Inessential Discriminant Divisors

Definition 4.3.1   A prime $p$ is an inessential discriminant divisor if $p\mid [\O_K : \mathbf{Z}[a]]$ for every $a\in\O_K$.

See Example 6.2.6 below for why it is called an inessential ``discriminant divisor'' instead of an inessential ``index divisor''.

Since $[\O_K : \mathbf{Z}[a]]^2$ is the absolute value of $\Disc (f(x))/\Disc (\O_K)$, where $f(x)$ is the characteristic polynomial of $f(x)$, an inessential discriminant divisor divides the discriminant of the characteristic polynomial of any element of $\O_K$.

Example 4.3.2 (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 SAGE to show that $2$ is an inessential discriminant divisor for $K$.

sage: K.<a> = NumberField(x^3 + x^2 - 2*x + 8); K
Number Field in a with defining polynomial x^3 + x^2 - 2*x + 8
sage: K.factor_integer(2)
(Fractional ideal (1/2*a^2 - 1/2*a + 1)) * 
(Fractional ideal (a^2 - 2*a + 3)) * 
(Fractional ideal (3/2*a^2 - 5/2*a + 4))

Thus $2\O_K=\mathfrak{p}_1\mathfrak{p}_2\mathfrak{p}_3$, with the $\mathfrak{p}_i$ distinct, and one sees directly from the above expressions that $\O_K/\mathfrak{p}_i\cong \mathbf{F}_2$ for each $i$. If $\O_K=\mathbf{Z}[a]$ for some $a\in\O_K$ with minimal polynomial $f$, then $\overline{f}(x)\in\mathbf{F}_2[x]$ must be a product of three distinct linear factors, which is impossible, since the only linear polynomials in $\mathbf{F}_2[x]$ are $x$ and $x+1$.