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In On $l$-independence of Algebraic Monodromy Groups in Compatible Systems of Representations, they stated such an example (Example 6.3):

Let $X$ be a connected normal scheme of finite type over $\text{Spec}(\mathbb{Z})$ and of dimension $\geq 1$. Let $K$ be the function field of $X$ and $\mathcal{G}=\text{Gal}(K^s/K)$. For every closed point $x$, let $K_x$ be the function field of the henselization of $X$ in $x$, and $K_{\bar{x}}$ that of the strict hensalization. We have canonical embeddings $K\hookrightarrow K_x\hookrightarrow K_{\bar{x}}$. Every extension $K^s\hookrightarrow K_x^s=K_{\bar{x}}^s$ of the embedding induces a homomorphism $j: \text{Gal}(K_x^s/K_x)\rightarrow \text{Gal}(K^s/K)$. Let $A$ be the set of all triples $\alpha=(x, j, F)$, where $F$ lies in the $j$-image of any representatives of Frobenius in $\text{Gal}(K_{\bar{x}}/K_x)$. By the density of Frobenius conjugacy classes, we have an $F$-group.

I could understand everything except the last sentence: what is the theory about the density of Frobenius conjugacy class? Is there a Chebotarev density theorem for a function field of higher transcendental degree? I did not find this on Internet, and appreciate every help or explanation! Thanks.

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Yes, the density theorems hold for arbitrary dimensional schemes X over a finite field and follow from basic point count estimates (Lang-Weil or RH). The idea is simple:

Suppose we have a galois cover f: Y to X with group G and we care about $\mathbb F_q$ points x on the base whose fiber is totally split. I claim that the number of such points is precisely $|Y(\mathbb F_q)|/|G|$ because x is totally split if and only if it is the image of a rational point along f and in this case there are |G| many such preimages. The density of such points is the 1/|G| because $|X(\mathbb F_q) |/|Y(\mathbb F_q)| \to 1$ as q grows.

If we want the fiber to induce a particular frobenius element g in G, we consider instead the twist of f by g and run the same argument through. Exercise for the reader.

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