Which statements about ideals of ℤ[x] are true?

(x) is a prime ideal that is not maximal

Which statements about finite fields are true?

F₁₆ contains a subfield with 4 elements

the multiplicative group of F₂₅ is cyclic

there is a field with 12 elements

F₁₆ contains a subfield with 4 elements

F₁₆ contains a subfield with 8 elements

the multiplicative group of F₂₅ is cyclic

The quotient ring ℤ₂[x]/(x³ + x + 1) is

a ring with zero divisors and 8 elements

Which statements are true?

every finite extension of a field is algebraic

no finite field is algebraically closed

the field of real algebraic numbers is algebraically closed

every finite extension of a field is algebraic

no finite field is algebraically closed

Rings, Ideals, Factorisation Domains, Polynomials and Fields

The ring and field half of Section 6 is a ladder of definitions, each with an example that sits on one rung and not the next. A commutative ring with 1 may have zero divisors; an integral domain has none; a Euclidean domain has a division algorithm; every Euclidean domain is a principal ideal domain, and every PID is a unique factorisation domain — and each inclusion is strict. Prime and maximal ideals are detected by their quotients: R/P is a domain, R/M is a field. For polynomial rings the working tools are Gauss’s lemma and Eisenstein’s criterion. Fields then enter through their extensions: degrees multiply in towers, finite extensions are algebraic, ℂ is algebraically closed, and there is exactly one finite field of each prime-power order, with a cyclic multiplicative group. The questions set here are counts — units and zero divisors of ℤₙ, ideals, degrees of extensions, primitive elements, irreducible polynomials over ℤₚ — and “which of these rings is a UFD” MSQs.

1. Rings, ideals, quotients and ℤₙ

An ideal I of a commutative ring R is an additive subgroup with rI ⊆ I; the quotient R/I is a ring, and kernels of ring homomorphisms are exactly the ideals. The ideals of ℤ are nℤ, and those of ℤₙ correspond to the divisors of n, so ℤ₃₆ has d(36) = 9 ideals. In ℤₙ every non-zero element is either a unit (gcd(a, n) = 1; there are φ(n) of them) or a zero divisor, so ℤ₂₄ has φ(24) = 8 units and 23 − 8 = 15 non-zero zero divisors. A finite integral domain is a field, which is why ℤₙ is a field exactly when n is prime.

2. Prime and maximal ideals

In a commutative ring with 1: P is prime iff R/P is an integral domain; M is maximal iff R/M is a field. So every maximal ideal is prime. In ℤ the non-zero prime ideals pℤ are maximal and (0) is prime but not maximal. The maximal ideals of ℤₙ are pℤₙ for the primes p | n. In F[x], (f) is maximal iff f is irreducible, and in a PID every non-zero prime ideal is maximal.

Ideals of ℤ[x] and ℝ[x], read from their quotients
Ideal	Quotient	Verdict
(x) in ℤ[x]	ℤ	prime, not maximal
(2, x) in ℤ[x]	ℤ₂	maximal, and not principal
(x² + 1) in ℤ[x]	ℤ[i]	prime, not maximal
(x² + 1) in ℝ[x]	ℂ	maximal
(x² − 1) in ℚ[x]	ℚ × ℚ	not prime: (x − 1)(x + 1) ∈ I, neither factor is

3. Euclidean domains, PIDs and UFDs

Euclidean domain ⊊ PID ⊊ UFD ⊊ integral domain. In a UFD every non-zero non-unit is a product of irreducibles, uniquely up to order and units, and irreducible elements are prime. Gauss: if R is a UFD, so is R[x]; hence ℤ[x] and F[x, y] are UFDs, although neither is a PID.

Where each standard ring sits
Ring	Euclidean	PID	UFD
ℤ, F[x], ℤ[i]	yes	yes	yes
ℤ[(1 + √−19)/2]	no	yes	yes
ℤ[x], F[x, y]	no	no: (2, x) or (x, y) is not principal	yes (Gauss)
ℤ[√−5]	no	no	no: 6 = 2 · 3 = (1 + √−5)(1 − √−5), four irreducibles, no two associate

4. Polynomial rings and irreducibility

A polynomial of degree 2 or 3 over a field is irreducible iff it has no root there; over ℚ, the rational root test limits the candidates to ±(divisors of a₀)/(divisors of aₙ).
Gauss’s lemma: a primitive polynomial in ℤ[x] that factors over ℚ factors over ℤ into polynomials of the same degrees.
Eisenstein: if a prime p divides a₀, …, aₙ₋₁, p ∤ aₙ and p² ∤ a₀, then aₙxⁿ + … + a₀ is irreducible over ℚ. So x⁵ + 6x³ + 9x + 3 (p = 3) and xⁿ − 2 (p = 2) are irreducible. After the shift x ↦ x + 1, Eisenstein at p proves the cyclotomic Φₚ(x) = xᵖ⁻¹ + … + 1 irreducible.

⚠️ No roots does not mean irreducible from degree 4 on

x⁴ + 4 has no rational root, yet x⁴ + 4 = (x² + 2x + 2)(x² − 2x + 2). And x⁴ + 1 is irreducible over ℚ (Eisenstein after x ↦ x + 1: x⁴ + 4x³ + 6x² + 4x + 2, p = 2) although it factors modulo every prime, so reducing modulo p can prove irreducibility but never disprove it.

5. Field extensions, algebraic extensions and finite fields

For fields F ⊆ K, the degree [K : F] is dim_F K, and in a tower F ⊆ K ⊆ L, [L : F] = [L : K][K : F]. α ∈ K is algebraic over F if it is a root of a non-zero polynomial; its minimal polynomial is the monic irreducible one, and [F(α) : F] equals its degree. Every finite extension is algebraic. [ℚ(∛2) : ℚ] = 3 (x³ − 2, Eisenstein); [ℚ(√2, √3) : ℚ] = 4; the splitting field of x³ − 2 is ℚ(∛2, ω), of degree 6, because ℚ(∛2) ⊂ ℝ cannot contain the non-real ω. A field is algebraically closed if every non-constant polynomial has a root in it: ℂ is (the fundamental theorem of algebra), as is the countable field of all algebraic numbers; ℝ and every finite field are not (x^q − x + 1 has no root in F_q).

A finite field has pⁿ elements, p its characteristic; for each prime power there is exactly one up to isomorphism, the splitting field of x^pⁿ − x over ℤₚ. There is no field with 6, 10 or 12 elements.
F_pⁿ contains a copy of F_pᵐ iff m | n: F₄ ⊂ F₁₆, but F₈ ⊄ F₁₆.
The multiplicative group F_q^× is cyclic of order q − 1, with φ(q − 1) generators: F₁₆ has φ(15) = 8 primitive elements.
ℤ₂[x]/(x³ + x + 1) is F₈, since x³ + x + 1 has no root in ℤ₂. The number of monic irreducible quadratics over ℤₚ is (p² − p)/2: all p² monic quadratics minus the p(p + 1)/2 that split, so 10 over ℤ₅.

Key takeaways

In ℤₙ, φ(n) units and n − 1 − φ(n) non-zero zero divisors; ideals correspond to divisors of n and maximal ideals to its prime divisors.
R/P is a domain for prime P and a field for maximal P; (x) in ℤ[x] is prime and not maximal, (2, x) is maximal and not principal.
ED ⊊ PID ⊊ UFD: ℤ[x] is a UFD and not a PID; ℤ[√−5] is not a UFD because 6 has two factorisations.
Eisenstein proves irreducibility over ℚ; degree ≤ 3 needs only a root test; x⁴ + 4 shows that no roots is not enough beyond that.
Degrees multiply in towers; one finite field of each order pⁿ, F_pᵐ ⊆ F_pⁿ iff m | n, and F_q^× is cyclic with φ(q − 1) generators.

Practice questions (14)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The number of units in the ring ℤ₃₆ is ____.
Numerical answer — type the value.
Show answer
Answer: 12
a is a unit of ℤₙ iff gcd(a, n) = 1, so the count is φ(36) = 36 · (1 − 1/2)(1 − 1/3) = 12.
The number of non-zero zero divisors in the ring ℤ₂₄ is ____.
Numerical answer — type the value.
Show answer
Answer: 15
In ℤₙ a non-zero a is a unit if gcd(a, n) = 1 and a zero divisor otherwise (a · n/gcd(a, n) = 0). There are 23 non-zero elements and φ(24) = 8 units, so 15 zero divisors.
The number of ideals of the ring ℤ₃₆ is ____.
Numerical answer — type the value.
Show answer
Answer: 9
Ideals of ℤₙ are the additive subgroups ⟨d⟩ with d | n, one for each divisor. 36 = 2²·3² has (2 + 1)(2 + 1) = 9 divisors. Only two of them, (2) and (3), are maximal.
Which statements about ideals of ℤ[x] are true?
1. (x) is a prime ideal that is not maximal
2. (2, x) is a maximal ideal
3. (2, x) is a principal ideal
4. (x² + 1) is a maximal ideal
Show answer
Answer: A — (x) is a prime ideal that is not maximal; B — (2, x) is a maximal ideal
(1) ℤ[x]/(x) ≅ ℤ, a domain that is not a field. (2) ℤ[x]/(2, x) ≅ ℤ₂, a field. (3) A generator f would divide 2 and x, forcing f = ±1, but 1 ∉ (2, x). (4) ℤ[x]/(x² + 1) ≅ ℤ[i], a domain but not a field (2 has no inverse), so the ideal is prime and not maximal.
Which of the following rings are unique factorisation domains?
1. ℤ[x]
2. ℤ[√−5]
3. ℚ[x, y]
4. ℤ[i]
Show answer
Answer: A — ℤ[x]; C — ℚ[x, y]; D — ℤ[i]
(1) and (3) by Gauss: ℤ and ℚ[x] are UFDs, hence so are ℤ[x] and ℚ[x][y]. (4) ℤ[i] is Euclidean under N(a + bi) = a² + b², hence a PID and a UFD. (2) fails: 6 = 2 · 3 = (1 + √−5)(1 − √−5) with all four factors irreducible (norms 4, 9, 6, 6, and no element has norm 2 or 3) and no two associate.
Which of the following rings is a principal ideal domain?
1. ℤ[i]
2. ℤ[x]
3. ℚ[x, y]
4. ℤ[√−5]
Show answer
Answer: A — ℤ[i]
ℤ[i] has a division algorithm with respect to the norm, so it is Euclidean and hence a PID. In ℤ[x] the ideal (2, x) and in ℚ[x, y] the ideal (x, y) are not principal; ℤ[√−5] is not even a UFD, and every PID is a UFD.
Which of the following polynomials are irreducible over ℚ?
1. x⁵ + 6x³ + 9x + 3
2. x⁴ + 4
3. x³ − 2
4. x⁴ + 1
Show answer
Answer: A — x⁵ + 6x³ + 9x + 3; C — x³ − 2; D — x⁴ + 1
(1) Eisenstein at p = 3: 3 | 6, 9, 3 (and the zero coefficients), 3 ∤ 1, 9 ∤ 3. (3) Eisenstein at 2, or: a cubic with no rational root (±1, ±2 fail). (4) Eisenstein at 2 after x ↦ x + 1. (2) is reducible: x⁴ + 4 = (x² + 2x + 2)(x² − 2x + 2), although it has no rational root.
The degree of the field extension ℚ(√2, √3) over ℚ is ____.
Numerical answer — type the value.
Show answer
Answer: 4
[ℚ(√2) : ℚ] = 2. √3 ∉ ℚ(√2): if √3 = a + b√2 with a, b rational, squaring forces ab = 0 and then 3 = a² or 3 = 2b², impossible. So x² − 3 stays irreducible over ℚ(√2) and the tower law gives 2 × 2 = 4, with basis 1, √2, √3, √6.
The degree over ℚ of the splitting field of x³ − 2 is ____.
Numerical answer — type the value.
Show answer
Answer: 6
The roots are ∛2, ∛2 ω, ∛2 ω², so the splitting field is ℚ(∛2, ω). [ℚ(∛2) : ℚ] = 3 by Eisenstein, and ω, a root of x² + x + 1, is not real, so it is not in ℚ(∛2) ⊂ ℝ: another factor of 2. Total 3 × 2 = 6.
The number of elements of the field F₁₆ that generate its multiplicative group is ____.
Numerical answer — type the value.
Show answer
Answer: 8
F₁₆^× is cyclic of order 15, and a cyclic group of order 15 has φ(15) = φ(3)φ(5) = 8 generators. Counting all non-zero elements (15) or all elements (16) ignores that most have smaller order.
Which statements about finite fields are true?
1. there is a field with 12 elements
2. F₁₆ contains a subfield with 4 elements
3. F₁₆ contains a subfield with 8 elements
4. the multiplicative group of F₂₅ is cyclic
Show answer
Answer: B — F₁₆ contains a subfield with 4 elements; D — the multiplicative group of F₂₅ is cyclic
(1) The order of a finite field is a prime power, and 12 is not. (2) F_2ᵐ ⊆ F_2⁴ iff m | 4; m = 2 works. (3) m = 3 does not divide 4 (also [F₁₆ : F₈] would be 4/3). (4) The multiplicative group of every finite field is cyclic, here of order 24.
The number of monic irreducible polynomials of degree 2 over the field ℤ₅ is ____.
Numerical answer — type the value.
Show answer
Answer: 10
There are 5² = 25 monic quadratics x² + bx + c. The reducible ones are (x − a)(x − b): 5 with a = b and C(5, 2) = 10 with a ≠ b, 15 in all. So 25 − 15 = 10 are irreducible, matching (p² − p)/2.
The quotient ring ℤ₂[x]/(x³ + x + 1) is
1. a field with 8 elements
2. a field with 6 elements
3. a ring with zero divisors and 8 elements
4. isomorphic to ℤ₈
Show answer
Answer: A — a field with 8 elements
x³ + x + 1 has no root in ℤ₂ (it takes the value 1 at 0 and at 1), so as a cubic it is irreducible and the ideal is maximal: the quotient is a field. Its elements are the 2³ = 8 remainders a + bx + cx². It has characteristic 2, so it is not ℤ₈.
Which statements are true?
1. ℂ is algebraically closed
2. the field of real algebraic numbers is algebraically closed
3. every finite extension of a field is algebraic
4. no finite field is algebraically closed
Show answer
Answer: A — ℂ is algebraically closed; C — every finite extension of a field is algebraic; D — no finite field is algebraically closed
(1) is the fundamental theorem of algebra. (2) is false: x² + 1 has no real root. (3) If [K : F] = n, the n + 1 powers 1, α, …, αⁿ are dependent, giving a polynomial with root α. (4) In F_q every a satisfies a^q = a, so x^q − x + 1 takes the value 1 everywhere and has no root.