Math-4129 · mathematics · Graph theory

Math-4129

Question 4.17 — Mazur

Status Partially solved · unsolvedmath.comLLM status Open unverified

Top Open Problems Not reported cataloguedPartially solvedOpen

Statement

Definitions and mathematical statement

Shared notation.
$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

Let $V$ be a smooth algebraic variety over $\mathbb Q$. Give $V(\mathbb R)$ its usual real topology and set $Z=\overline{V(\mathbb Q)}^{\,V(\mathbb R)}$. The condition that $V(\mathbb Q)$ is Zariski dense means it is not contained in a proper Zariski-closed subset of $V$.

Statement recorded in the supplied source.
If $V(\mathbb Q)$ is Zariski dense in $V$, must $Z$ be a union of connected components of $V(\mathbb R)$? [S2] The cited survey records a negative answer to this general question: Colliot-Thélène, Skorobogatov and Swinnerton-Dyer constructed a smooth surface with Zariski-dense rational points whose real closure is not a union of components. The survey subsequently distinguishes the separate abelian-variety case. [S2]

Short English statement

If rational points are algebraically dense, must their real closure consist of whole real connected components?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 0

No attempts yet. Significant partial results and substantial reductions are welcome. Contribute an attempt.