In ℤ[√−5], the number 6 factors as both 2·3 and (1+√−5)(1−√−5), yet the class number of ℤ[√−5] is 2, not 4. What does the class number actually count?
AThe number of distinct element factorizations of every integer in the ring
BThe number of prime elements that remain irreducible in the ring
CThe number of equivalence classes of fractional ideals modulo the principal ideals
DThe number of generators needed to describe the ring as a module
The class number h(K) is the order of the ideal class group — the set of fractional ideals modulo the subgroup of principal ideals. Two ideals are equivalent when their ratio is principal. In ℤ[√−5], h = 2 means there are exactly two such classes: the class of principal ideals, and one non-principal class. It does not directly count element factorizations or prime elements.
Question 2 Multiple Choice
A number ring has class number h(K) = 1. Which property does this guarantee?
AEvery element of the ring can be factored into at least two irreducibles
BEvery ideal of the ring of integers is generated by a single element (is principal)
CThere are no non-trivial ideals in the ring
DThe ring contains only finitely many prime elements
Class number 1 means the ideal class group is trivial — every fractional ideal is equivalent to the trivial class, which means every ideal is principal. In a Dedekind domain (which every ring of integers is), being a principal ideal domain is equivalent to having unique factorization of elements. So h(K) = 1 is both necessary and sufficient for elements to factor uniquely.
Question 3 True / False
In ℤ[√−5], the factorizations 2·3 and (1+√−5)(1−√−5) represent genuinely different factorizations even at the level of ideals.
TTrue
FFalse
Answer: False
This is the key insight of ideal theory. Both element factorizations decompose into the same four prime ideal factors: (2, 1+√−5), (2, 1−√−5), (3, 1+√−5), and (3, 1−√−5). The apparent ambiguity at the element level vanishes at the ideal level — unique factorization into prime ideals holds in every ring of integers, even when element factorization fails. The class group captures this: it measures how the ideal factorization can fail to lift to element factorization.
Question 4 True / False
The ideal class group of any number field is always a finite group.
TTrue
FFalse
Answer: True
Finiteness of the class group is a fundamental theorem in algebraic number theory, proved using the Minkowski bound — an explicit upper bound on which prime ideals need to be checked to determine the class group. This bound depends only on the degree and discriminant of the number field, guaranteeing that the search terminates. Finiteness means the class number h(K) is a well-defined positive integer and a key invariant of the field.
Question 5 Short Answer
Why do algebraic number theorists work with ideals rather than elements when studying factorization in rings like ℤ[√−5]?
Think about your answer, then reveal below.
Model answer: Elements in rings like ℤ[√−5] can fail to factor uniquely, as 6 = 2·3 = (1+√−5)(1−√−5) demonstrates. But ideals in any ring of integers always factor uniquely into prime ideals — this is guaranteed by the Dedekind domain structure. The ideal class group quantifies exactly how far elements fall short: class number 1 means every ideal is principal (generated by a single element), so ideal factorizations lift to element factorizations and unique factorization holds. Larger class numbers measure how many non-principal ideal classes exist, capturing the precise failure of unique factorization at the element level.
The historical motivation was Kummer's attempt on Fermat's Last Theorem, which failed when he assumed unique factorization in cyclotomic rings. Correcting this error led him to develop ideal theory. The ideal class group is the invariant that makes precise what Kummer's assumption missed.