# Discriminants

Suppose are a basis for a number field , which we view as a -vector space. Let be the embedding , where are the distinct embeddings of into . Let be the matrix whose rows are . The quantity depends on the ordering of the , and need not be an integer.

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 .

Lemma 10.2.1   The trace pairing is nondegenerate.

Proof. If the trace pairing is degenerate, then there exists such that for every we have . In particularly, taking we see that , which is absurd. Definition 10.2.2 (Discriminant)   Suppose is any -basis of . The of is The of an order in is the discriminant of any basis for . The of the number field is the discrimimant of .

Note that the discriminants defined above are all nonzero by Lemma 10.2.1.

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.

Proposition 10.2.3   Suppose is an order in . Then Proof. Let be a matrix whose rows are the images via of a basis for , and let be a matrix whose rows are the images via of a basis for . Since has finite index, there is an integer matrix such that , and . Then  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.

Example 10.2.4   Consider the ring of integers of . The discriminant of the basis is Let be the order generated by . Then has basis , so William Stein 2004-05-06