4
$\begingroup$

This question is motivated by a computational issue. Suppose $R$ is a product of orders in numberfields such that there is no ring homomorphism $R \to \mathbb Z$, then can one write an algorithm that actually proves that there is no ring homomorphism $R \to \mathbb Z$ if one is only able to compute $R/pR$ for primes $p$. Sidenote: in this question one should think of $R$ as terribly complicated such that it is not possible to directly work with it in characteristic 0. For so far the motivation now to the mathematics :)

If for a prime $p$ there there is no map $R/p \to \mathbb F_p$ then there is also no map $R \to \mathbb Z$. So restricting for simplicity to the case where $R$ is an order instead of a product of orders one then naturally arrives at the question in the title.

Does there exist an order $R\neq \mathbb Z$ that has a map to $\mathbb F_p$ for all primes p.

Now I already have done a bit of thinking about this question but have not been able to solve it yet. However using galois theory one can reduce it to a question of finite groups.

For simplicity let's assume $R=O_K$ is a maximal order in a numberfield $K$ and also ignore the primes in $\mathbb Z$ above which ramification occurs. Now let $L$ be the galois closure of $K$, $G=Gal(L/\mathbb Q)$ and $H=Gal(L/K)$. Let $q \mid pO_K$ be a prime, $q' \subset O_L$ a prime lying above $q$ and let $D_{q'} \subset G$ (this is step is where I ignore the ramified primes for simplicity) be the decomposition group of $q'$, then $$O_K/q \cong \mathbb F_p \Leftrightarrow f(q)=1 \Leftrightarrow D_{q'} \subset H.$$ So one sees that there is a map $O_K \to \mathbb F_p$ if and only if $H \cap \text{Frob}_p \neq \emptyset$ where $\text{Frob}_p \subset G$ denotes the conjugacy class of frobenius. By Chebotarevs density theorem all conjugacy classes will occur as $\text{Frob}_p$ for a positive density of the primes. So this leads to the following group theoretic question:

Let $H \subset G$ be two finite groups with $H \cap C \neq \emptyset$ for all conjugacy classes $C$ of $G$ then does this imply that $G=H$?

The above at least feels like the answer might be yes, because $H$ should be at least very big if it is to contain an element of every conjugacy class. Using a small computer search I proved that the answer is yes if $G \subset S(9)$. So an order as in my first question has to have at least rank 10 over $\mathbb Z$.

$\endgroup$
5
  • 7
    $\begingroup$ Yes, a proper subgroup of a finite group $G$ cannot intersect every conjugacy class. This has appeared here at least once before: mathoverflow.net/questions/26979 $\endgroup$ Jun 14, 2013 at 1:55
  • 3
    $\begingroup$ That every conjugacy class is a Frobenius conjugacy class for a positive density of primes is generally attributed to Chebotarev rather than Dirichlet. $\endgroup$
    – KConrad
    Jun 14, 2013 at 1:57
  • 1
    $\begingroup$ Related topic: if a monic irred. polynomial in ${\mathbf Z}[x]$ has a root mod $p$ for all but finitely many primes $p$ then the poly. has degree 1. This is proved by the "$H = G$" conjugation result you ask about. Its connection to your question is that if $\alpha$ is a root of that irred. polynomial and $R = {\mathbf Z}[\alpha]$, then finding a ring homomorphism $R \rightarrow {\mathbf F}_p$ amounts to finding a root of the polynomial mod $p$. If your question had "all but finitely many $p$" then the order would be in $\mathbf Q$, and ${\mathbf Z}[1/N]$ maps to ${\mathbf F}_p$ if $(p,N)=1$. $\endgroup$
    – KConrad
    Jun 14, 2013 at 2:06
  • $\begingroup$ Thanks Noam, it seems that when looking for duplicates of this question I looked to much at the number theoretic side of things. $\endgroup$ Jun 14, 2013 at 2:16
  • $\begingroup$ @KConrad: This fact can also be proved (Chudnovsky) by showing that the power series $y=(1+x)^\alpha$, solution of the differential equation $y'=\alpha y/(1+x)$ is an algebraic function. Hence $\alpha$ is rational. $\endgroup$
    – ACL
    Jun 14, 2013 at 4:38

1 Answer 1

1
$\begingroup$

Since the answer is already in the comments I'm just putting it here as community wiki.

As Noam Elkies noted:

Yes, a proper subgroup of a finite group G cannot intersect every conjugacy class. This has appeared here at least once before: mathoverflow.net/questions/2697

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.