For any $a = [a_0; \dots; a_n]\in \mathbb{P}^n(\mathbb{Q})$, the corresponding Galois group $G_a$ of $f(X) = a_n X^n + \cdots + a_1 X + a_0\in \mathbb{Q}[X]$ is a subgroup of $S_n$. For a given group $G \subset S_n$, what exactly can be said about the size of the set of points $a$ with $G_a = G$? For $G = S_n$, the answer from here is that $G = S_n$ iff the resolvent of $f$ is irreducible, and this happens for $a$ in a Zariski-dense set by Hilbert's irreducibility theorem. Generalizing from that, is there a specific sense in which $A(X) = \{a\in \mathbb{P}^n(\mathbb{Q}):\, G_a\subset X\}$ has $A(H)\subset A(G)$ "small" in $A(G)$ for $H < G$? That is, is there a result invoking some notion like the dimension of a variety (although we're necessarily working over $\mathbb{Q}$ here), a thin set in the sense of Serre, etc. that makes this vague idea of "smallness" more precise?
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I'm far from an expert, but this seems related to the inverse galois problem. This has been extensively studied, so I'm sure there are some partial results that might interest you. It's also famously open, so I'm not sure if you'll be able to get a fully satisfactory answer. – HallaSurvivor Jul 02 '21 at 18:54
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1@HallaSurvivor: The idea of rigid sets was the original impetus for this question, but I'm looking for something more quantitative or specific than just the nontriviality of the sets $A(G)$. – anomaly Jul 02 '21 at 22:13