, r = a + ⟨2 -i⟩, a ∈ ℤ. Futhermore, 2 = i ⇒ 4 = i2 = -1 or 5 = 0.

5 + 6i + ⟨2 -i⟩ = 17 + ⟨2 -i⟩ = 2 + 5 + 5 + 5 + ⟨2 -i⟩ = 2 + ⟨2 -i⟩. It follows that ℤ[i]/⟨2 -i⟩ is likely to be {0 + ⟨2 -i⟩ = ⟨2 -i⟩, 1 + ⟨2 -i⟩, 2 + ⟨2 -i⟩, 3 + ⟨2 -i⟩, 4 + ⟨2 -i⟩}.

Are these five cosets distinct? 5(1 + ⟨2 -i⟩) =[(a + I) + (b + I) = (a + b) + I] 5 + ⟨2 -i⟩ = 0 + ⟨2 -i⟩. So 1 + ⟨2 -i⟩ has order 5 or 1.

Let’s suppose for the sake of contradiction 1 + ⟨2 -i⟩ has order 1, i.e., 1 + ⟨2 -i⟩ = 0 + ⟨2 -i⟩ ⇒ 1 ∈ ⟨2 -i⟩ ⇒ ∃a + bi ∈ ℤ[i] such that 1 = (2 - i)(a + bi) = 2a + b + (-a + 2b)i ⇒ 1 = 2a + b, 0 = -a + 2b ⇒ a = 2/5, b =1/5 ⊥ Hence, ℤ[i]/⟨2 -i⟩ = {0 + ⟨2 -i⟩ = ⟨2 -i⟩, 1 + ⟨2 -i⟩, 2 + ⟨2 -i⟩, 3 + ⟨2 -i⟩, 4 + ⟨2 -i⟩} ≋[|ℤ[i]/⟨2 -i⟩| = 5, 5 is prime] 5.

• Let R = {f: ℝ → ℝ| f is a continuous function}, I = {f ∈ ℝ | f(0) = f(1) = 0}.
1. Subring. ∀f, g ∈ I, f(0) = f(1) = 0 and g(0) = g(1) = 0 ⇒ (f - g)(0) = f(0) - g(0) = (f - g)(1) = f(1) - g(1) = 0 ⇒ f - g ∈ I. (f · g)(0) = f(0) · g(0) = (f · g)(1) = f(1) · g(1) = 0 ⇒ f · g ∈ I.
2. Absorption. ∀f ∈ I, h ∈ R, f(0) = f(1) = 0 ⇒ (f·h)(0) = f(0)·h(0) = 0, (f·h)(1) = f(1)·h(1) = 0 ⇒ f·h ∈ I.
• Let ℝ[x] be the ring of polynomials with real coefficient, let ⟨x2 + 1⟩ be the principal ideal generated by x2 + 1.

⟨x2 + 1⟩ = {f(x)(x2 + 1) | f(x) ∈ ℝ[x]}.

ℝ[x]/⟨x2 + 1⟩ = {g(x) + ⟨x2 + 1⟩ | g(x) ∈ ℝ[x]}. Let g(x) ∈ ℝ[x] ⇒ [By dividing g(x) by x2 + 1] g(x) = q(x)(x2 + 1) + r(x) where q(x) is the quotient, r(x) is the remainder, degree of r(x) < 2 ⇒ r(x) = ax + b, a, b ∈ ℝ.

g(x) + ⟨x2 + 1⟩ = q(x)(x2 + 1) + r(x) + ⟨x2 + 1⟩ = r(x) + ⟨x2 + 1⟩ where r(x) = ax + b, a, b ∈ ℝ because the ideal ⟨x2 + 1⟩ absorbs the term q(x)(x2 + 1). Therefore, ℝ[x]/⟨x2 + 1⟩ = {g(x) + ⟨x2 + 1⟩ | g(x) ∈ ℝ[x]} = {ax + b + ⟨x2 + 1⟩ | a, b ∈ ℝ} where x2 + ⟨x2 + 1⟩ = -1 + ⟨x2 + 1⟩.

Observe that x2 + 1 + ⟨x2 + 1⟩ = 0 + ⟨x2 + 1⟩, and so when dealing with coset representatives x2 + 1 = 0 or x2 = -1. Example: (x + 7 + ⟨x2 + 1⟩)·(3x + 4 + ⟨x2 + 1⟩) = 3x2 + 25x + 28 + ⟨x2 + 1⟩ =[28 -3] 25x + 25 + ⟨x2 + 1⟩. The quotient ring R[X] / ⟨X2 + 1⟩ is naturally isomorphic to the field of complex numbers ℂ, with the class [X] playing the role of the imaginary unit i.

The evaluation map Φ: ℝ[x] → ℂ, f(x)→ f(i) is onto, Ker(Φ) = ⟨x2 +1⟩ ⇒By the first isomorphism theorem, ℝ[x]/⟨x2+1⟩ ≋ ℂ.

# Prime and Maximal ideals

A prime ideal A of a commutative ring R is a proper ideal of R such that if a, b are two elements of R, and whenever their product ab is an element of A, then either a is in A, b is in A or both, i.e., a, b ∈ R, ab ∈ A ⇒ a ∈ A or b ∈ A. Example: ⟨x⟩ ⊆ ℤ[x] is a prime ideal, but ⟨x2⟩ is not, because x2 ∈ ⟨x2⟩ but x ∉ ⟨x2⟩.

This generalizes Euclid’s lemma: if p is a prime number and if p divides a product ab of two integers, then p divides a or p divides b. We have basically exchanged or abstracted divisibility with being an element of an ideal.

A maximal ideal A of a commutative ring R is a proper ideal of R that is maximal with respect to set inclusion amongst all proper ideals, i.e., whenever B is an ideal of R that lives between R and A, A ⊆ B ⊆ R, then B = A or B = R. In other words, we cannot fit any proper ideals between the ring and a maximal ideal.

Examples:

• The prime ideals of ℤ are {0} and pℤ where p is prime.

Proof.

1. Suppose ab ∈ {0} ⇒ ab = 0 ⇒[ ℤ is an integral domain] a = 0 or b = 0 ⇒ a ∈ {0} or b ∈ {0}.
2. Suppose I ⊆ ℤ, I ≠ {0}, I is a prime ideal ⇒[Proposition. Every ideal of ℤ is a principal ideal.] I = nℤ = ⟨n⟩ = {···-2n, -n, 0, n, 2n, ···} for some n ∈ ℕ. A prime ideal is a proper ideal, so n is not allowed to be 1 (I = ⟨1⟩ ⇒ I = ℤ), n ≥ 2.
3. Let’s prove that n is prime. Let’s suppose a, b, such that 1 ≤ a, b ≤ n, ab = n ∈ I ⇒[I is prime] a ∈ I or b ∈ I. Without loss of generality, we could assume that a ∈ I ⇒ a ∈ nℤ = ⟨n⟩ = {···-2n, -n, 0, n, 2n, ···}, 1 ≤ a ≤ n ⇒ a = n ⇒ b = 1 ⇒ n is prime.
• In the ring of integers, the maximal ideals are the principal ideals generated by a prime number.
• ⟨2⟩ and ⟨3⟩ are maximal ideals for ℤ36 and ℤ24. ⟨2⟩, ⟨3⟩, and ⟨5⟩ are maximal ideals for ℤ30. ⟨2⟩ is the maximal ideal for ℤ8.

The ideals of ℤn are additive subgroups of ℤn, and we know that additive subgroups of ℤn have the form ⟨d⟩ where d divides n. Therefore, the ideals in ℤn are precisely the sets of the form ⟨d⟩ where d divides n.

⟨2⟩ = {0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34}
⟨3⟩ = {0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 36}

If gcd(k, 36) = 1, then ⟨k⟩ = ℤ36, and observe that every integer not coprime with 36 is listed above, therefore ⟨2⟩ and ⟨3⟩ are our maximal ideals.  • The ideal I = ⟨x2 + 1⟩ is a maximal ideal in ℝ[x]. ⟨x2 + 1⟩ denotes the principal ideal generated by x2 + 1, i.e., ⟨x2 + 1⟩ = {f(x)(x2 + 1) | f(x) ∈ ℝ[x]} and ℝ[x] denotes the ring of polynomials with real coefficients.

Let A be an ideal of ℝ[x] that properly contains ⟨x2 + 1⟩, i.e., I = ⟨x2 + 1⟩ ⊂ A ⊆ ℝ[x] ⇒ ∃f(x) ∈ A, but f(x) ∉ ⟨x2 + 1⟩ ⇒ f(x) = q(x)(x2 + 1) + r(x), deg(r(x)) < 2, so r(x) = ax + b, where a and b are not both simultaneously equal to zero (a2 + b2 ≠ 0) because f(x) ∉ ⟨x2 + 1⟩.

ax + b = r(x) = f(x) -q(x)(x2 + 1) ∈ A ⇒ [By assumption, A is an ideal, (ax -b) ∈ ℝ[x]] a2x2 -b2 = (ax + b)(ax -b) ∈ A and [x2 + 1 ∈ ⟨x2+1⟩ ⊆ A, a2 ∈ ℝ[x]] a2(x2 + 1) ∈ A ⇒ 0 ≠ a2 + b2 =[We have already demonstrated that a2(x2 + 1) = (a2x2 + a2) ∈ A and a2x2 -b2 ∈ A] (a2x2 +a2) - (a2x2 -b2) ∈ A, hence 0 ≠ a2 + b2 ∈ A.

Therefore, ∃c ∈ ℝ, c ≠ 0, c ∈ A ⇒ 1 = (1/c)c ∈ A ⇒ [∀f(x) ∈ ℝ[x], f(x) = 1f(x) ∈ A] A = ℝ[x] ∎

• The ideal I = ⟨x2 + 1⟩ is not prime in ℤ2[x], ⟨x2 + 1⟩ = {f(x)(x2 + 1) | f(x) ∈ ℤ2[x]}, since it contains (x + 1)2 = x2 + 2x + 1 = x2 + 1, but x + 1 ∉ ⟨x2 + 1⟩. Alternatively, x2 + 1 =[1 =2 -1] x2 - 1 = (x-1)(x+1), but neither x -1, nor x + 1 are in ⟨x2 + 1⟩.

Recall the definition of prime ideal in commutative rings, ab ∈ A ⇒ a ∈ A or b ∈ A

Theorem. Let R be a commutative ring with unity, and let A be an ideal of R. Then, A is a prime ideal iff R/A is an integral domain.

Proof.

⇐) Suppose that R/A is an integral domain. is A prime? ∀a, b ∈ R, such that ab ∈ A ⇒ a ∈ A or b ∈ A?

(a + A)(b + A) = ab + A = [ab ∈ A, aH = H ↭ a ∈ H] A, the zero element of the ring R/A. However, by assumption, R/A is an integral domain ⇒ it cannot have zero divisors ⇒ a + A or b + A is zero (R/A) ⇒ a + A = 0 + A = A ↭ a ∈ A or b + A = 0 + A = A ↭ b ∈ A

⇒) Suppose A is prime. What does it mean to be an integral domain? It means to be a commutative ring with multiplicative identity and no zero divisors. R/A is already a commutative ring with multiplicative identity for any proper ideal A. We claim that R/A has no zero-divisors.

Let’s suppose that there exist a + A, b + A, such that (a + A)(b + A) = 0 + A = A ⇒ ab ∈ A ⇒ [A is a prime ideal] a ∈ A or b ∈ A ⇒ a + A = A or b + A = A, i.e., a + A or b + A is the zero coset, i.e., the zero element in R/A ∎

Theorem. Let R be a commutative ring with unity and let A be an ideal of R. Then, R/A is a field iff A is a maximal ideal.

Proof

⇐) Suppose that A is a maximal ideal. We need to show that every non-zero element of R/A has a multiplicative inverse (all other properties follow quite trivially).

Let’s take a non-zero element of R/A, say b + A ∈ R/A, b ∉ A (Otherwise, if b ∈ A ⇒ b + A = A = 0 + A). Consider the following change of ideals, A ⊂ ⟨b⟩ + A ⊂ R ⇒[By assumption, A is a maximal ideal, and we also know that b ∉ A] ⟨b⟩ + A = R ⇒[By assumption, R is a commutative ring with unity] 1 ∈ ⟨b⟩ + A ⇒ 1 = br + a’ for some r ∈ R, a’ ∈ A ⇒[1 -br ∈ A, a + H = b + H ↭ a-1b ∈ H] 1 + A = br + A ⇒ (b + A)(r + A) = br + A = 1 + A, and therefore we have just found the multiplicative inverse of b + A ∎

Recall Given a ring R with ideals I, J ⊆ R, I + J = {i + j | i ∈ I, j ∈ J}, IJ = {i1j1 + ··· + injn | ik ∈ I, jk ∈ J, ∀k: 1 ≤ k ≤ n} I ∩ J, I + J, IJ are all ideals. In particular, I + J is an ideal and ⟨b⟩ + A is an ideal.

⇒) Suppose R/A is a field. is A maximal?

Let B an ideal of R properly containing A, A ⊆ B, A ≠ B ⇒ ∃b ∈ B, b ∉ A ⇒ b + A is an non zero element of R/A ⇒ [R/A is a field, there are multiplicative inverses for all non zero elements] ∃c + A ∈ R/A such that 1 + A = (b + A)·(c + A) = bc + A ⇒ 1 - bc ∈ A ⊂ B ⇒ 1 - bc ∈ B ⇒[bc ∈ B because b ∈ B, c ∈ R] 1 - bc + bc ∈ B ⇒ 1 ∈ B ⇒ ∀r ∈ R, 1.r = r ∈ B ⇒ B = R ⇒[B is an arbitrary ideal of R containing A] A is maximal.

Corollary. Let R be a commutative ring with unity, let A be a maximal ideal ⇒ R/A is a field, and in particular R/A is an integral domain ⇒[Theorem. Let R be a commutative ring with unity, and let A be an ideal of R. Then, A is a prime ideal iff R/A is an integral domain] A is prime.

# Examples

• Let ℤ[x] be the ring of polynomials with integer coefficients, let I = ⟨x⟩. If f(x) ∈ I, f(x) = x(anxn+···+a1x + a0), it satisfies f(0) = 0. I = ⟨x⟩ = {f(x) ∈ ℤ[x] | f(0) = 0}. Conversely, {f(x) ∈ ℤ[x] | f(0) = 0} ⊆ ⟨x⟩ because f(x) = anxn+···+a1x = x(anxn-1+···+a1) ∈ ⟨x⟩

The ideal ⟨x⟩ = {f(x) ∈ ℤ[x]| f(0) = 0} ⊆ ℤ[x] is prime because ℤ[x]/⟨x⟩ ≋ ℤ is an integral domain [Recall, let R be a commutative ring with unity, and let A be an ideal of R. Then, A is a prime ideal iff R/A is an integral domain], but not maximal because ℤ is not a field [Recall. Let R be a commutative ring with unity and let A be an ideal of R. Then, R/A is a field iff A is a maximal ideal].

Let Φ: ℤ[x] → ℤ, Φ(f(x)) = f(0). Φ is clearly surjective. Ker(Φ) = ⟨x⟩ = {f(x) ∈ ℤ[x]| f(0) = 0}. By the First Isomorphism Theorem for rings, ℤ[x]/⟨x⟩ ≋ ℤ.

Another way of demonstrating ⟨x⟩ is prime is as follows, If g(x)h(x) ∈ ⟨x⟩ = {f(x) ∈ ℤ[x] | f(0) = 0}, then g(0)h(0) = 0 ⇒ [g(0), h(0) ∈ ℤ and ℤ is an integral domain -integral domains are generalizations of the ring of integers-] g(0)=0 or h(0) = 0 ⇒ g(x) ∈ ⟨x⟩ or h(x) ∈ ⟨x⟩. Therefore, ⟨x⟩ is a prime ideal in ℤ[x].

• ⟨x, 2⟩ is maximal ↭ ℤ[x]/⟨x, 2⟩ ≋ ℤ/2ℤ is a field.

⟨x⟩ is not maximal because ⟨x⟩ ⊂ ⟨x, 2⟩ ⊂ ℤ[x].

An element in ⟨x, 2⟩ has the form f(x) = xg(x) + 2h(x) ⇒ f(0) = 2h(0), so I = ⟨x, 2⟩ ⊆ {f(x)∈ℤ[x] | f(0) is an even integer}.

Let f(x) ∈ {f(x)∈ℤ[x] | f(0) is an even integer} ⇒ f(x) = anxn+···+a1x+ 2k =[Recall that the ideal generated by the family a1, a2, ···, an is ⟨a1, a2, ···, an⟩ = {r1a1 + r2a2 + ··· + rnan | ri ∈ R}] x(anxn-1+···+a1) + 2k ∈ ⟨x, 2⟩, therefore {f(x)∈ℤ[x] | f(0) is an even integer} ⊆ I = ⟨x, 2⟩, hence I = ⟨x, 2⟩ = {f(x)∈ℤ[x] | f(0) is an even integer}.

ℤ[x]/⟨x, 2⟩ is a field because every element of ℤ[x]/⟨x, 2⟩ can be expressed as

1. anxn+···+a1x + 2k + ⟨x, 2⟩ = 0 + ⟨x, 2⟩
2. anxn+···+a1x + (2k + 1) + ⟨x, 2⟩ =[anxn+···+a1x + 2k is absorbed by ⟨x, 2⟩] 1 + ⟨x, 2⟩

Therefore, ℤ[x]/⟨x, 2⟩ ≋ ℤ2 ⇒ ℤ[x]/⟨x, 2⟩ is a field ⇒ ⟨x, 2⟩ is both prime and maximal.

Another way of proving ℤ[x]/⟨x, 2⟩ ≋ ℤ2 is as follows. Let Φ: ℤ[x] → ℤ, Φ(f(x)) = f(0), ψ: ℤ → ℤ2, ψ(x) = x mod 2. Consider the composite homomorphism ψ ∘ Φ: ℤ[x] → ℤ2. ψ ∘ Φ is obviously onto. Ker(ψ ∘ Φ) = {f(x)∈ℤ[x] | f(0) is an even integer} = ⟨x, 2⟩. By the First Isomorphism Theorem for Rings, ℤ[x]/⟨x, 2⟩ ≋ ℤ2.

• Let Z[i] = {a + bi|a, b ∈ ℤ} be the Gaussian integers. I = ⟨3 -i⟩ ⊆ ℤ[i] is not a prime ideal. (1 + i)(1 - 2i) = (3 -i), but 1 + i ∉ ⟨3 -i⟩ and (1 -2i) ∉ ⟨3 -i⟩
1 + i ∉ ⟨3 -i⟩. Suppose 1 +i = (a +bi)(3 -i) = (3a +b) + (3b -a)i ⇒ 1 = 3a +b, 1 = 3b -a ⇒[1 = 3b -a ⇒ a = 3b -1] 1 = 3a + b = 3(3b -1) + b = 10b -3, 1 = 10b -3 ⇒ b = 4/10, b ∈ ℤ ⊥ Similarly, (1 -2i) ∉ ⟨3 -i⟩.

# Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.
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