Let $K$ be an algebraic number field, $K=\Bbb{Q}(\alpha)$. Let $f \in \mathbb Z[X]$ be the minimal polynomial of $\alpha$.

If $p$ does not divide $\operatorname{disc}(f)$, then $f$ mod $p$ is a separable polynomial in $(\Bbb{Z}/p\Bbb{Z})[X]$. Moreover the roots of $f \bmod Q$ are distinct, where $Q$ is a prime ideal of $\mathcal O_K$ (the ring of integers of $K$) above $p$.

Are these assertions true?


The answer to both questions is yes. To see this, denote by $\overline{\phantom{x}}$ the natural map $\mathbf Z[X] \to (\mathbf Z / p \mathbf Z)[X]$. As the discriminant can be expressed as a polynomial in the coefficients of $f$ and $\overline{\phantom{x}}$ is a morphism of rings, we have $\overline{\operatorname{disc}(f)} = \operatorname{disc}(\overline f)$.
Thus if $p$ does not divide the discriminant of $f$, then $\operatorname{disc}(\overline f) \neq 0$. But the discriminant is also given by $$ \operatorname{disc}(\overline f) = c \cdot \prod_{i \neq j} (\alpha_i - \alpha_j), $$ where $c \in \mathbf Z / p\mathbf Z$ and $\alpha_1,\dotsc,\alpha_r$ are the roots of $\overline f$ in some splitting field. Since the discriminant $\operatorname{disc}(\overline f)$ is nonzero, we conclude that the $\alpha_i$ have to be distinct. Thus $f$ is separable.

If $f \bmod Q$ would have a multiple root, then $\overline f \in (\mathbf Z/ p \mathbf Z)[X]$ would have a multiple root in a splitting field, contradicting the separability.

  • $\begingroup$ Is the discriminant of f same as the discriminant of $K$? $\endgroup$ – MathStudent Jan 4 '15 at 14:05
  • 1
    $\begingroup$ @dulalnarugopal No, in general this is not true. This happens if and only if $\mathbf Z[\alpha] = \mathcal O_K$. You can see this already for quadratic fields. Consider for example $\alpha = \sqrt{-3}$. $\endgroup$ – Hans Giebenrath Jan 4 '15 at 14:09

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