A PAC counterexample and the projectivity boundary

Definitions and mathematical statement

The AIM question asks whether the absolute Galois group of $K$ must be topologically generated by one element if every $\mathcal O$-acyclic variety over $K$ has a rational point. The extracted $\operatorname{Gal}(K/K)$ is interpreted as $G_K=\operatorname{Gal}(K^{\mathrm{sep}}/K)$. Procyclic means having a dense cyclic subgroup.

The supplied report explicitly uses the standard smooth projective geometrically connected test class with \[ H^0(X,\mathcal O_X)=K,\qquad H^i(X,\mathcal O_X)=0\quad(i>0), \] compatibly after algebraic closure. The original excerpt does not define $\mathcal O$-acyclic or its variety class. Omitting the degree-zero/geometric condition would change the question, since finite extension spectra have no higher coherent cohomology but need not have $K$-points.

Under the stated test convention the report gives a negative answer: there exists a characteristic-zero pseudo-algebraically closed field $K$ with \[ G_K\cong\widehat F_2, \] the free profinite group on two generators. A PAC field has a rational point on every nonempty geometrically integral variety, hence on every variety in this smooth projective connected test class. Yet $\widehat F_2$ surjects onto $C_2\times C_2$ and is therefore not procyclic. The existence/realization theorem for this field is invoked without a full citation or explicit field construction.

The report also claims a necessary boundary. If $K$ has the stated $\mathcal O$-acyclic point property, every finite separable extension $L/K$ has it: use Weil restriction, whose geometric base change is a product of conjugate varieties, preserving the cohomology condition, and whose $K$-points are the original $L$-points. Applying this to Severi--Brauer varieties then forces \[ \operatorname{Br}(L)=0\quad\text{for every finite separable }L/K. \] For perfect $K$, the report concludes that $G_K$ is projective as a profinite group, equivalently that its finite embedding problems admit weak solutions. This necessary projectivity condition does not bound the number of topological generators. The report does not prove a converse characterizing all fields with the $\mathcal O$-acyclic point property, and exact Weil-restriction/cohomology and Galois criteria require verification.

Short English statement

Ask whether rational points on every suitably defined $\mathcal O$-acyclic variety force a procyclic absolute Galois group. The report uses a PAC field with free profinite group of rank two to give a negative answer. It separately derives vanishing Brauer groups at finite separable levels and, for perfect fields, projectivity rather than a generator bound.

Sources

• Local registry: data/unsolvedmath/problems.json, permanent ID $20000462$, source code catalogue entry; original question and supplied research summary.
• AIM workshop "Rational and integral points on higher dimensional varieties", Question 29: https://aimath.org/WWN/qptsurface2/qptsurface2.pdf. Canonical location aim-algebraic-number-theory-notes.json, notes[113]; no original remarks or literature citation. No PDF was opened here.
• The summary invokes a classical PAC-field realization with free profinite group of rank two and a Brauer/projectivity criterion without supplying authoritative bibliographic references or an explicit construction.
• Supplied machine-generated research attempt 1: counterexample; status refuted at the run, registry solved and classification SOLVED-IN-LITERATURE; rigor metadata "proved", novelty confidence low. The non-peer-reviewed synthesis requires independent verification before citation.