Fix a number field $k$ and some $\alpha \in \mathcal{O}_k \setminus \mathbb{Z}$. Let $p$ be a rational prime which splits completely in $k$, so that $p \mathcal{O}_k = P_1 \cdots P_m$ for $m = [k : \mathbb{Q}]$ and some prime ideal $P_1, \ldots, P_m$ of $\mathcal{O}_k$. Now it is easy to see that for any $i=1,\ldots,m$ there exists an unique integer $0 \leq \alpha_i < p$ such that $\alpha \equiv \alpha_i$ (mod $P_i$). For no reason we can say that $\alpha_1, \ldots, \alpha_m$ are all distinct integers, indeed this is not certainly the case if $p < m$.

My question is: Is it true that for "most" of the primes $p$ which split completely in $k$ we have that $\alpha_1, \ldots, \alpha_m$ are all distinct?

Here by "most", I mean, for example, a set of relatively positive density, or even density $1$... I let to you give a precise meaning.

My guess is that the answer should be YES. Indeed, assuming that each $\alpha_i$ is a random non-negative integer less than $p$, the probability that $\alpha_1, \ldots, \alpha_m$ are all distinct is $$\left(1 - \frac1{p - 1}\right)\left(1 - \frac1{p - 2}\right)\cdots\left(1 - \frac1{p - m + 1}\right) \to 1 ,$$ as $p \to +\infty$. Moreover, in the special case $m = 2$, we have that $\alpha_1 = \alpha_2$ means that $\alpha \in \alpha_1 + p \mathcal{O}_k \subseteq \mathbb{Z} + p \mathcal{O}_k$ and it seems that the set $\mathbb{Z} + p \mathcal{O}_k$ is "small" for large $p$, so $\alpha$ should be not in it.

Thanks for any idea/reference.


1 Answer 1


Let $f\in\mathbb{Z}[X]$ be the characteristic polynomial of $\alpha$. If the reduction of $f$ mod $p$ has any repeated factors in $\mathbb{F}_p[X]$, then $p$ divides the discriminant of $f$. This is true regardless whether all the factors are linear or not.

Therefore the $\alpha_i$ will be distinct for all but finitely many of the completely split $p$ when $\alpha$ generates $k$ over the rationals (in which case its characteristic polynomial is also its minimal polynomial).

But no $p$ at all will work when $\alpha$ lies in a proper subfield between $k$ and $\mathbb{Q}$, since $f$ is then a proper power of the minimal polynomial.

  • $\begingroup$ Please can you expand a bit? In particular the last sentence: Do you mean that in such a case $\alpha_1 = \cdots = \alpha_m$? $\endgroup$
    – sercej
    Oct 26, 2016 at 13:52
  • $\begingroup$ They won't necessarily all be equal, but each value will necessarily occur repeatedly, with multiplicity at least the degree of $k$ over $\mathbb{Q}(\alpha)$. $\endgroup$
    – GNiklasch
    Oct 26, 2016 at 14:35
  • $\begingroup$ I still don't get why this should happened. What is the link between the $\alpha_i$'s and the characteristic polynomial? $\endgroup$
    – sercej
    Oct 26, 2016 at 14:56
  • $\begingroup$ The $\alpha_i$ are the roots of the reduction mod $p$ of the characteristic polynomial. $\endgroup$
    – GNiklasch
    Oct 26, 2016 at 15:10
  • $\begingroup$ OK, thanks. Do you have a reference for this fact? $\endgroup$
    – sercej
    Oct 26, 2016 at 22:30

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