Ideal class group

If ${\displaystyle d}$ is a square-free integer (a product of distinct primes) other than 1, then ${\displaystyle \mathbf {Q} ({\sqrt {d}}\,)}$ is a quadratic extension of Q. If ${\displaystyle d<0}$, then the class number of the ring ${\displaystyle R}$ of algebraic integers of ${\displaystyle \mathbf {Q} ({\sqrt {d}}\,)}$ is equal to 1 for precisely the following values of ${\displaystyle d}$: ${\displaystyle d=-1,-2,-3,-7,-11,-19,-43,-67,-163}$. 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 ${\displaystyle \mathbf {Q} ({\sqrt {d}}\,)}$ 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.[2]

For d < 0, the ideal class group of ${\displaystyle \mathbf {Q} ({\sqrt {d}}\,)}$ is isomorphic to the class group of integral binary quadratic forms of discriminant equal to the discriminant of ${\displaystyle \mathbf {Q} ({\sqrt {d}}\,)}$. 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 ${\displaystyle \mathbf {Q} ({\sqrt {d}}\,)}$.[3]

For real quadratic integer rings, the class number is given in OEIS A003649; for the imaginary case, they are given in OEIS A000924.

The quadratic integer ring R = Z[√−5] is the ring of integers of Q(√−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 + √−5)

is not principal, which can be proved by contradiction as follows. ${\displaystyle R}$ has a norm function ${\displaystyle N(a+b{\sqrt {-5}})=a^{2}+5b^{2}}$, which satisfies ${\displaystyle N(uv)=N(u)N(v)}$, and ${\displaystyle N(u)=1}$ if and only if ${\displaystyle u}$ is a unit in ${\displaystyle R}$. First of all, ${\displaystyle J\neq R}$, because the quotient ring of ${\displaystyle R}$ modulo the ideal ${\displaystyle (1+{\sqrt {-5}})}$ is isomorphic to ${\displaystyle \mathbf {Z} /6\mathbf {Z} }$, so that the quotient ring of ${\displaystyle R}$ modulo ${\displaystyle J}$ is isomorphic to ${\displaystyle \mathbf {Z} /2\mathbf {Z} }$. If J were generated by an element x of R, then x would divide both 2 and 1 + √−5. Then the norm ${\displaystyle N(x)}$ would divide both ${\displaystyle N(2)=4}$ and ${\displaystyle N(1+{\sqrt {-5}})=6}$, so N(x) would divide 2. If ${\displaystyle N(x)=1}$ then ${\displaystyle x}$ is a unit and ${\displaystyle J=R}$, a contradiction. But ${\displaystyle N(x)}$ cannot be 2 either, because R has no elements of norm 2, because the Diophantine equation ${\displaystyle b^{2}+5c^{2}=2}$ has no solutions in integers, as it has no solutions modulo 5.

One also computes that J² = (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 × 3 = (1 + √−5) × (1 − √−5).