• The rings \Z, \Z[i], and \Z[\omega], respectively the
integers,
Gaussian integers, and
Eisenstein integers, are all principal ideal domains (and in fact are all
Euclidean domains), and so have class number 1: i.e., they have trivial ideal class groups. • If K is a field, then the
polynomial ring K[x_1,x_2,x_3,\dots] is an integral domain. It has a countably infinite set of ideal classes.
Class numbers of quadratic fields If d is a
square-free integer (a product of distinct primes) other than 1, then \Q(\sqrt{d}) is a
quadratic extension of \Q. If d , then the class number of the ring R of algebraic integers of \Q(\sqrt{d}) is equal to 1 for precisely the following values of d: d = -1,-2,-3,-7,-11, -19, -43, -67, -163. This result was first
conjectured by
Gauss and proven by
Kurt Heegner, although Heegner's proof was not believed until
Harold Stark gave a later proof in 1967 (see
Stark–Heegner theorem). This is a special case of the famous
class number problem. If, on the other hand, d > 0, then it is unknown whether there are infinitely many fields \Q(\sqrt{d}) with class number 1. Computational results indicate that there are a great many such fields. However, it is not even known if there are infinitely many
number fields with class number 1. For d , the ideal class group of \Q(\sqrt{d}) is isomorphic to the class group of integral
binary quadratic forms of
discriminant equal to the discriminant of \Q(\sqrt{d}). For d > 0, the ideal class group may be half the size since the class group of integral binary quadratic forms is isomorphic to the
narrow class group of \Q(\sqrt{d}). For
real quadratic integer rings, the class number is given in OEIS A003649; for the
imaginary case, they are given in OEIS A000924.
Example of a non-trivial class group The quadratic integer ring R=\Z[\sqrt{-5}] is the ring of integers of \Q(\sqrt{-5}). It does not possess unique factorization; in fact the class group of R is
cyclic of order 2. Indeed, the ideal : J=(2,1+\sqrt{-5}) is not principal, which can be
proved by contradiction as follows: R has a
multiplicative norm function defined by N(a + b \sqrt{-5}) = a^2 + 5 b^2 , which satisfies N(u) = 1 if and only if u is a unit in R. Firstly, J \neq R, because the
quotient ring of R modulo the ideal (1 + \sqrt{-5}) is
isomorphic to \Z / 6\Z, so that the quotient ring of R modulo J is isomorphic to \Z / 3\Z. Now if J=(a) were principal (that is, generated by an element a of R), then a would divide both 2 and 1+\sqrt{-5}. Then the norm N(a) would divide both N(2) = 4 and N(1 + \sqrt{-5}) = 6, so N(a) would divide 2. If N(a) = 1 then a is a unit and so J = R, a contradiction. But N(a) cannot be 2 either, because R has no elements of norm 2, because the
Diophantine equation b^2 + 5 c^2 = 2 has no solutions in integers, as it has no solutions
modulo 5. One also computes that J^2=(2), which is principal, so the class of J in the ideal class group has order two. Showing that there aren't any other ideal classes requires more effort. The fact that this J is not principal is also related to the fact that the element 6 has two distinct factorisations into
irreducibles: :6 = 2\times 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}). == Connections to class field theory ==