Chebotarëv data over number fields

Define a Chebotarëv datum over a number field $K$ to be a finite group $G$ together with a map $\mathfrak{p}\mapsto\gamma_{\mathfrak{p}}$ from a cofinite set of primes of $K$ into the set of conjugacy classes of $G$ such that for every conjugacy class $c\subset G$, the proportion of $\mathfrak{p}$ with $\gamma_{\mathfrak{p}}=c$ is $\operatorname{Card}c/\operatorname{Card}G$.

Two Chebotarëv data $(G,\gamma)$, $(G',\gamma')$ over the same number field are said to be equivalent if there is an isomorphism $\varphi:G\to G'$ such that $\varphi(\gamma_{\mathfrak{p}})=\gamma'_{\mathfrak{p}}$ for almost all $\mathfrak{p}$. If so, we identify the two.

Every finite galoisian extension $L$ of $K$ gives rise to a Chebotarëv datum $(\operatorname{Gal}(L|K),\gamma_{L|K})$ (Chebotarëv's density theorem).

Moreover, if $L_1$ and $L_2$ are two finite galoisian extensions of $K$ for which the associated Chebotarëv data $(\operatorname{Gal}(L_i|K),\gamma_{L_i|K})$ are equivalent, then $L_1=L_2$ (see Lemma 1, p. 174, of Mazur's recent article).

Question. Does every Chebotarëv datum over a given number field $K$ arise from some finite galoisian extension $L$ of $K$ ?

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+1 for the perfect transliteration of Chebotarëv – Georges Elencwajg Jan 25 2011 at 12:53
I had flirted with the idea of writing Чеботарев...  – Chandan Singh Dalawat Jan 25 2011 at 13:21
@Chandan: If you are really careful about spellings, you should spell it Чеботарёв, not Чеботарев. (The same applies to Чебышёв = Chebyshëv, by the way.) – Dmitri Pavlov Jan 26 2011 at 4:18
@Dmitiri: Many thanks. My Cyrillic needs brushing up. – Chandan Singh Dalawat Jan 26 2011 at 4:56
Given David's answer, it seems natural to replace "for almost all $\mathfrak{p}$" by "for all $\mathfrak{p}$ outside a set of density 0" in the definition of equivalence of Chebotarëv data. Is the answer known in this case (at least conjecturally) ? – François Brunault Jan 26 2011 at 8:57

I think this is false for cardinality reasons. Take some exceedingly thin, but infinite, set of primes P. Then you can take any valid Chebotarev datum and change it arbitrarily at each prime in P, and it still remains a valid datum, since P is too thin to change the densities. If the group has at least two conjugacy classes, then there'll be uncountably many ways of changing it like this. Since the set of number fields is clearly countable, that's a contradiction.

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 David, I shouldn't have gone home for tea. – Chandan Singh Dalawat Jan 25 2011 at 12:08

In addition to David's answer, one can provide explicit counterexamples, at least in specific cases.

For simplicity, take $K=\mathbf{Q}$ and $G={\pm 1}$. Define the Chebotarëv datum $\gamma$ by $\gamma(p) = -1$ if $p \equiv 1 \pmod{4}$ and $\gamma(p)=1$ if $p \equiv 3 \pmod{4}$. If $\gamma$ comes from a quadratic field $L$ of discriminant $D$, then there exists a quadratic character $\chi : (\mathbf{Z}/D\mathbf{Z})^* \to {\pm 1}$ such that $\gamma(p) = \chi(p)$ for almost all primes $p$. We now have a contradiction by looking at primes $p \equiv 1 \pmod{4D}$ (there are infinitely many such primes).

In fact, by Dirichlet's theorem on primes in arithmetic progressions, the Chebotarëv datum $\gamma$ cannot be modified on a subset of natural density zero in order to come from a quadratic field.

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 Very nice indeed. – Chandan Singh Dalawat Jan 27 2011 at 9:21