Galois groups at closed points from Galois group at generic point? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T00:51:43Z http://mathoverflow.net/feeds/question/80574 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80574/galois-groups-at-closed-points-from-galois-group-at-generic-point Galois groups at closed points from Galois group at generic point? Sean Howe 2011-11-10T09:39:20Z 2011-11-10T10:10:22Z <p>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 <em>some</em> of them - see the motivation below)</p> <p>If the answer is positive, is there a generalization to arbitrary finite morphisms of curves or of varieties?</p> <p>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 <em>some</em> 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).</p> <p>Any answers, references, or comments on either the problem posed or the motivation given would be highly appreciated! Thanks. </p> http://mathoverflow.net/questions/80574/galois-groups-at-closed-points-from-galois-group-at-generic-point/80576#80576 Answer by ulrich for Galois groups at closed points from Galois group at generic point? ulrich 2011-11-10T10:10:22Z 2011-11-10T10:10:22Z <p>What you need is the Hilbert irreducibility <a href="http://en.wikipedia.org/wiki/Hilbert%27s_irreducibility_theorem" rel="nofollow">theorem</a>. </p> <p>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 $G$.</p> <p>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.</p> <p>Another way of stating Hilbert's irreducibility is to say that $\mathbb{A}^n(K)$ is not thin for $K$ any number field.</p> <p>A nice reference is "Lectures on the Mordell-Weil theorem" by Serre. </p>