gpt 6 astra pro attempt on Math-214

Brauer-Manin obstruction for number-field curves: Hasse-principle form

Definitions and mathematical statement

Let $C/k$ be a smooth projective geometrically connected curve of genus at least two over a number field. Its Brauer--Manin set is \[ C({\mathbb{A}}_k)^{\operatorname{Br}}= \{(P_v):\sum_v\operatorname{inv}_v(\alpha(P_v))=0 \text{ for every }\alpha\in H^2_{\mathrm{et}}(C,\mathbf G_m)\}. \] The local invariants take values in ${\mathbb{Q}}/{\mathbb{Z}}$ and the sum has only finitely many nonzero terms for fixed $\alpha,(P_v)$. The selected assertion is nonemptiness of this set implies $C(k)\ne\varnothing$. This is only the Hasse-principle form; density of rational points in the adelic set is not added. No rational divisor class of degree one or finiteness assumption on a Tate--Shafarevich group is permitted as an unstated premise.

Formulation and concept references: [S1].

Short English statement

For a number-field curve of genus at least two, do adelic points passing every Brauer–Manin test guarantee an actual rational point?

Research attempt

Current frontier.

Stoll identifies the full Brauer set of a curve with its finite abelian descent set. His Corollary 7.4, assuming the Jacobian's Tate--Shafarevich group has no nonzero divisible subgroup, both treats the absence of a degree-one divisor class and describes the Brauer set using the closure of the Mordell--Weil group. His main theorem also proves the assertion for curves admitting suitable nonconstant maps to rank-zero abelian varieties. [S1] Creutz's September 2026 revision proves an adelic Mordell--Lang statement over global function fields with a nonisotriviality hypothesis; its analogous number-field results retain additional conjectural input. It is not the unrestricted number-field theorem requested here.[E1]

Attack route.

The tempting compactness argument produces a profinite Mordell--Weil point, not an actual rational point. I try to repair it by coupling the finite-place sieve to an archimedean height budget. Two precise mechanisms are examined: bounded-height representatives of increasingly many congruences, and a product-formula inequality forcing a sufficiently small representative onto the curve. The second yields an unconditional separation lemma over all number fields; obtaining representatives below its threshold is the unresolved arithmetic step.

Attempt.

1. Do not assume the missing base point. Write $J=\operatorname{Jac}(C)$. Temporarily impose the explicit hypothesis \begin{equation}\label{wsref-0-33-30-30-30-36-37-39-31-3a-73-68-61} \operatorname{Sha}(k,J)_{\mathrm{div}}=0. \end{equation} Under (30006791:sha), Stoll's Corollary 7.4 says that $C({\mathbb{A}}_k)^{\operatorname{Br}}\ne\varnothing$ forces a $k$-rational divisor class of degree one. Such a class gives a closed embedding $i:C\hookrightarrow J$, and the same corollary gives \[ i\left(C({\mathbb{A}}_k)^{\operatorname{Br}}_\bullet\right) =i\left(C({\mathbb{A}}_k)_\bullet\right)\cap\overline{J(k)}. \] The subscript means passage to connected components at infinity. [S1] In particular, any fixed Brauer tuple can be approximated at any finite collection of finite places by elements of $J(k)$. None of this conclusion is being invoked without (30006791:sha). A rational divisor class is sufficient for the embedding; it has not been replaced by an assumed rational point on $C$.

2. A height-tightness lemma. Fix a projective embedding $J\hookrightarrow{\mathbb{P}}^r_k$ and its absolute logarithmic height $h$. Enumerate the good finite places as $v_1,v_2,\ldots$. Suppose there are $a_n\in J(k)$ whose reductions at $v_1,\ldots,v_n$ lie on the corresponding reduction of $i(C)$ and for which $h(a_n)\le H$, with one finite $H$ independent of $n$. Then $C(k)$ is nonempty.

To prove this, the number-field degree is fixed and Northcott finiteness makes $\{a_n:n\ge1\}$ finite. At least one value $a$ occurs for unbounded $n$, so it reduces onto $i(C)$ at every good finite place. If $a\notin i(C)$, choose a homogeneous polynomial $F$ over $k$ vanishing on $i(C)$ but with $F(a)\ne0$. Outside a finite set of places, coordinates for $a$ and coefficients for $F$ are integral and a coordinate of $a$ is a unit. A nonzero algebraic number $F(a)$ can vanish modulo only finitely many such prime ideals. This contradicts all the reductions of $a$. Hence $a=i(P)$ for a $k$-point $P$ of $C$, since $i$ is a closed $k$-embedding.

This proof uses only ordinary height finiteness and elementary reduction; it does not require an effective height bound, consistent with the noneffective existence target. It also explains why compactness in the adelic topology alone is inadequate: compactness would need to take place inside a finite set of bounded-height rational points.

3. A quantitative height-versus-sieve lemma over any number field. The following gives a stronger stopping test. Put $D=[k:{\mathbb{Q}}]$ and use absolute values extending the standard ones on ${\mathbb{Q}}$, with weights $n_v=[k_v:{\mathbb{Q}}_v]/D$. Thus $\sum_v n_v\log|b|_v=0$ for $b\in k^\times$. For projective coordinates $x=(x_0,\ldots,x_r)$ set \[ \|x\|_v=\max_i|x_i|_v,\qquad h([x])=\sum_v n_v\log\|x\|_v. \] Let $F\in k[X_0,\ldots,X_r]$ be homogeneous of degree $e\ge1$. At nonarchimedean $v$ let $c_v(F)$ be the maximum of $1$ and its coefficient absolute values; at archimedean $v$ use the maximum of $1$ and their sum. The finite constant $c(F)=\sum_v n_v\log c_v(F)$ gives $|F(x)|_v\le c_v(F)\|x\|_v^e$ everywhere.

Let $T$ be a finite set of good prime ideals where these coefficients are integral and where the projective reduction of $[x]$ satisfies $F=0$. Write $M_T=\prod_{\mathfrak p\in T}N\mathfrak p$. If $F(x)\ne0$, then \begin{equation}\label{wsref-0-33-30-30-30-36-37-39-31-3a-73-69-65-76-65} \log M_T\le D\left(eh([x])+c(F)\right). \end{equation} Indeed, after locally scaling a coordinate to a unit, divisibility by $\mathfrak p$ gives \[ \frac{|F(x)|_v}{\|x\|_v^e}\le p^{-1/e_v} \quad(v\leftrightarrow\mathfrak p\mid p), \qquad n_v\log(p^{-1/e_v})=-D^{-1}\log N\mathfrak p. \] At every remaining place the same normalized expression is bounded by $c_v(F)$. Sum its logarithm. The product formula makes the left sum $-eh([x])$, so $-eh([x])\le c(F)-D^{-1}\log M_T$, proving (30006791:sieve). This argument is invariant under rescaling $x$, and so does not assume that the number field has principal ideal class group.

Choose finitely many homogeneous equations $F_1,\ldots,F_s$ cutting out $i(C)$ inside the projective model of $J$, and enlarge the finite excluded set of primes so that reduction respects these equations. Put $e=\max_j\deg F_j$ and $c=\max_jc(F_j)$. Every $a\in J(k)\setminus i(C)$ which survives the sieve at $T$ must satisfy \begin{equation}\label{wsref-0-33-30-30-30-36-37-39-31-3a-73-65-70-61-72-61-74-69-6f-6e} h(a)\ge\frac{D^{-1}\log M_T-c}{e}. \end{equation} To see this, select an equation not vanishing at $a$ and apply (30006791:sieve), using $h(a)\ge0$. Conversely a surviving rational point with $D(eh(a)+c)<\log M_T$ lies on $i(C)$ and yields a rational point on $C$. This is a finite, exact certificate, not a density heuristic.

4. The missing selection theorem. The attempted completion would prove the following statement for the Jacobians and embeddings obtained in part 1:

For each Brauer tuple there are increasing finite sets $T_n$ of good places and $a_n\in J(k)$ matching its reductions at $T_n$, such that $\log M_{T_n}\to\infty$ and, for some $\eta>0$, \[ h(a_n)\le(1-\eta)\frac{\log M_{T_n}}{De} \quad\text{for arbitrarily large }n. \]

For those $n$, eventually $D(eh(a_n)+c)<\log M_{T_n}$, and the separation lemma gives a point of $C(k)$. Bounded-height selection is a special case. Neither the existence of these representatives nor the height saving is proved here. Weak approximation in the closure supplies compatible congruences, not this archimedean control.

5. A concrete failure of the compactness shortcut. In $\widehat{{\mathbb{Z}}}=\prod_p{\mathbb{Z}}_p$ take $z_2=-1$ and $z_p=0$ for every odd prime. Every finite collection of its prime-power congruences has an integer representative by the Chinese remainder theorem, and the congruence sets form a compatible inverse system. There is no integer representing $z$: an integer divisible by $3^m$ for every $m$ is zero, but zero does not have $2$-adic coordinate $-1$.

More quantitatively, let $a_m$ simultaneously satisfy $a_m\equiv-1\pmod{2^m}$ and $a_m\equiv0\pmod{3^m}$. Such representatives cannot remain bounded, because any bounded subsequence repeating one integer would contradict the preceding argument. The generic Chinese remainder representative bound is exponential in the logarithmic modulus; viewing these representatives as Mordell--Weil coordinates gives no reason for their quadratic canonical heights to meet part 4. This is an abstract model of the obstruction, not an adelic point on a particular curve violating Brauer--Manin.

Stress test.

The full Brauer group enters through Stoll's exact comparison theorem; it has not been replaced by a finite list of Brauer classes or by local solubility alone. The degree-one class is deduced only under the displayed Tate--Shafarevich hypothesis. Without that hypothesis, the torsor $\operatorname{Pic}^1(C)$ and divisible obstruction cannot be discarded. Even granting it does not establish the conjecture: the new height-selection step remains independent and unproved.

The norm inequality sums over all archimedean and nonarchimedean places, with the field-degree weights written explicitly. Thus the argument is not restricted to $k={\mathbb{Q}}$ or to rational points having globally primitive integral coordinates. Finite exceptional primes and the coefficients of the defining equations contribute the constant $c$; they are not ignored. The strict height inequality is essential: equality in a divisibility bound cannot force a polynomial value to be zero.

Outcome.

Outcome: Conditional reduction.

The unconditional output is the sieve separation bound (30006791:separation) and the bounded-height lifting lemma. Together with both vanishing of the relevant divisible Tate--Shafarevich subgroup and the explicit height-saving selection statement, they imply the original Hasse-principle assertion for all number fields. Both additional global hypotheses remain unproved. No curve is constructed with a nonempty full Brauer set and no rational point, and no unconditional resolution is claimed. The product-formula lemma is a notebook derivation, not a claim of previously unknown arithmetic geometry.

Sources

• S1 Michael Stoll, Finite descent obstructions and rational points on curves, arXiv math/0606465v3. https://arxiv.org/pdf/math/0606465v3. Catalogue formulation source.
• E1 Brendan Creutz, Adelic Mordell--Lang and the Brauer--Manin Obstruction, arXiv:2510.25931v2 (22 September 2026), Theorems 1.1--1.2 and Sections 4.1 and 5. https://arxiv.org/abs/2510.25931. Consulted 27 September 2026. The unrestricted number-field assertion is not an unconditional consequence of this preprint.