{"id":"63370f3d-221b-49d3-ad84-1b16985bb5fd","arxiv_id":"2608.01884","paper_version":1,"verdict":"ACCEPT","confidence":"HIGH","novelty_score":8.0,"correctness_risk":"low","formal_verification":"none","parameter_count":0,"one_line_summary":"For every odd prime p, a uniform family of finite left braces of order p^{2p+1} has powerful multiplicative group of class two but nonterminating right series, disproving the Shalev-Smoktunowicz conjecture in every odd characteristic.","lead":"For every odd prime p, the paper constructs a finite left brace, a set with two compatible group operations, whose multiplicative group is powerful but which is not right nilpotent. The construction disproves the Shalev-Smoktunowicz conjecture and produces new irretractable solutions of the Yang-Baxter equation.","discovery_kind":"new_method","skeptic_critique":{"model":"deepseek-v4-flash","headline":"No significant objection identified","rationale":"The reader's ACCEPT is justified. I re-examined the construction end to end. Lemma 3.1 is indeed the most assumption-heavy step, but the Gröbner basis computation is valid: the only nontrivial S-polynomial among the first two generators is y^{p+2}, and the remaining S-polynomials reduce to zero, so the standard monomials are as claimed. The subsequent lemmas use only standard properties of finite local algebras, derivations with D^2=0, and the kernel-graph regular subgroup construction. I checked the brace formula a∗b=ab−ε(a)D(b)−ε(a)aD(b), the invariance of ℓ, the p-power/commutator identities in Proposition 6.1, the persistence of T under right multiplication, the left series A^n=J^n, and the socle computation. All are internally consistent. The paper honestly states the remaining open cases (p=2, minimal order). Since no gap affects the central claim, the verdict should remain unchanged.","tokens_in":11790,"tokens_out":29705,"duration_ms":300576,"concrete_test":"Run an independent Gröbner basis computation in SageMath for p=3 and p=5 on the ideal (xy, x^{p+1}-y^{p+1}) in F_p[x,y] with lexicographic order x>y, verifying that {xy, x^{p+1}-y^{p+1}, y^{p+2}} is a Gröbner basis and the quotient has F_p-dimension 2p+2 with J^{p+2}=0. If the computation confirms the basis and dimension, the foundation of the construction is stable.","verdict_should_be":"UNCHANGED","load_bearing_attack":"I find no load-bearing flaw. The most delicate point is Lemma 3.1's monomial basis, and it is correct: with lex order x>y, the S-polynomial of xy and x^{p+1}-y^{p+1} is exactly y^{p+2}, and the remaining S-polynomials reduce to zero, so {xy, x^{p+1}-y^{p+1}, y^{p+2}} is a Gröbner basis. The standard monomials are 1,x,...,x^p,y,...,y^{p+1}; after replacing y^{p+1} by z=x^{p+1}, dim J=2p+1, J^{p+1}=F_p z, and J^{p+2}=0 all follow. The derivation D, the invariant character ℓ with ℓ(1+z)=1, the regular affine subgroup G_ℓ, and the identity T∗A_p=T all check out. The later claims G'=G^p=U^p≅C_p^2, cl(G)=2, exp(G)=p^2, and Soc(A_p)=0 follow from the displayed commutators and annihilation arguments. No internal inconsistency or omitted verification touches the central counterexample.","agreement_with_reader":"agree"},"referee_report":{"model":"deepseek-v4-flash","summary":"For each odd prime p, the paper constructs a finite left brace A_p of order p^{2p+1} whose additive group is elementary abelian and whose multiplicative group G_p is powerful with G_p' = G_p^p ≅ C_p^2, cl(G_p) = 2, exp(G_p) = p^2, and Soc(A_p) = 0. The brace is left nilpotent but not right nilpotent: it contains a 3-dimensional trivial ideal T with T ∗ A_p = T, and both T and A_p/T are right nilpotent, so right nilpotence is not extension-closed. The construction starts from the local algebra B = F_p[x,y]/(xy, x^{p+1}-y^{p+1}), a square-zero derivation D, and an invariant character ℓ, producing a regular affine subgroup. This disproves the Shalev–Smoktunowicz conjecture in every odd characteristic and yields finite irretractable involutive Yang–Baxter solutions whose permutation groups are powerful p-groups of class two.","tokens_in":11867,"tokens_out":9433,"duration_ms":103215,"significance":"This is a substantial negative answer to a conjecture recorded in [1] and [10]. The proof is self-contained and highly explicit: the Gröbner basis computation, derivation identities, group invariants, and the persistent right-series ideal are all concrete and checkable by hand. The construction is uniform in p and uses no numerical or black-box steps. I verified the central computations: the basis and dimension in Lemma 3.1, D^2 = 0 and D(a)a = 0 in Lemma 3.2, U^p = 1+W and U ≅ C_{p^2}^2 × C_p^{2p-3} in Lemma 4.1, the invariant character in Lemma 4.2, and the commutator/power computations in Proposition 6.1. The paper is a strong contribution to the theory of braces and set-theoretic Yang–Baxter solutions.","major_comments":[],"minor_comments":[{"comment":"The Gröbner basis argument is correct but compressed into two sentences. Since every later computation (dimension of J, J^{p+1} = F_p z, J^{p+2} = 0, and the basis of U) depends on this lemma, please expand the S-polynomial reduction: show that the S-polynomial of xy and x^{p+1}-y^{p+1} reduces to y^{p+2}, and that all remaining S-polynomials reduce to zero.","section":"Lemma 3.1"},{"comment":"Immediately after defining the affine action on J, the text says 'Since G_ℓ acts regularly, the restriction of this action to G_ℓ is faithful.' At that point regularity has not yet been proved; the following paragraph proves the orbit map is bijective. Reorder the argument so that regularity is established before the faithfulness statement, or phrase it as 'the orbit map below shows regularity, hence faithfulness.'","section":"§2.2, Proposition 2.2"},{"comment":"In the commutator calculation, the notation g_x and g_y should be defined explicitly, via the identification a ↦ g_{1+a} in Definition 5.1. It is because ℓ(1+x) = ℓ(1+y) = 0 that g_x = M_{1+x} and g_y = M_{1+y}; this is worth stating to avoid ambiguity.","section":"§6, Proposition 6.1"},{"comment":"The sentence 'Since the solution has more than one element and its first retraction is itself, it is not a multipermutation solution' is terse. A one-sentence reminder that a non-singleton irretractable solution is not a multipermutation solution would help readers not immersed in the Yang–Baxter terminology.","section":"§9, Corollary 9.1"}],"recommendation":"accept","confidential_remarks":null},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Colleague,\n\nThe short version: this paper settles the Shalev–Smoktunowicz conjecture for odd primes in the negative, with a single uniform construction that is explicit enough to check by hand. I went through the main computations (the Gröbner basis in Lemma 3.1, the derivation identities, the invariant character, the regular subgroup, the persistent ideal T, the left series, Soc=0) and they hold. The stress-test note's verification of the S-polynomial for xy and x^{p+1}-y^{p+1} is right, and the later claims follow. The reader's ACCEPT verdict is fair.\n\nWhat's new: previous non-right-nilpotent braces of order p^4 had non-powerful multiplicative groups for odd p; positive results in [8] and [14] live in complementary regimes (large characteristic, low class). This family has multiplicative group powerful of class two with derived subgroup C_p^2, exponent p^2, yet right series does not terminate. The obstruction is a three-dimensional trivial ideal T with T*A_p=T, and both T and A_p/T are right nilpotent, so extension closure fails. That's a clean and surprising package.\n\nSoft spots, all minor. Lemma 3.1's Gröbner basis proof is a two-sentence sketch; it's correct, but a referee will want a sentence showing the remaining S-polynomials reduce to zero. Proposition 7.4's annihilation statement is also compressed. The paper explicitly leaves p=2 and minimal order open; that's honest, not a flaw. The AI-assistance declaration looks standard and the author states all claims were verified.\n\nCitation pattern is appropriate. The Oberwolfach report and Vendramin's survey are the right places for the conjecture; the positive results are cited and contextualized. No self-citation inflation.\n\nWho is this for: people working on braces, skew braces, powerful p-groups, and set-theoretic Yang–Baxter solutions. It also gives a nice example of extension failure for a nilpotence property, useful beyond the immediate topic.\n\nRecommendation: send this to a serious referee. It deserves a regular review, not a desk reject. I'd ask the referee to expand the Gröbner basis verification and double-check the quotient brace in Corollary 7.5; otherwise accept.\n\nBest,","headline":"Uniform counterexample family for every odd prime; verifies cleanly; a few compressed proofs but no load-bearing gaps.","tokens_in":12579,"tokens_out":4621,"would_cite":true,"duration_ms":44080,"reading_group":"yes","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["16T25","20D15","16N40"],"pacs":[],"model":"deepseek-v4-flash","headline":"For every odd prime p, there is a finite left brace A_p with powerful multiplicative group of class two that is nevertheless not right nilpotent, disproving a conjecture that powerfulness forces right nilpotence.","keywords":["left brace","powerful p-group","right nilpotence","Yang–Baxter equation","regular subgroup","skew brace","nilpotency class","elementary abelian group"],"falsifier":"Compute a Gröbner basis for the ideal (xy, x^{p+1}-y^{p+1}) in F_p[x,y] for a small odd prime such as p=3, and check whether the standard monomials are exactly 1,x,...,x^p,y,...,y^p,z and whether the product x^p*x equals z in the quotient. If z can be expressed as a combination of x^p and y^p, or if J^{p+2}≠0, then the dimension claim fails and with it the entire construction; conversely verifying these identities for p=3 would confirm the paper's core computational claim.","tokens_in":11450,"feed_emoji":"🧩","tokens_out":7306,"duration_ms":67900,"temperature":0.7,"pith_summary":"Powerful p-groups behave like abelian p-groups in many ways, so a natural question was whether a brace whose multiplicative group is powerful must be right nilpotent. This paper answers that question negatively for every odd prime p by constructing a left brace A_p of order p^{2p+1} with elementary abelian additive group. Its multiplicative group G_p is powerful of nilpotency class two, with G_p'=G_p^p≅C_p^2 and exponent p^2, yet the right series A^{(n)}_p never reaches zero. The obstruction is an explicit three-dimensional trivial ideal T inside A_p for which T*A_p=T, while the left series is just the ideal-power filtration of a nilpotent commutative algebra and terminates. The construction is uniform and yields finite irretractable Yang–Baxter solutions with powerful permutation groups, so the failure is not an isolated curiosity.","feed_headline":"Every odd prime yields a brace defeating right nilpotence","feed_subtitle":"A three-dimensional ideal keeps the right series alive; powerful multiplicative groups do not save it.","key_machinery":"The key mechanism is the kernel-graph construction (Proposition 2.2): given a finite-dimensional nilpotent commutative F_p-algebra J, a derivation D with D^2=0 and (D(J))^2=0, and a (1+D)-invariant character ℓ on the principal-unit group U=1+J, the set G_ℓ={M_u Φ^{-ℓ(u)} : u∈U} is a subgroup of GL(B) that acts regularly on J; identifying U with J yields a left brace with λ_a(b)=(1+a)(1-ℓ(1+a)D)(b). Together with Lemma 2.3, the powerfulness criterion, this gives control over the multiplicative group: if φ=1+D acts trivially on U/U^p and the generators of U^p lie in the kernel of ℓ, then G_ℓ is powerful. The specific algebra B then supplies the persistent ideal T, whose products under ∗ reprod","core_discovery":"The paper establishes a uniform counterexample to the conjecture that finite braces of abelian type with powerful multiplicative group must be right nilpotent. For each odd prime p, the brace A_p is built from the local algebra B = F_p[x,y]/(xy, x^{p+1}-y^{p+1}) with radical J. The derivation D defined by D(x)=y^p, D(y)=-x^p satisfies D^2=0 and (D(J))^2=0, so φ=1+D is an automorphism of order p. A carefully chosen φ-invariant character ℓ of the principal-unit group U=1+J, taking value 1 on 1+z with z=x^{p+1}=y^{p+1}, produces a regular affine subgroup via the kernel-graph construction; the resulting brace has operation a∘b = a+(1+a)φ^{-ℓ(1+a)}(b). The paper shows that G_p=(A_p,∘) satisfies G","pith_inferences":["If the monomial-basis computation is correct, the same kernel-graph mechanism may yield counterexamples in other families, for instance by varying the relation x^{p+1}-y^{p+1} to higher powers or adding more variables, potentially approaching the paper's open question of the minimal order of such a brace.","The explicit failure of extension closure suggests that right nilpotence is sensitive to the entire multiplicative action; testing whether stronger conditions on the brace's own products (such as requiring the right series to vanish after one step for all ideals) restore closure would clarify the boundary.","For p=2, where powerfulness means G'≤G^4, the construction does not directly apply; testing the analogous algebra over F_2 would either produce a dyadic counterexample or reveal a genuinely different obstruction.","The construction may be modifiable to yield braces with nonzero socle while retaining non-right-nilpotence, which would test the role of trivial socle in the irretractability of the associated Yang–Baxter solutions."],"forward_implications":["The conjecture that a finite skew brace of abelian type with powerful multiplicative group must be right nilpotent is false in every odd characteristic.","Right nilpotence is not closed under extensions: the brace A_p contains an ideal T such that both T and A_p/T are right nilpotent but A_p is not.","The simultaneous conditions cl(G)=2, exp(G)=p^2, and |G'|=p^2 are insufficient to force right nilpotence.","For every odd prime p there exists a finite irretractable involutive set-theoretic solution of the Yang–Baxter equation of cardinality p^{2p+1} whose permutation group is powerful of class two and exponent p^2.","The examples have order p^{2p+1}, so they lie outside the previously known positive range p>n+1."],"fun_headline_variants":["Powerful groups don't force right nilpotence in braces","Odd primes yield braces with powerful groups but no right nilpotence","Right nilpotence not forced by powerful groups in braces","Shalev–Smoktunowicz conjecture disproved for odd primes","Counterexample: powerful multiplicative groups fail in braces"],"cache_read_input_tokens":2304,"weakest_assumption_plain":"The whole construction rests on the claim in Lemma 3.1 that the quotient algebra B has basis 1,x,...,x^p,y,...,y^p,z with z=x^{p+1}=y^{p+1} a new element independent of x^p and y^p, so that dim J=2p+1 and J^{p+2}=0; if that dimension were wrong or z collapsed into the span of x^p and y^p, the brace's order, the ideal T, and the invariant character would all fail.","fun_headline_variants_meta":{"raw":{"variants":["Powerful groups don't force right nilpotence in braces","Odd primes yield braces with powerful groups but no right nilpotence","Right nilpotence not forced by powerful groups in braces","Shalev–Smoktunowicz conjecture disproved for odd primes","Counterexample: powerful multiplicative groups fail in braces"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000184,"raw_usage":{"total_tokens":1217,"prompt_tokens":870,"completion_tokens":347,"prompt_tokens_details":{"cached_tokens":256},"prompt_cache_hit_tokens":256,"prompt_cache_miss_tokens":614,"completion_tokens_details":{"reasoning_tokens":261}},"tokens_in":614,"tokens_out":347,"duration_ms":4644,"temperature":1.0,"reasoning_tokens":261,"cache_read_input_tokens":256,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-04T18:57:06.255076+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Compute a Gröbner basis for the ideal (xy, x^{p+1}-y^{p+1}) in F_p[x,y] for a small odd prime such as p=3, and check whether the standard monomials are exactly 1,x,...,x^p,y,...,y^p,z and whether the product x^p*x equals z in the quotient. If z can be expressed as a combination of x^p and y^p, or if J^{p+2}≠0, then the dimension claim fails and with it the entire construction; conversely verifying these identities for p=3 would confirm the paper's core computational claim.","supporting_citations":[],"review_version":1}