Regular Inverse Galois Problem over Q

Regular Inverse Galois Problem over Q

Definitions and mathematical statement

A finite extension $F/{\mathbb{Q}}(T)$ is regular over ${\mathbb{Q}}$ when $F\cap\overline{{\mathbb{Q}}}={\mathbb{Q}}$ inside an algebraic closure, so there are no new algebraic constants. The question is \[ \forall G\text{ finite}\ \exists F/{\mathbb{Q}}(T)\text{ finite Galois and regular}:\quad \operatorname{Gal}(F/{\mathbb{Q}}(T))\simeq G. \] Geometrically, this asks for a geometrically integral Galois cover of ${\mathbb{P}}^1_{{\mathbb{Q}}}$ with group $G$. A realization only over ${\mathbb{Q}}$ or with a nontrivial constant-field extension does not meet the regular target.

Formulation and concept references: [S1].

Short English statement

Can every finite group be realized by a regular Galois cover of the projective line defined over the rationals?

Research attempt

Current frontier.

The formulation source supplies new regular realizations for certain $\operatorname{PSL}_2({\mathbb{F}}_{p^2})$ groups subject to arithmetic conditions; it also distinguishes regular realizations from realizations only over ${\mathbb{Q}}$. Its introduction and main results, checked in the December 2025 version, do not give all finite groups. [S1] The constructions below are independent elementary deductions illustrating what an all-groups argument must preserve: geometric connectedness and no new constants. A complex cover, a number-field extension, or a cover of a higher-genus base is not silently substituted.

Attack routes.

A branch-cycle approach can start with topological covers over ${\mathbb{C}}$, but needs a rational moduli point and valid descent to ${\mathbb{Q}}$. A generic permutation action can produce every finite group over an invariant function field, but must reduce that base to ${\mathbb{Q}}(T)$. A constructive closure approach combines already realizable groups. We prove the closure operations carefully, then examine the first operation that fails to yield a universal construction.

Attempt I: a fully checked regular example.

Consider the splitting field $F$ of \[ f(X)=X^3-X-T\in{\mathbb{Q}}(T)[X]. \] Over any characteristic-zero field $k$, the inclusion $k(T)\subset k(x)$ with $T=x^3-x$ has degree three: the rational map ${\mathbb{P}}^1_x\to{\mathbb{P}}^1_T$ has degree three, or equivalently the pole order of $T$ at infinity is three. Thus $f$ is irreducible over both ${\mathbb{Q}}(T)$ and $\overline{\mathbb{Q}}(T)$. Its discriminant is \[ \operatorname{disc}_X(f)=4-27T^2. \] This has two simple zeros over $\overline{\mathbb{Q}}$, so is not a square in $\overline{\mathbb{Q}}(T)$. An irreducible cubic has transitive Galois group in $S_3$; if that group were $A_3$, its alternating product of root differences would be fixed, making the discriminant a square. Conversely that product detects the sign character. Hence the geometric and arithmetic groups are both $S_3$.

Let $k_0=F\cap\overline{\mathbb{Q}}$. The geometric group is the kernel of the arithmetic action on $k_0$ and here already has order six, the order of the full arithmetic group. Thus $k_0={\mathbb{Q}}$. This proves a regular $S_3$ realization with the catalogue's exact constant-field requirement. The discriminant identity is also verified symbolically by the companion script; no finite specialization is used to infer regularity.

Attempt II: closure under quotients and products.

If $F/{\mathbb{Q}}(T)$ is regular Galois with group $G$ and $N\triangleleft G$, then $F^N/{\mathbb{Q}}(T)$ is Galois with group $G/N$, and $F^N\cap\overline{\mathbb{Q}}\subseteq F\cap\overline{\mathbb{Q}}={\mathbb{Q}}$. Thus regular realizability is closed under quotients.

It is also closed under finite direct products. To see this, take two regular Galois covers with groups $G_1,G_2$. Their finite branch sets in ${\mathbb{P}}^1(\overline{\mathbb{Q}})$ can be made disjoint by applying a rational Möbius transformation to one base. Indeed, for each pair of branch points the condition that the transformation send one to the other is a proper algebraic condition on $\operatorname{PGL}_2$. Finitely many such conditions do not exhaust its rational points: on a rational affine chart their product is a nonzero polynomial, and ${\mathbb{Q}}$ is infinite. This also handles branch points at infinity by using suitable charts.

After base extension to $\overline{\mathbb{Q}}(T)$, the intersection of the two Galois function fields is Galois and can ramify only in the intersection of their branch sets. It is therefore unramified everywhere. The Riemann--Hurwitz formula for a connected unramified degree-$d$ cover of ${\mathbb{P}}^1$ gives $2g-2=-2d$, impossible for $d>1$. The intersection is trivial. The two fields are consequently linearly disjoint and their geometric compositum has group $G_1\times G_2$.

The arithmetic compositum embeds into the same product, whereas its geometric subgroup already has the full product order. Equality follows, with constant field ${\mathbb{Q}}$. Iteration proves finite-product closure. For example, every quotient of a finite product of copies of the displayed $S_3$ extension has a regular realization. This is an actual family, but not a family known to contain every finite group.

Attempt III: why the evident subgroup construction stops.

Every finite group embeds in a symmetric group. Suppose we have a regular $S_m$ extension $F/{\mathbb{Q}}(T)$ and a subgroup $H\leq S_m$. The extension $F/F^H$ has group $H$, but its base is $F^H$, not generally ${\mathbb{Q}}(T)$. Its corresponding curve need not have genus zero. Regularity of $F/{\mathbb{Q}}$ ensures that the base has no extra algebraic constants; it says nothing about that curve being rational. The operation therefore does not prove closure under arbitrary subgroups in the required one-variable sense. The preceding quotient argument works precisely because the base stays fixed.

A universal version of the same issue begins with independent variables $\{x_g:g\in G\}$. Let $G$ act by left translation on their labels. The action on $L={\mathbb{Q}}(x_g:g\in G)$ is faithful, and the elementary fixed field theorem for finite automorphism groups yields \[ L/L^G\text{ Galois with group }G. \] Both fields have constant field ${\mathbb{Q}}$: an element of a purely transcendental field algebraic over ${\mathbb{Q}}$ belongs to ${\mathbb{Q}}$. However $\operatorname{trdeg}_{{\mathbb{Q}}}L^G=|G|$, so this is not a realization over ${\mathbb{Q}}(T)$. Setting many variables to rational functions of one parameter can identify branches or reduce the Galois group.

Here is an exact sufficient geometric target. Restrict the linear permutation representation to its open set $U$ of points with trivial stabilizer. The quotient map $U\to Y=U/G$ is a finite étale $G$-torsor in characteristic zero. It would suffice to find, for each $G$, a nonempty open subset $V\subset{\mathbb{P}}^1_{{\mathbb{Q}}}$ and a morphism $V\to Y$ whose pulled-back torsor is geometrically connected. The function field of that pullback is then regular Galois over ${\mathbb{Q}}(T)$ with group $G$: connectedness after base change makes the geometric monodromy transitive on a regular $G$-fiber, hence equal to $G$. This is a sufficient rational-curve-and-monodromy condition, not a claim that every regular realization must arise from this particular model.

Rational points of $Y$ alone are insufficient. Their fibers give finite extensions over ${\mathbb{Q}}$, not one geometrically connected cover varying over $V$. Likewise a rational curve in $Y$ whose image lies in a locus of smaller monodromy does not suffice. These distinctions are exactly where a naive specialization argument runs out of justification.

Stress test.

The product argument used disjoint branch sets geometrically, not merely disjoint sets of rational branch points. It used characteristic zero to exclude nontrivial everywhere-unramified covers of the geometric projective line; it did not claim an analogous argument for arbitrary higher-genus bases. The $S_3$ example checks geometric irreducibility as well as the arithmetic discriminant, preventing an unnoticed constant extension.

The missing rational-curve condition is substantially stronger than having many complex covers or a rational point on some moduli space. No descent obstruction has been proved to vanish for every $G$, and no group has been shown to have an unavoidable obstruction for every possible cover. In particular, failure of this chosen invariant-field strategy would not disprove the original conjecture. A positive solution would also imply the ordinary inverse Galois problem over ${\mathbb{Q}}$ through Hilbert irreducibility, as the formulation source explains; this consequence underscores the depth of the unproved step. [S1]

Outcome.

Outcome: Exploratory approach with unresolved gap.

The notebook proves a regular $S_3$ example and the quotient/product closure operations, then isolates a sufficient geometrically connected pullback condition for a universal construction. The direct subgroup and generic-invariant-field shortcuts fail to preserve the required base. No all-groups construction or counterexample has been obtained; the elementary lemmas are not claimed to be new to the literature.

Sources

• S1 Demeio and Gvirtz-Chen, Galois Realisations of PSL2(Fp^2). https://arxiv.org/html/2512.23615v1. Catalogue formulation source.