Discriminants

Suppose $w_1,\ldots,w_n$ are a basis for $\O_K$ as a $\mathbf {Z}$-module, which we view as a $\mathbf {Q}$-vector space. Let $\sigma:K\hookrightarrow \mathbf{C}^n$ be the embedding $\sigma(a)=(\sigma_1(a),\ldots,\sigma_n(a))$, where $\sigma_1,\ldots, \sigma_n$ are the distinct embeddings of $K$ into  $\mathbf{C}$. Let $A$ be the matrix whose rows are $\sigma(w_1), \ldots, \sigma(w_n)$. The quantity $\det(A)$ depends on the ordering of the $w_i$, and need not be an integer.

If we consider $\det(A)^2$ instead, we obtain a number that does not depend on ordering; moreover, as we will see, it is an integer. Note that

$$\det(A)^2 = \det(AA) = \det(A)\det(A)=\det(A)\det(A^t)= \det(A A^t)$$

$$= \det\left(\sum_{k=1,\ldots,n} \sigma_k(w_i)\sigma_k(w_j)\right) = \det\left(\sum_{k=1,\ldots,n} \sigma_k(w_i w_j)\right)$$

$$= \det(\Tr (w_i w_j)_{1\leq i,j\leq n}),$$

so $\det(A)^2$ can be defined purely in terms of the trace without mentioning the embeddings $\sigma_i$. Also, changing our choice of basis for $\O_K$ is the same as left multiplying $A$ by an integer matrix $U$ of determinant $\pm 1$; this does not change the squared determinant, since $\det(UA)^2 = \det(U)^2\det(A)^2 = \det(A)^2$. Thus $\det(A)^2\in \mathbf{Z}$ is well defined as a quantity associated to $\O_K$.

If we view $K$ as a $\mathbf {Q}$-vector space, then $(x,y)\mapsto \Tr (xy)$ defines a bilinear pairing $K\times K \to \mathbf{Q}$ on $K$, which we call the trace pairing. The following lemma asserts that this pairing is nondegenerate, so $\det(\Tr (w_i w_j))\neq 0$ hence $\det(A)\neq 0$.

Lemma 6.2.1   The trace pairing is nondegenerate.

Proof. If the trace pairing is degenerate, then there exists $a\in K$ such that for every $b\in K$ we have $\Tr (ab)=0$. In particularly, taking $b=a^{-1}$ we see that $0=\Tr (a a^{-1})=\Tr (1)=[K:\mathbf{Q}]>0$, which is absurd. $\qedsymbol$

Definition 6.2.2 (Discriminant)   Suppose $a_1,\ldots, a_n$ is any $\mathbf {Q}$-basis of $K$. The discriminant of $a_1,\ldots, a_n$ is

$$\Disc (a_1,\ldots,a_n) = \det(\Tr (a_i a_j)_{1\leq i,j\leq n})\in\mathbf{Q}.$$

The discriminant $\Disc (\O)$ of an order $\O$ in $\O_K$ is the discriminant of any basis for $\O$. The discriminant $d_K=\Disc (K)$ of the number field $K$ is the discriminant of $\O_K$. 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 $\alpha\in \O_K$ with $\mathbf{Z}[\alpha]$ of finite index in $\O_K$, and $f$ is the minimal polynomial of $\alpha$, then $\Disc (f)=\Disc (\mathbf{Z}[\alpha])$. To see this, note that if we choose the basis $1,\alpha,\ldots,\alpha^{n-1}$ for $\mathbf{Z}[\alpha]$, 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 $\Disc (K)$ is defined to be the discriminant of the polynomial you happened to use to define $K$.

> 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 $\O$ in $\O_K$ is bigger than $\disc (\O_K)$ by a factor of the square of the index.

Proposition 6.2.5   Suppose $\O$ is an order in $\O_K$. Then

$$\Disc (\O) = \Disc (\O_K)\cdot [\O_K:\O]^2.$$

Proof. Let $A$ be a matrix whose rows are the images via $\sigma$ of a basis for $\O_K$, and let $B$ be a matrix whose rows are the images via $\sigma$ of a basis for $\O$. Since $\O\subset \O_K$ has finite index, there is an integer matrix $C$ such that $CA=B$, and $\vert\det(C)\vert= [\O_K:\O]$. Then

$$\Disc (\O) = \det(B)^2 = \det(CA)^2 = \det(C)^2\det(A)^2 = [\O_K:\O]^2 \cdot \Disc (\O_K).$$

$\qedsymbol$

Example 6.2.6   Let $K$ be a number field and consider the quantity

$$D(K) = \gcd\{\Disc (\alpha) : \alpha \in \O_K$$    and $$[\O_K:\mathbf{Z}{}[\alpha]] < \infty\}.$$

One might hope that $D(K)$ is equal to the discriminant $\Disc (\O_K)$ of $K$, but this is not the case in general. Recall Example 4.3.2, in which we considered the field $K$ generated by a root of $f = x^3 + x^2 - 2x+8$. In that example, the discriminant of $\O_K$ is $-503$ with $503$ prime:

sage: K.<a> = NumberField(x^3 + x^2 - 2*x + 8)
sage: factor(K.discriminant())
-1 * 503

For every $\alpha\in \O_K$, we have $2\mid [\O_K:\mathbf{Z}[\alpha]]$, since $\O_K$ fails to be monogenic at $2$. By Proposition 6.2.5, the discriminant of $\mathbf{Z}[\alpha]$ is divisible by $4$ for all $\alpha$, so $\Disc (\alpha)$ is also divisible by $4$. This is why $2$ is called an ``inessential discriminant divisor''.

Proposition 6.2.5 gives an algorithm for computing $\O_K$, albeit a slow one. Given $K$, find some order $\O\subset K$, and compute $d=\Disc (\O)$. Factor $d$, and use the factorization to write $d=s\cdot f^2$, where $f^2$ is the largest square that divides $d$. Then the index of $\O$ in $\O_K$ is a divisor of $f$, and we (tediously) can enumerate all rings $R$ with $\O\subset R\subset K$ and $[R:\O] \mid f$, 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 $p$-maximal order for each prime divisor of $f$, then add these $p$-maximal orders together.

Example 6.2.7   Consider the ring $\O_K = \mathbf{Z}[(1+\sqrt{5})/2]$ of integers of $K=\mathbf{Q}(\sqrt{5})$. The discriminant of the basis $1,a=(1+\sqrt{5})/2$ is

$$\Disc (\O_K) = \left\vert \left( \begin{matrix}2&1 1&3 \end{matrix}\right) \right\vert = 5.$$

Let $\O=\mathbf{Z}[\sqrt{5}]$ be the order generated by $\sqrt{5}$. Then $\O$ has basis $1,\sqrt{5}$, so

$$\Disc (\O) = \left\vert \left( \begin{matrix}2&0 0&10 \end{matrix}\right) \right\vert = 20 = [\O_K:\O]^2\cdot 5,$$

hence  $[\O_K:\O] = 2$.

Example 6.2.8   Consider the cubic field $K=\mathbf{Q}(\sqrt[3]{2})$, and let $\O$ be the order $\mathbf{Z}[\sqrt[3]{2}]$. Relative to the base $1,\sqrt[3]{2}, (\sqrt[3]{2})^2$ for $\O$, the matrix of the trace pairing is

$$A = \left( \begin{matrix}3&0&0 0&0&6 0&6&0 \end{matrix}\right).$$

Thus

$$\disc (\O) = \det(A)= 108 = 2^2\cdot 3^3.$$

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

$$\Disc (\O_K)\cdot [\O_K:\O]^2 = 2^2\cdot 3^3,$$

so $3\mid \disc (\O_K)$, and $[\O_K:\O] \mid 6$. Thus to prove $\O=\O_K$ it suffices to prove that $\O$ is $2$-maximal and $3$-maximal, which could be accomplished as described in Section 4.3.3.