Consider the finite map $\mathbb{A}^1_\mathbb{Q}\rightarrow \mathbb{A}^1_\mathbb{Q}$ given by $z\mapsto z^5-z$. The fiber over generic point is the field extension $\mathbb{Q}(t)[z]/(z^5-z-t)$ over $\mathbb{Q(t)}$ whose normalization has Galois group equal to the full symmetric group $ S_5 $ (can prove this by tensoring with $\mathbb{C}$ and calculating the monodromy action - but how to obtain this is unimportant for the question). What I want to know is, can we use this information about the generic point to deduce anything about the analogous Galois group at closed points whose fiber is a field extension? (I don't expect a statement at all such points, just something about some of them - see the motivation below)

If the answer is positive, is there a generalization to arbitrary finite morphisms of curves or of varieties?

I am thinking about this problem after spending some time trying to ferret out the exact difference between what is given by two different proofs of the insolubility of the quintic - the first by demonstrating that the Galois group of $z^5 - z - 1$ (or any other specific quintic over $\mathbb{Q}$) is the full $S_5$, the second by demonstrating that the "generic" quintic, i.e. the quintic with indeterminate coefficients has Galois group the full $S_5$ over the field obtained by adjoining its symmetric polynomials to $\mathbb{Q}$. In particular, I am interested to know if you can deduce that this fact must hold for some polynomial over $\mathbb{Q}$ from the fact that it holds for the "generic" polynomial (it is less interesting to me that it holds for independent complex transcendentals over $\mathbb{Q}$, even though this is enough to deduce that there is no "quintic formula" for quintics with complex coefficients).

Any answers, references, or comments on either the problem posed or the motivation given would be highly appreciated! Thanks.


1 Answer 1


What you need is the Hilbert irreducibility theorem.

This implies that the Galois group for "most" rational points (i.e. outside a thin subset) is the full symmetric group. More generally one can consider dominant generically finite Galois morphisms $f:X \to Y$ over a number field $K$, where $Y$ is a rational variety (this is important) and if the generic Galois group is $G$ then the Galois group outside a thin subset of $Y(K)$ will also be $G$.

A thin set is defined to be any subset of a finite union of sets of two types: 1) $Z(K)$ where $Z$ is a proper subvariety of $Y$ and 2) $f(X(K))$ for $f:X \to Y$ a quasifinite morphism over $K$ with $X$ irreducible and $f$ not having a rational section.

Another way of stating Hilbert's irreducibility is to say that $\mathbb{A}^n(K)$ is not thin for $K$ any number field.

A nice reference is "Lectures on the Mordell-Weil theorem" by Serre.


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.