Discriminants

Suppose $w_1,\ldots,w_n$ are a basis for a number field $K$, 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 ${\mathrm{Det}}(A)$ depends on the ordering of the $w_i$, and need not be an integer.

If we consider ${\mathrm{Det}}(A)^2$ instead, we obtain a number that is a well-defined integer which can be either positive or negative. Note that

$${\mathrm{Det}}(A)^2 = {\mathrm{Det}}(AA) = {\mathrm{Det}}(A A^t)$$

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

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

so ${\mathrm{Det}}(A)^2$ can be defined purely in terms of the trace without mentioning the embeddings $\sigma_i$. Also, changing the basis for $\O _K$ is the same as left multiplying $A$ by an integer matrix $U$ of determinant $\pm 1$, which does not change the squared determinant, since ${\mathrm{Det}}(UA)^2 = {\mathrm{Det}}(U)^2{\mathrm{Det}}(A)^2 = {\mathrm{Det}}(A)^2$. Thus ${\mathrm{Det}}(A)^2$ is well defined, and does not depend on the choice of basis.

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 . The following lemma asserts that this pairing is nondegenerate, so ${\mathrm{Det}}(\Tr (w_i w_j))\neq 0$ hence ${\mathrm{Det}}(A)\neq 0$.

Lemma 10.2.1   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 10.2.2 (Discriminant)   Suppose $a_1,\ldots, a_n$ is any $\mathbf{Q}$-basis of $K$. The of $a_1,\ldots, a_n$ is

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

The $\Disc (\O )$ of an order $\O$ in $\O _K$ is the discriminant of any basis for $\O$. The $d_K=\Disc (K)$ of the number field $K$ is the discrimimant of $\O _K$.

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

Warning: In $\Disc (K)$ is defined to be the discriminant of the polynomial you happened to use to define $K$, 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 $\O$ in $\O _K$ is bigger than $\disc (\O _K)$ by a factor of the square of the index.

Proposition 10.2.3   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 $\vert{\mathrm{Det}}(C)\vert= [\O _K:\O ]$. Then

$$\Disc (\O ) = {\mathrm{Det}}(B)^2 = {\mathrm{Det}}(CA)^2 = {\mathrm{Det}}(C)^2{\mathrm{Det}}(A)^2 = [\O _K:\O ]^2 \cdot \Disc (\O _K).$$

$\qedsymbol$

This result is enough to give an algorithm for computing $\O _K$, albeit a potentially 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.

Example 10.2.4   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

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