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

$\displaystyle {\mathrm{Det}}(A)^2$ $\displaystyle = {\mathrm{Det}}(AA) = {\mathrm{Det}}(A A^t)$    
  $\displaystyle = {\mathrm{Det}}\left(\sum_{k=1,\ldots,n} \sigma_k(w_i)\sigma_k(w_j)\right)$    
  $\displaystyle = {\mathrm{Det}}(\Tr (w_i w_j)_{1\leq i,j\leq n}),$    

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

$\displaystyle \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

$\displaystyle \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

$\displaystyle \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

$\displaystyle \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

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

William Stein 2004-05-06