The scandal: 6 factors two ways in Z[√-5]
Z is a unique factorization domain: every integer factors into primes in essentially one way. We would love O_K to be a UFD too. It usually isn't. The standard cautionary example is O_K = Z[√-5] (here -5 ≡ 3 mod 4, so this is the full ring of integers). The number 6 has two factorizations into irreducibles.
In O_K = Z[sqrt(-5)], norm N(a+b sqrt(-5)) = a^2 + 5 b^2.
N is multiplicative; an element is a unit iff N = 1, so units = {+1,-1}.
6 = 2 * 3 = (1 + sqrt(-5)) * (1 - sqrt(-5)).
Norms:
N(2) = 4
N(3) = 9
N(1 + sqrt(-5)) = 1 + 5 = 6
N(1 - sqrt(-5)) = 6
Are these irreducible? A proper factor would have norm 2 or 3.
But a^2 + 5 b^2 = 2 and = 3 have NO integer solutions.
So 2, 3, 1±sqrt(-5) are all irreducible, and none is an associate
of another (their norms differ / units are only ±1).
==> 6 = 2*3 = (1+sqrt(-5))(1-sqrt(-5)) are TWO distinct
factorizations into irreducibles. UFD FAILS.Dedekind's fix: factor ideals, not elements
Kummer's insight, made rigorous by Dedekind: stop factoring elements and factor ideals instead. The element 2 splits no further, but the ideal (2) does — into a prime ideal squared. Write p = (2, 1+√-5). One checks p² = (2), and similarly the ideals over 3 give q = (3, 1+√-5), q' = (3, 1-√-5) with (3) = qq'. Then both factorizations of 6 refine to the *same* ideal factorization.
Set p = (2, 1+sqrt(-5))
q = (3, 1+sqrt(-5))
q' = (3, 1-sqrt(-5))
Claim: (2) = p^2, (3) = q q'.
Ideal factorizations of the two sides of 6 = 2*3 = (1+s)(1-s):
(6) = (2)(3) = p^2 q q'
(6) = (1+s)(1-s) = (p q)(p q')
because (1+sqrt(-5)) = p q, (1-sqrt(-5)) = p q'.
Both give (6) = p^2 q q' -- the SAME prime-ideal factorization.
Uniqueness is restored: the ambiguity was only in how the prime
ideals clumped together into principal ideals (elements).Why it always works: O_K is a Dedekind domain
The rescue is not a lucky trick for Z[√-5]; it is structural. A Dedekind domain is an integral domain that is (1) Noetherian, (2) integrally closed in its fraction field, and (3) of dimension one — every nonzero prime ideal is maximal. The ring of integers O_K of any number field satisfies all three, so O_K is always Dedekind.
The payoff theorem: in a Dedekind domain, every nonzero ideal factors uniquely as a product of prime ideals. That is unique factorization of ideals — the central structural result of the subject. Locally, each prime gives a discrete valuation ring, so factorization is governed by a clean order-of-vanishing at each prime. The next guide turns the *gap* between ideals and elements into a finite invariant: the class group.