{"id":"bed76a3d-df87-4be0-bb14-5e1211200a4f","arxiv_id":"2501.18774","paper_version":1,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":7.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":3,"one_line_summary":"For any quadratic extension K/F, there exists an abelian variety over F of positive rank whose rank is unchanged after base change to K, yielding undecidability of integer polynomial equations over every number field.","lead":"This paper proves that for every quadratic extension of number fields, some abelian variety has positive rank that does not grow when the base field is enlarged. It uses this to conclude that Hilbert's tenth problem has a negative answer over the ring of integers of every number field.","discovery_kind":"new_method","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Lemma 2.4's silent-prime step uses inertness in K instead of in F(√(qℓ)), so the Selmer equality is unproved for infinitely many allowed twists.","rationale":"The central claim of the paper is Theorem 1.1, and the proof's key step is the assertion that Selφ(J_{qℓr²t²})=0 for a flexible set of t. The only justification for this vanishing when t introduces primes outside S is the silent-prime lemma. As written, that lemma is applied to the wrong quadratic extension: Σ is defined via primes inert in K, while the ambient field F(√(qℓr²t²)) equals F(√(qℓ)), not K unless √ℓ∈F. The supplied proof does not prove or assume this equality, and it cannot be guaranteed by the initial Weil-restriction reduction. For p inert in K but split in F(√(qℓ)), T_p is nonzero and the curves can have different local conditions; since such p are allowed in Σ, the rank-zero conclusion is unsupported. This is the load-bearing point: if Lemma 2.4 fails, the construction of an A with rank stability collapses, and Corollary 1.2 no longer follows from this method. The reader's verdict identified the same gap; I agree, though the suggested repair 'choose ℓ≡1 mod4 so √ℓ∈F' is not a valid sufficient condition (ℓ≡1 mod4 does not put √ℓ in F). The concrete test above gives an explicit instance in F=Q(ζ_3) where the hypotheses of Lemma 2.4 are met and the local conditions are expected to be visibly different. Since the gap may be repairable by redefining Σ with respect to F(√(qℓ)) and reproving Proposition 2.5 with the corresponding ray class character, CONDITIONAL rather than REJECT is appropriate.","tokens_in":9857,"tokens_out":34637,"duration_ms":334468,"concrete_test":"Let F=Q(ζ_3), ℓ=3, K=F(√2) (q=2), r=1, and let P be a prime of F above 19. In the residue field F_19, 2 is a nonsquare and 6 is a square, so P is inert in K but split in F(√6). The rational integer 19 is a Σ-unit (both primes above 19 are inert in K) and 19≡1 mod 9, hence a cube in F_{P_3} for P_3|3, so it satisfies the hypotheses of Lemma 2.4. Compute the local 3-isogeny Selmer conditions W_P for J_6 and J_{6·19²} at P. The second curve has bad reduction at P while the first has good reduction, and T_P has nonzero order because √6∈F_P; if W_P(J_6)≠W_P(J_{6·19²}), then Lemma 2.4's asserted equality is false for an admissible t.","verdict_should_be":"UNCHANGED","load_bearing_attack":"In §2.3, Σ is defined as S ∪ S_inert with S_inert = {p : p inert in K}, where K=F(√q). Lemma 2.4 then asserts Selφ(J_{qℓr²t²})=0 for any Σ-unit t that is an ℓ-th power at all p∈S. The proof invokes Lemma 2.2 for primes 'inert or ramified in K'. But Lemma 2.2 gives T_p=0 only when p is inert or ramified in F(√n)/F. Here n=qℓr²t², so F(√n)=F(√(qℓ)) because r,t∈F. This equals K exactly when √ℓ∈F; that condition is not stated and cannot be guaranteed (only finitely many rational primes are squares in a given number field). For primes p inert in K but split in F(√(qℓ)), the Galois module J[φ] has nonzero invariants at p, so T_p≠0, and the good-reduction argument in the 'all other cases' clause does not apply because t may have nonzero valuation at p (such p are in Σ). Hence the local conditions W_p for J_{qℓr²} and J_{qℓr²t²} are not shown to coincide; the crucial rank-zero conclusion of Lemma 2.4 is unsupported. This is load-bearing: it is the only step forcing rank J_{qℓr²t²}(F)=0 in the final construction.","agreement_with_reader":"agree"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper proves Theorem 1.1: for any quadratic extension K/F of number fields, there exists an abelian variety A/F with rank A(F) = rank A(K) > 0. By a theorem of Shlapentokh (in the form given in [MRS24]), this implies Corollary 1.2: Hilbert's tenth problem has a negative answer over the ring of integers of every number field. The proof works with the Jacobians J_n of the curves C_n: y^2 = x^l + n over a field F containing the l-th roots of unity, and studies the (1-zeta)-Selmer groups. It first uses a theorem of Yu to choose n = q l r^2 with vanishing Selmer group, then exploits silent primes to show that multiplying n by t^2 for suitable Sigma-units t does not change the Selmer group, and finally uses solutions of a Sigma-unit equation a + 2rb = 1, supplied by Mitsui's and Kai's results, to produce a rational point on the relevant quadratic twist. The construction is assembled in Section 4.","tokens_in":10158,"tokens_out":25542,"duration_ms":254200,"significance":"If the proof is completed, this is a landmark result: it would resolve, unconditionally and uniformly, a major open problem of the last few decades. The strategy is attractive and conceptually clean, combining Selmer-group methods (Mazur-Rubin, Yu) with additive combinatorics in number fields (Mitsui, Kai). The paper is also honest in attributing the reduction from rank stability to Diophantine undecidability to Shlapentokh, and the main claim is independently supported by the recent Koymans-Pagano proof of the corollary. The construction is parameter-free in the sense that it relies on external theorems rather than fitted data, and the main theorem is exactly the hypothesis needed for Shlapentokh's theorem, so the logical structure is transparent.","major_comments":[{"comment":"The proof of the equality of local conditions W_p is incomplete for primes in S_inert. For n = q l r^2 t^2, we have F(√n) = F(√(q l)), so Lemma 2.2 yields T_p = 0 only when p is inert or ramified in F(√(q l))/F. However, S_inert is defined as the set of primes inert in K = F(√q). If p is inert in K but l is not a square in F_p^×, then p is split in F(√(q l)), so T_p is nontrivial; since p is in Sigma, t may have nonzero valuation at p, and the subsequent 'good reduction' clause cannot be invoked. Hence the assertion that Sel_φ(J_{q l r^2 t^2}) = Sel_φ(J_{q l r^2}) is not established. This is the only step forcing rank J_{q l r^2 t^2}(F) = 0 in Section 4, so the proof of Theorem 1.1 depends on repairing it. A natural repair is to choose l ≡ 1 (mod 4) before the Weil-restriction step in Section 4, so that √l ∈ F(ζ_l) and therefore F(√(q l)) = K for the reduced quadratic extension; with this hypothesis, every p inert in K is silent for J_{q l r^2 t^2}. Please state this condition and amend Lemma 2.4 and the proof accordingly.","section":"§2.3, Lemma 2.4"},{"comment":"The sentence 'Since a ≠ 0, this is not a torsion point' is not by itself a valid reason: on a curve of positive genus, a point with nonzero x-coordinate can represent a torsion class (for example on an elliptic curve). The intended argument must use the equality J_{r^2 a^{l-1} b^2}(F)_tors = J_{r^2 a^{l-1} b^2}[φ](F) obtained from Lemma 2.6 for all but finitely many t_{a,b}: because a ≠ 0, P is not fixed by the automorphism (x,y) ↦ (ζx,y), so the divisor class of P − ∞ is not fixed by ζ and hence is not in J[φ](F). Please make this reasoning explicit, since the positivity of rank in the final construction depends on it.","section":"§2.4, Proposition 2.7"}],"minor_comments":[{"comment":"The term 'S_inert-unit' is used in the proof of Proposition 2.5 but is not formally defined; please define it explicitly, for instance as an element whose prime support is contained in S_inert.","section":"§2.3, notation"},{"comment":"The parenthetical justification involving Yu's assumption Gal(f) ≃ S_n is terse and potentially confusing: for f(x) = x^l + n over F(ζ_l), the Galois group is cyclic, not S_n. Please state precisely which theorem of Yu is being invoked and what condition on q corresponds to it.","section":"§2.3, Lemma 2.3"},{"comment":"The proof of Proposition 3.1 is presented as a sketch with references to Mitsui and Kai; since this proposition is a key input, please indicate the precise theorem or section of Mitsui (or of Kai) that supplies the three-prime result in the stated ideal-class form.","section":"§3, Proposition 3.1"},{"comment":"The quotation of Simon and Garfunkel's 'The Sound of Silence' is extraneous to a mathematics research paper; I recommend removing it.","section":"After Section 1"},{"comment":"The uniform bound on torsion and the choice of the integer N would be clearer if the text specified explicitly that N depends only on l and F, not on n or t.","section":"§2.4, Lemma 2.6"}],"recommendation":"major_revision","confidential_remarks":"This is a potentially landmark paper, but the current version contains a genuine gap in the silent-prime argument of Lemma 2.4. The repair (e.g., requiring l ≡ 1 mod 4 so that √l lies in the base field) appears straightforward, and the main corollary is independently supported by Koymans-Pagano. I therefore recommend major revision rather than rejection. Please also ask the authors to clarify the precise range of applicability of Yu's theorem cited in Lemma 2.3, since the Galois-group remark there is confusing."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Dear colleague,\n\nThe headline: this paper has a genuinely new result in Theorem 1.1 — for any quadratic extension K/F there exists an abelian variety A/F with positive rank that does not grow upon base change to K. The ℓ-isogeny Selmer method used here is a fresh approach, and the authors are honest that the H10 corollary is not new, since Koymans and Pagano independently proved it. That disclosure is commendable.\n\nThe architecture is coherent: Yu gives a rank-zero twist J_{qℓr²}; silent primes are introduced so that multiplying by t does not change the Selmer group; a Σ-unit equation produces a rational point on the quadratic twist. The additive combinatorics input (Mitsui or Kai) is appropriate, and the paper is mostly clear.\n\nBut there is a load-bearing gap in Lemma 2.4. The proof says that for p inert or ramified in K, T_p = 0 by Lemma 2.2. Lemma 2.2 requires p to be inert or ramified in F(√n)/F, and here n = qℓr²t² or qℓr², so F(√n) = F(√(qℓ)) in both cases. That equals K only if √ℓ ∈ F, which is neither guaranteed nor stated. For a prime p inert in K but split in F(√(qℓ)), T_p ≠ 0. Since such p lie in Σ, t may have nonzero valuation at p, so the “all other cases” clause (good reduction for both curves) also fails. The equality of local conditions, and hence the conclusion Selφ(J_{qℓr²t²}) = 0, is not proved as written. This is not a minor typo; it is the step that forces the rank-zero Jacobian in the final construction.\n\nThe likely fix is to redefine the silent-prime set using F(√(qℓ)) instead of K, then adapt the Σ-unit construction in Proposition 2.5 accordingly. That seems feasible, but it requires real work. I do not think the main idea is wrong, but this version is incomplete.\n\nVerdict: send it to a serious referee. The result is important, the method is new, and an expert can judge whether the gap is patchable. I would not accept as is, but I would definitely engage with it.\n\nBest.","headline":"Strong new rank-stability theorem, but Lemma 2.4 has a real gap where inertness in K is used instead of inertness in F(√(qℓ)); the proof as written is incomplete.","tokens_in":10701,"tokens_out":7893,"would_cite":false,"duration_ms":73203,"reading_group":"yes","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["11U05","11G10","14G05","11R04"],"pacs":[],"model":"deepseek-v4-flash","headline":"For every quadratic extension $K/F$ of number fields, the paper constructs an abelian variety $A/F$ of positive rank with $\\operatorname{rank} A(F)=\\operatorname{rank} A(K)$, yielding a negative answer to Hilbert's tenth problem over the…","keywords":["Hilbert's tenth problem","rank stability","abelian varieties","Selmer groups","silent primes","quadratic extensions","diophantine models","Fermat curves"],"falsifier":"For a concrete instance, take $F=\\mathbb{Q}(\\zeta_3)$, $\\ell=3$, and $K=F(\\sqrt{2})$, choose $r$ as in Yu's theorem, and compute $T_{\\mathfrak{p}}=H^1(F_{\\mathfrak{p}}, J_{6r^2}[\\phi])$ at a prime $\\mathfrak{p}$ inert in $K$; if $T_{\\mathfrak{p}}\\neq 0$ for any such $\\mathfrak{p}$, the silent-prime lemma does not apply and the proof of Lemma 2.4 fails.","tokens_in":9579,"feed_emoji":"🔢","tokens_out":18435,"duration_ms":150875,"temperature":0.7,"pith_summary":"The paper proves that for every quadratic extension $K/F$ of number fields there exists an abelian variety $A/F$ with positive rank whose rank does not grow when the base field is extended from $F$ to $K$: $\\operatorname{rank} A(F)=\\operatorname{rank} A(K)>0$. This is exactly the hypothesis that earlier work had shown would imply a negative answer to Hilbert's tenth problem over the ring of integers of every number field, so the paper completes that reduction. The proof produces the abelian variety as the Jacobian of a hyperelliptic curve $C_n:y^2=x^\\ell+n$, controls it through its $(1-\\zeta)$-Selmer group, and uses solutions to a $\\Sigma$-unit equation to put a rational point on the right quadratic twist. If the proof is correct, no algorithm can decide whether a multivariable polynomial equation over $\\mathcal{O}_K$ has a solution, for any number field $K$.","feed_headline":"Rank-stable abelian varieties settle Hilbert's tenth problem","feed_subtitle":"Stable positive rank over quadratic extensions answers Hilbert's tenth problem negatively for every number field.","key_machinery":"The machinery is the $(1-\\zeta)$-Selmer group of the Jacobian $J_n$ of $C_n:y^2=x^\\ell+n$, where $\\ell$ is an odd prime and $F$ contains $\\zeta_\\ell$; here $\\phi=1-\\zeta$ is a self-isogeny of degree $\\ell$. A prime $\\mathfrak{p}$ is silent when the local cohomology group $T_{\\mathfrak{p}}=H^1(F_{\\mathfrak{p}},J_n[\\phi])$ vanishes, and Lemma 2.2 shows this happens whenever $\\mathfrak{p}$ is inert or ramified in $F(\\sqrt{n})$. Silence is what makes the Selmer group unchanged when $n$ is multiplied by $t^2$ for a $\\Sigma$-unit $t$, so the rank-zero property of $J_{q\\ell r^2}$ survives for $J_{q\\ell r^2t^2}$. The paper combines this with the quadratic-twist rank identity $\\operatorname{rank} J_{r^2a^{\\ell-1}b^2}(K)=\\operatorname{rank} J_{q\\ell r^2a^{\\ell-1}b^2}(F)+\\operatorname{rank} J_{r^2a^{\\ell-1}b^2}(F)$, in which the first summand vanishes.","core_discovery":"The central claim is Theorem 1.1: for any quadratic extension $K/F$ of number fields, there exists an abelian variety $A/F$ such that $\\operatorname{rank} A(F)=\\operatorname{rank} A(K)>0$. By the rank formula for quadratic twists, it is enough to find a Jacobian $J_{r^2a^{\\ell-1}b^2}$ of positive rank over $F$ whose $K$-quadratic twist $J_{q\\ell r^2a^{\\ell-1}b^2}$ has rank zero. The paper obtains the rank-zero side from the $(1-\\zeta)$-Selmer group: Yu's theorem supplies $r$ with $\\operatorname{Sel}_\\phi(J_{q\\ell r^2})=0$, and the silent-prime lemma lets the Selmer group remain zero after multiplying the parameter by $t^2$, where $t=a^{(\\ell-1)/2}b$ comes from a solution to the $\\Sigma$-unit equation $a+2rb=1$. A twisted Fermat-curve cover then produces an $F$-rational point on $J_{r^2a^{\\ell-1}b^2}$, and Lemma 2.6 guarantees it is non-torsion, so the rank is positive. Since $A:=J_{r^2a^{\\ell-1}b^2}$ has the same positive rank over $F$ and $K$, Corollary 1.2 follows via the prior diophantine-stability result [MRS24].","pith_inferences":["A natural extension is to cyclic extensions of prime degree: the only local input is that primes inert in $F(\\sqrt{n})$ are silent, so replacing the quadratic character by a cyclic character and keeping the same Fermat-curve Selmer setup may produce rank stability for such extensions. The authors prove the quadratic case only.","The proof is non-effective because Mitsui's theorem is a qualitative infinitude statement; an effective number-field circle method would convert the construction into an explicit family of abelian varieties with controlled conductor, which the paper does not address.","For fields $F$ that already contain $\\zeta_3$ and for which $3$ is unramified in $K/F$, choosing $\\ell=3$ makes the constructed abelian variety an elliptic curve; this is a direct dimension count from the genus formula, though the paper states the general abelian-variety version."],"forward_implications":["Hilbert's tenth problem has a negative answer over $\\mathcal{O}_K$ for every number field $K$: no algorithm can decide whether a multivariable polynomial equation over $\\mathcal{O}_K$ has a solution.","$\\mathbb{Z}$ has a diophantine model over $\\mathcal{O}_K$ for every number field $K$, meaning the integers are existentially definable in the ring of integers in the sense used for undecidability transfers.","The result is unconditional; it does not depend on the finiteness of Tate-Shafarevich groups or any other unproved conjecture.","The specific Jacobians $J_{r^2a^{\\ell-1}b^2}$ exhibit diophantine stability over $K/F$: their rational points do not grow under base change, matching the Mazur-Rubin notion of diophantine stability.","Together with the independent proof of Koymans and Pagano, the paper closes the last case of Hilbert's tenth problem over rings of integers of number fields."],"supporting_citations":[{"why":"Supplies the twist-counting method that forces the $(1-\\zeta)$-Selmer rank to be zero.","marker":"[MR10]"},{"why":"Provides Theorem 4, the existence of $r$ with $\\operatorname{Sel}_\\varphi(J_{q\\ell r^2})=0$.","marker":"[Yu16]"},{"why":"Defines the $\\varphi$-Selmer group and the local conditions $W_{\\mathfrak{p}}$ used in the descent.","marker":"[PS97]"},{"why":"Gives everywhere-potential good reduction and the torsion bound used in Lemma 2.6.","marker":"[ST68]"},{"why":"Generalizes Vinogradov's circle method to number fields, producing the infinite family of $\\Sigma$-unit solutions.","marker":"[Mit60]"},{"why":"Is the classical circle-method source behind the ternary-Goldbach-type equation $p_1+\\beta p_2=p_3$.","marker":"[Vin04]"},{"why":"Proves that rank stability for every quadratic extension implies the negative answer to Hilbert's tenth problem over $\\mathcal{O}_K$, which the paper invokes for Corollary 1.2.","marker":"[MRS24]"}],"fun_headline_variants":["Rank-stable abelian varieties prove undecidability over number fields","Hilbert's tenth problem solved negatively for every number field","Rank stability in quadratic extensions negates Hilbert's tenth problem","Stable rank abelian varieties solve Hilbert's tenth problem"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The load-bearing premise is that the primes left silent in Lemma 2.4—those inert or ramified in $K=F(\\sqrt{q})$—are precisely the primes that Lemma 2.2 proves silent for the twists $J_{q\\ell r^2t^2}$, because Lemma 2.2 only guarantees silence for primes inert or ramified in $F(\\sqrt{q\\ell})$, and the paper does not impose an extra hypothesis, such as $\\ell \\equiv 1 \\bmod 4$ with $\\sqrt{\\ell}\\in F$, that would make these two quadratic extensions coincide.","fun_headline_variants_meta":{"raw":{"variants":["Rank-stable abelian varieties prove undecidability over number fields","Hilbert's tenth problem solved negatively for every number field","Rank stability in quadratic extensions negates Hilbert's tenth problem","Stable rank abelian varieties solve Hilbert's tenth problem"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.001192,"raw_usage":{"total_tokens":4919,"prompt_tokens":949,"completion_tokens":3970,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":565,"completion_tokens_details":{"reasoning_tokens":3899}},"tokens_in":565,"tokens_out":3970,"duration_ms":24946,"temperature":1.0,"reasoning_tokens":3899,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-09T22:33:08.888182+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"For a concrete instance, take $F=\\mathbb{Q}(\\zeta_3)$, $\\ell=3$, and $K=F(\\sqrt{2})$, choose $r$ as in Yu's theorem, and compute $T_{\\mathfrak{p}}=H^1(F_{\\mathfrak{p}}, J_{6r^2}[\\phi])$ at a prime $\\mathfrak{p}$ inert in $K$; if $T_{\\mathfrak{p}}\\neq 0$ for any such $\\mathfrak{p}$, the silent-prime lemma does not apply and the proof of Lemma 2.4 fails.","supporting_citations":[],"review_version":1}