Discriminants

Suppose are a basis for as a -module, 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 does not depend on ordering; moreover, as we will see, it is an integer. Note that

so can be defined purely in terms of the trace without mentioning the embeddings . Also, changing our choice of basis for is the same as left multiplying  by an integer matrix of determinant ; this does not change the squared determinant, since . Thus is well defined as a quantity associated to .

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

Lemma 6.2.1   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 6.2.2 (Discriminant)   Suppose is any -basis of . The discriminant of is

The discriminant of an order in is the discriminant of any basis for . The discriminant of the number field  is the discriminant of . Note that these discriminants are all nonzero by Lemma 6.2.1.

Remark 6.2.3   It is also standard to define the discriminant of a monic polynomial to be the product of the differences of the roots. If with of finite index in , and  is the minimal polynomial of , then . To see this, note that if we choose the basis for , then both discriminants are the square of the same Vandermonde determinant.

Example 6.2.4   In SAGE, we compute the discriminant of a number field or order using the discriminant command:
sage: K.<a> = NumberField(x^2 - 5)
sage: K.discriminant()
5


This also works for orders (notice the square factor below, which will be explained by Proposition 6.2.5):

sage: R = K.order([7*a]); R
Order in Number Field in a with defining polynomial x^2 - 5
sage: factor(R.discriminant())
2^2 * 5 * 7^2


Warning: In is defined to be the discriminant of the polynomial you happened to use to define .

> K := NumberField(x^2-5);
> Discriminant(K);
20

This is an intentional choice done for efficiency reasons, since computing the maximal order can take a long time. Nonetheless, it conflicts with standard mathematical usage, so beware.

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

Proposition 6.2.5   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

Example 6.2.6   Let be a number field and consider the quantity

and

One might hope that is equal to the discriminant of , but this is not the case in general. Recall Example 4.3.2, in which we considered the field generated by a root of . In that example, the discriminant of is with prime:
sage: K.<a> = NumberField(x^3 + x^2 - 2*x + 8)
sage: factor(K.discriminant())
-1 * 503

For every , we have , since fails to be monogenic at . By Proposition 6.2.5, the discriminant of is divisible by  for all , so is also divisible by . This is why is called an inessential discriminant divisor''.

Proposition 6.2.5 gives an algorithm for computing , albeit a 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. A much better algorithm is to proceed exactly as just described, except use the ideas of Section 4.3.3 to find a -maximal order for each prime divisor of , then add these -maximal orders together.

Example 6.2.7   Consider the ring of integers of . The discriminant of the basis is

Let be the order generated by . Then  has basis , so

hence  .

Example 6.2.8   Consider the cubic field , and let be the order . Relative to the base for , the matrix of the trace pairing is

Thus

Suppose we do not know that the ring of integers is equal to . By Proposition 6.2.5, we have

so , and . Thus to prove it suffices to prove that  is -maximal and -maximal, which could be accomplished as described in Section 4.3.3.

William Stein 2012-09-24