Editing Ideal class group (section)

== Examples of ideal class groups ==

* The rings <math>\Z, \Z[i]</math>, and <math>\Z[\omega]</math>, respectively the [[Integer#Algebraic properties|integers]], [[Gaussian integers]], and [[Eisenstein integers]], are all principal ideal domains (and in fact are all [[Euclidean domain]]s), and so have class number 1: i.e., they have trivial ideal class groups.
 
*If <math>K</math> is a field, then the [[polynomial ring]] <math>K[x_1,x_2,x_3,\dots]</math> is an integral domain. It has a countably infinite set of ideal classes.

=== Class numbers of quadratic fields ===
If <math>d</math> is a [[square-free integer]] (a product of distinct primes) other than 1, then <math>\Q(\sqrt{d})</math> is a [[quadratic field|quadratic extension]] of <math>\Q</math>. If <math>d < 0</math>, then the class number of the ring <math>R</math> of algebraic integers of <math>\Q(\sqrt{d})</math> is equal to 1 for precisely the following values of <math>d</math>: <math>d = -1,-2,-3,-7,-11, -19, -43, -67, -163</math>. This result was first [[conjecture]]d by [[Carl Friedrich Gauss|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, <math>d > 0</math>, then it is unknown whether there are infinitely many fields <math>\Q(\sqrt{d})</math> 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 field]]s with class number 1.<ref>{{harvnb|Neukirch|1999}}</ref>

For <math>d < 0</math>, the ideal class group of <math>\Q(\sqrt{d})</math> is isomorphic to the class group of integral [[binary quadratic form]]s of [[discriminant of a quadratic form|discriminant]] equal to the discriminant of <math>\Q(\sqrt{d})</math>. For <math>d > 0</math>, 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 <math>\Q(\sqrt{d})</math>.<ref>{{harvnb|Fröhlich|Taylor|1993|loc=Theorem 58}}</ref>

For [[real number|real]] [[quadratic integer]] rings, the class number is given in [https://oeis.org/A003649 OEIS A003649]; for the [[imaginary number|imaginary]] case, they are given in [https://oeis.org/A000924 OEIS A000924].

==== Example of a non-trivial class group ====

The quadratic integer ring <math>R=\Z[\sqrt{-5}]</math> is the ring of integers of <math>\Q(\sqrt{-5})</math>. It does not possess unique factorization; in fact the class group of <math>R</math> is [[cyclic group|cyclic]] of order 2. Indeed, the ideal
: <math>J=(2,1+\sqrt{-5})</math>
is not principal, which can be [[proof by contradiction|proved by contradiction]] as follows: <math>R</math> has a [[multiplicative function|multiplicative]] [[field norm|norm]] function defined by <math>N(a + b \sqrt{-5}) =  a^2 + 5 b^2 </math>, which satisfies <math>N(u) = 1</math> if and only if <math>u</math> is a unit in <math>R</math>. 

Firstly, <math> J \neq R</math>, because the [[quotient ring]] of <math>R</math> modulo the ideal <math>(1 + \sqrt{-5})</math> is [[ring isomorphism|isomorphic]] to <math>\Z / 6\Z</math>, so that the quotient ring of <math>R</math> modulo <math>J</math> is isomorphic to <math>\Z / 3\Z</math>.  Now if <math>J=(a)</math> were principal (that is, generated by an element <math>a</math> of <math>R</math>), then <math>a</math> would divide both <math>2</math> and <math>1+\sqrt{-5}</math>. Then the norm <math>N(a)</math> would divide both <math>N(2) = 4</math> and <math>N(1 + \sqrt{-5}) = 6</math>, so <math>N(a)</math> would divide 2. If <math>N(a) = 1</math> then <math>a</math> is a unit and so <math>J = R</math>, a contradiction. But <math>N(a)</math> cannot be 2 either, because <math>R</math> has no elements of norm 2, because the [[Diophantine equation]] <math>b^2 + 5 c^2 = 2</math> has no solutions in integers, as it has no solutions [[modular arithmetic|modulo 5]].

One also computes that <math>J^2=(2)</math>, which is principal, so the class of <math>J</math> in the ideal class group has order two. Showing that there aren't any other ideal classes requires more effort. <!-- well, I can't think of any method apart from the minkowski bound off the top of my head -->

The fact that this <math>J</math> is not principal is also related to the fact that the element <math>6</math> has two distinct factorisations into [[irreducible element|irreducibles]]:
:<math>6 = 2\times 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})</math>.