Discriminants

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
.

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); 20This 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.

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 * 503For 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

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.

William Stein 2012-09-24