Essential Discriminant Divisors

Definition 8.1.5   A prime $ p$ is an if $ p\mid [\O _K : \mathbf{Z}[a]]$ for every $ a\in\O _K$.

Since $ [\O _K : \mathbf{Z}[a]]$ is the absolute value of $ \Disc (f(x))/\Disc (\O _K)$, where $ f(x)$ is the characteristic polynomial of $ f(x)$, an essential discriminant divisor divides the discriminant of the characteristic polynomial of any element of $ \O _K$.

Example 8.1.6 (Dedekind)   Let $ K=\mathbf{Q}(a)$ be the cubic field defined by a root $ a$ of the polynomial $ f = x^3 + x^2 - 2x+8$. We will use , which implements the algorithm described in the previous section, to show that $ 2$ is an essential discriminant divisor for $ K$.
   > K<a> := NumberField(x^3 + x^2 - 2*x + 8);
   > OK := MaximalOrder(K);
   > Factorization(2*OK);
   [
   <Prime Ideal of OK
   Basis:
   [2 0 0]
   [0 1 0]
   [0 0 1], 1>,
   <Prime Ideal of OK
   Basis:
   [1 0 1]
   [0 1 0]
   [0 0 2], 1>,
   <Prime Ideal of OK
   Basis:
   [1 0 1]
   [0 1 1]
   [0 0 2], 1>
   ]
Thus $ 2\O _K=\mathfrak{p}_1\mathfrak{p}_2\mathfrak{p}_3$, with the $ \mathfrak{p}_i$ distinct. Moreover, one can check that $ \O _K/\mathfrak{p}_i\cong \mathbf{F}_2$. If $ \O _K=\mathbf{Z}[a]$ for some $ a\in\O _K$ with minimal polynomial $ g$, then $ \overline{g}(x)\in\mathbf{F}_2[x]$ must be a product of three distinct linear factors, which is impossible.



William Stein 2004-05-06