In ℤ[√−5], we have 6 = 2·3 = (1+√−5)(1−√−5). What does this tell us about the ideal class group of ℤ[√−5]?
AThe class group is trivial (h = 1), because 6 has a factorization
BThe class group is nontrivial (h > 1), because element factorization is not unique
CThe class group is nontrivial (h > 1), because ideal factorization also fails in this ring
DNothing — the class group is defined for fields, not rings
The non-unique factorization of 6 as an element signals that ℤ[√−5] is not a unique factorization domain, hence not a principal ideal domain, hence h > 1. The class group for ℤ[√−5] is ℤ/2ℤ, so h = 2. Note that option C is wrong: ideal factorization into prime ideals is always unique — that is precisely what ideals restore. The class group measures the failure at the element level via the structure of ideals, not a failure of ideal factorization.
Question 2 Multiple Choice
A number ring has class number h(K) = 1. Which conclusion follows?
AEvery nonzero element factors uniquely into irreducibles
BThe ring has no prime ideals
CEvery ideal is principal, so the ring is a PID with unique factorization
DThe ring contains no zero divisors
h(K) = 1 means every fractional ideal is principal — the class group is trivial. For rings of algebraic integers, being a PID is equivalent to being a UFD, so h = 1 is the precise condition for unique factorization to hold. Option A sounds correct but is the consequence, not the definition: the PID property is the direct interpretation of h = 1, and unique factorization follows from it.
Question 3 True / False
In a ring of algebraic integers with h(K) > 1, most element fails to factor uniquely into irreducibles.
TTrue
FFalse
Answer: False
This is the most common misconception about the ideal class group. When h > 1, the ring is not a UFD, meaning unique factorization fails for *some* elements — but many elements still factor uniquely. The class group quantifies the global obstruction to unique factorization; it is not a statement that every element is problematic. In ℤ[√−5] with h = 2, for example, the element 5 = (√−5)² factors uniquely, while 6 = 2·3 = (1+√−5)(1−√−5) does not.
Question 4 True / False
If two ideals I and J satisfy I = (α)J for some element α, then I and J represent the same element of the ideal class group.
TTrue
FFalse
Answer: True
The ideal class group is defined by declaring two fractional ideals equivalent when they differ by multiplication by a principal ideal (α). This equivalence relation defines the classes: I ~ J iff I = (α)J for some nonzero α. The group operation is ideal multiplication, and the identity element is the class of principal ideals. The fact that principal ideals form the trivial class is exactly why h = 1 means 'everything is principal.'
Question 5 Short Answer
Why do ideals restore unique factorization in rings where elements do not factor uniquely, and what does the class group measure about this restoration?
Think about your answer, then reveal below.
Model answer: In a Dedekind domain (which rings of algebraic integers are), every nonzero ideal factors uniquely into prime ideals — even when elements do not factor uniquely. The obstruction to element-level unique factorization is that some ideals are not principal: a 'missing' element factorization corresponds to a factorization of ideals that cannot be expressed as products of principal ideals. The class group — fractional ideals modulo principal ideals — measures exactly how many 'non-principal' equivalence classes exist. h(K) = 1 means every ideal is principal, collapsing ideal factorization and element factorization into the same thing.
The key insight is the two-level structure: ideals always factor uniquely (Dedekind domain property), but elements factor uniquely only when every ideal is principal. The class group is the gap between these two levels. A class number of 2 means there is one non-trivial equivalence class of ideals that lacks a principal representative — and this is what causes element factorization to fail in specific cases like 6 in ℤ[√−5].