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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 $\mathbb Z$.

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Yes, a proper subgroup of a finite group $G$ cannot intersect every conjugacy class. This has appeared here at least once before: – Noam D. Elkies Jun 14 '13 at 1:55
That every conjugacy class is a Frobenius conjugacy class for a positive density of primes is generally attributed to Chebotarev rather than Dirichlet. – KConrad Jun 14 '13 at 1:57
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 $(p,N)=1$. – KConrad Jun 14 '13 at 2:06
Thanks Noam, it seems that when looking for duplicates of this question I looked to much at the number theoretic side of things. – Maarten Derickx Jun 14 '13 at 2:16
@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. – ACL Jun 14 '13 at 4:38
up vote 1 down vote accepted

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:

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