Discriminants

If we consider instead, we obtain a number that is a well-defined integer which can be either positive or negative. Note that

so can be defined purely in terms of the trace without mentioning the embeddings . Also, changing the basis for is the same as left multiplying by an integer matrix of determinant , which does not change the squared determinant, since . Thus is well defined, and does not depend on the choice of basis.

If we view as a
-vector space, then
defines a bilinear pairing
on , which we call
the *. The following lemma asserts that this
pairing is nondegenerate, so
hence
.
*

Warning: In is defined to be the discriminant of the polynomial you happened to use to define , which is (in my opinion) a poor choice and goes against most of the literature.

The following proposition asserts that the discriminant of an order in is bigger than by a factor of the square of the index.

This result is enough to give an algorithm for computing , albeit a potentially slow one. Given , find some order , and compute . Factor , and use the factorization to write , where is the largest square that divides . Then the index of in is a divisor of , and we (tediously) can enumerate all rings with and , until we find the largest one all of whose elements are integral.

William Stein 2004-05-06