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

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

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

$\displaystyle \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()

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);
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

$\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\det(C)\vert= [\O_K:\O]$. Then

$\displaystyle \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

$\displaystyle D(K) = \gcd\{\Disc (\alpha) : \alpha \in \O_K$    and $\displaystyle [\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

$\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,

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

$\displaystyle A = \left(
\begin{matrix}3&0&0 0&0&6 0&6&0


$\displaystyle \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

$\displaystyle \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.

William Stein 2012-09-24