Editing Ideal class group (section)

==== 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>.