Questions: Failure of Unique Factorization in Algebraic Number Fields

Question 1

In ℤ[√−5], the element 2 divides (1 + √−5)(1 − √−5) = 6, but 2 divides neither (1 + √−5) nor (1 − √−5). What does this demonstrate about 2 in ℤ[√−5]?

• A. 2 is a prime element in ℤ[√−5], since it divides a product
• B. 2 is a unit in ℤ[√−5], since it divides every element
• C. 2 is irreducible but not prime in ℤ[√−5], and this is why unique factorization fails
• D. 2 is neither irreducible nor prime in ℤ[√−5], because it has the factorization 2 = (1 + √−5)(1 − √−5)/3

Answer: (C)

An element p is prime if p | ab implies p | a or p | b. Here 2 | (1 + √−5)(1 − √−5) but 2 divides neither factor, so 2 is not prime. Yet 2 is irreducible: the norm N(2) = 4, and there are no elements of norm 2 in ℤ[√−5] (a² + 5b² = 2 has no integer solutions), so 2 cannot be factored into non-units. In ℤ (and any PID/UFD), irreducible implies prime — the divergence of these concepts is precisely what breaks unique factorization in ℤ[√−5].

Question 2

Kummer and Dedekind's resolution to the failure of unique factorization in rings like ℤ[√−5] was to:

• A. Prove that the two factorizations 2 × 3 and (1 + √−5)(1 − √−5) are actually associate, by finding a unit relating them
• B. Restrict to a subring of ℤ[√−5] where unique factorization holds
• C. Factor ideals rather than elements, recovering unique factorization at the level of ideal products
• D. Add new 'ideal numbers' to ℤ[√−5] to make it into a PID

Answer: (C)

The key insight was to shift the object of factorization from elements to ideals. Even though the element 2 is irreducible-but-not-prime, the ideal (2) factors as 𝔭² for a prime ideal 𝔭 = (2, 1 + √−5). When you express both factorizations of 6 as products of prime ideals, they agree — uniqueness is restored at the ideal level. This is not about restricting the ring (option B) or adjoining new elements (option D); it is about recognizing that ideals, not elements, are the right objects for unique factorization in general algebraic number rings.

Question 3

In any integral domain, every prime element is irreducible.

• T. True
• F. False

Answer: True

This implication always holds. Suppose p is prime and p = ab. Then p | ab, so p | a or p | b. Say p | a, so a = pc. Then p = pcb, giving 1 = cb (since integral domains have cancellation), so b is a unit. This shows every factorization of a prime p has one factor being a unit — exactly the definition of irreducible. The converse (irreducible implies prime) fails in rings without unique factorization, like ℤ[√−5]. In UFDs, the converse holds, which is what makes them well-behaved.

Question 4

The two factorizations 6 = 2 × 3 and 6 = (1 + √−5)(1 − √−5) in ℤ[√−5] are considered equivalent because the elements are related by units of the ring.

• T. True
• F. False

Answer: False

This is false, and verifying it is essential to understanding the failure. Two factorizations are equivalent (associate) if corresponding factors differ only by units (±1 in ℤ[√−5]). The elements 2, 3, (1 + √−5), and (1 − √−5) are genuinely non-associate: e.g., 2 × (±1) gives only ±2, which is neither (1 + √−5) nor (1 − √−5). The factorizations are genuinely distinct — not a notational variant — which is what makes ℤ[√−5] fail to be a UFD.

Question 5

What does the ideal class group of ℤ[√−5] measure, and how does its order of 2 encode the failure of unique factorization in that ring?

Model answer: The ideal class group measures the obstruction to every ideal being principal (generated by a single element). If the class group is trivial, every ideal is principal, and the ring is a UFD. The class group of ℤ[√−5] has order 2, meaning there are two distinct ideal classes: one containing all principal ideals, and one containing 'non-principal' ideals like 𝔭 = (2, 1 + √−5). The order 2 corresponds directly to the two-fold ambiguity in element factorization — the fact that 6 has exactly two distinct irreducible factorizations.

The class group is the precise algebraic invariant measuring how far a ring is from being a UFD. Order 1 = UFD. Order 2 means exactly the kind of 'two-way' ambiguity seen in ℤ[√−5]. This is not a coincidence: each element factorization corresponds to a product of prime ideals, and the non-uniqueness at the element level arises exactly when a prime ideal is non-principal (so the ideals in its class cannot be 'named' by a single element). Kummer's original insight — later made precise by Dedekind — was that ideal factorization is always unique even when element factorization fails, and the class group measures the gap.