{"id":"0536022d-23d0-483d-97c1-9dcad04d4594","arxiv_id":"1908.06365","paper_version":1,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":6.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"In an arbitrary-rank valued field, Rν[α] is integrally closed exactly when every repeated residue factor gives a Euclidean remainder of minimal positive Gaussian valuation, or no factor repeats.","lead":"This math paper finds exact conditions for a ring built from a polynomial root to be integrally closed, the algebraic equivalent of having no missing fractions, over any valued field. It extends Dedekind's classical criterion to arbitrary-rank valuations and gives a simple remainder-based test plus a full description of how valuations split.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"The proof of Theorem 2.5(ii) contains an invalid Euclidean-division step when adjusting lifts of simple factors; as stated, the sufficiency argument for the main Dedekind criterion is not established.","rationale":"The reader identified the separability hypothesis behind Lemma 2.1 as the weakest assumption. That is a legitimate concern, but the most load-bearing problem I find is internal to the proof of Theorem 2.5(ii): the Euclidean-division construction used to normalize lifts of simple factors is not valid as written. The concrete 5-adic example shows that the claimed quotient and remainder do not satisfy the degree condition for Euclidean division. This matters because the sufficiency proof of the main criterion explicitly reduces to the case νG(r_i) = σ for every i and then applies Lemma 2.4 to all i. The theorem may still be true and the gap may be repairable by performing the remaining division step, but the current manuscript does not establish the central claim. This does not warrant rejection, since the results are plausible and the flaw is localized; it does require a corrected proof before the paper can be accepted. Thus the reader's conditional verdict remains appropriate, though for a different reason than the one stated in the reader's weakest_assumption.","tokens_in":11403,"tokens_out":40930,"duration_ms":428907,"concrete_test":"Recompute the Euclidean division in the stated example: with f(X) = X^3 + X^2 + 25, φ_i = X+1, and φ_i** = X+6, divide f by φ_i** and compare the quotient and remainder with the paper's q_i** and r_i**. The paper's asserted remainder r_i** = 25X−5 has degree 1, equal to deg(φ_i**), whereas the actual remainder is the constant −155. This single computation settles that the proof step is false. To test whether the theorem itself survives, repeat the normalization step with the correctly reduced remainder: here the true remainder has valuation σ, so the construction may be fixable; if for some lift every corrected remainder has valuation > σ, the sufficiency proof would require a substantially different argument.","verdict_should_be":"CONDITIONAL","load_bearing_attack":"In Section 2, proof of Theorem 2.5(ii), the paragraph beginning 'By making an appropriate choice of a lifting...' claims that for i not in I with νG(r_i) > σ, setting φ_i** = φ_i + π, q_i** = q_i − πq_i*, and r_i** = r_i − πr_i* + π^2q_i* gives the Euclidean quotient and remainder of f by φ_i**. This is false when deg(q_i*) ≥ deg(φ_i), because then deg(r_i**) can equal or exceed deg(φ_i**). Concretely, take K = Q with the 5-adic valuation (σ = 1, π = 5) and f(X) = X^3 + X^2 + 25. This f is irreducible over Q and reduces modulo 5 to X^2(X+1), so the simple factor φ_i = X+1 has l_i = 1. Here q_i = X^2, q_i* = X−1, r_i* = 1, φ_i** = X+6, q_i** = X^2−5X+5, and r_i** = 25X−5. But the actual Euclidean division of f by X+6 gives quotient X^2−5X+30 and remainder −155, not r_i**. Thus the stated construction does not produce the remainder. Since the rest of the sufficiency proof uses Lemma 2.4 for every i, including i not in I, this is a load-bearing gap in the proof of the central criterion. The gap may be repairable—in this example the true remainder −155 has value 1 = σ—but the proof as written does not justify the reduction to the case νG(r_i) = σ for all i.","agreement_with_reader":"disagree"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper proves a generalized Dedekind criterion for integral closedness of Rν[α] in a separable finite simple extension L=K(α) of an arbitrary-rank valued field (K,ν). Under Assump's (f monic irreducible separable in Rν[X], with reduction factorized as ∏ φ_i^{l_i}), Theorem 2.5 characterizes integral closedness by the condition νG(r_i)=min(Γ_ν^+) for every i with l_i≥2, where r_i is the remainder of f modulo φ_i. Theorem 2.9 gives an equivalent characterization in terms of the number of extensions of ν to L and the values l_iω_i(φ_i(α)). The paper also derives ramification and residue-degree data (Corollary 2.10), an Eisenstein-style irreducibility lemma, and applications (Corollaries 3.1–3.3, Examples 1–2).","tokens_in":11751,"tokens_out":12664,"duration_ms":105487,"significance":"If the results are correct, the paper gives a clean and useful generalization of Ershov's and Khanduja–Kumar's Dedekind criteria to arbitrary-rank valued fields, removing the Henselian assumption, and it computes the ramification and residue-degree decomposition in the integrally closed case. The manuscript is largely self-contained: Lemma 2.2, generalizing a previous result of the authors, is proved in full, and Lemma 2.4 contains a sharp and correct identity l_iω(φ_i(α))=σ. The main theorems are falsifiable and illustrated with concrete examples. However, the proof of the sufficiency direction of Theorem 2.5 contains two gaps that need repair before the result can be accepted.","major_comments":[{"comment":"The claim that the polynomials q_i** and r_i** defined there are the Euclidean quotient and remainder of f by φ_i** is not valid in general. For example, take K=Q with the 5-adic valuation (π=5, σ=1), f(X)=X^3+X^2+25, φ_i=X+1. Then q_i=X^2, q_i*=X−1, r_i*=1, φ_i**=X+6, q_i**=X^2−5X+5, and r_i**=25X−5. Here deg(r_i**)=deg(φ_i**)=1, so r_i** is not the remainder; the true remainder is −155. Since this step is used to justify reducing to the case νG(r_i)=σ for all i, and hence to apply Lemma 2.4 to every i in the final argument, the proof as written is incomplete. The gap is repairable: for i∉I one has l_i=1, and in the final contradiction one may take m_i=0, so the precise value of ω(φ_i(α)) is not needed; alternatively one must prove that the true remainder after further Euclidean division still has Gaussian value σ.","section":"Section 2, proof of Theorem 2.5(ii), paragraph beginning 'By making an appropriate choice of a lifting...'"},{"comment":"The assertion that after writing g = S_i φ_i^{m_i} + T_i with φ_i ∤ S_i one has νG(T_i) ≥ σ is not a consequence of the definition of m_i. If m_i is the highest power of φ_i dividing g in kν[X], the remainder T_i may have a coefficient of value 0, so in general only νG(T_i) ≥ 0 holds. The subsequent equality ω(g(α)) = m_iσ/l_i relies on the stronger bound and is therefore not established. The desired contradiction can still be recovered: since ω(T_i(α)) ≥ νG(T_i) ≥ 0 and m_iσ/l_i ≥ 0, the minimum is 0 < σ, giving ω(g(α)) = 0 < σ and hence ω(θ) < 0. The proof should be corrected to use this weaker but sufficient bound.","section":"Section 2, proof of Theorem 2.5(ii), final paragraph"}],"minor_comments":[{"comment":"The phrase 'r_i is zero' is used where the argument only shows that the reduction of r_i modulo Mν is zero (equivalently νG(r_i)>0). This should be reworded to avoid a literal false statement.","section":"Lemma 2.3 and Corollary 2.7"},{"comment":"The sentence 'on does not explicitly state the assumption that f is separable, both references crucially use, which in turn assumes the separability of f' is garbled and needs to be rewritten.","section":"Page 1, after the abstract"},{"comment":"In the line 'deg(h) ≤ deg(h)', the notation should clarify that h denotes the reduction of h modulo Mν.","section":"Theorem 2.5(i), proof"},{"comment":"The notation 'R[θv/π u]' should be typeset as Rν[θ^v/π^u], and 'A = av/πnu' as A = a^v/π^{nu}.","section":"Corollary 3.3"},{"comment":"Reference [1] is listed as 'to appear'; if the paper is still unpublished, please update the status or provide a preprint reference.","section":"References"}],"recommendation":"major_revision","confidential_remarks":"The manuscript demonstrates good command of valuation theory and the main theorems are likely true, but the proof of Theorem 2.5(ii) needs repair. The gaps are localized and repairable, so I recommend major revision rather than rejection. I also recommend a careful proofreading pass, as there are several garbled sentences and typographical errors."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"I read this paper closely, and I think you should know two things up front. First, it does contain new and citable results: Theorem 2.9, characterizing integral closedness of Rν[α] in terms of s distinct extensions and the equality liω_i(φ_i(α)) = min(Γ+ν), and Corollary 2.10, computing ramification indices and residue degrees as li and deg(φ_i) when the order is integrally closed. These are genuine advances over the Khanduja–Kumar framework, and the paper correctly notes that the Henselian hypothesis is not needed. Second, the proof of Theorem 2.5(ii) has a real gap, though I think it is repairable.\n\nThe main problem is in the paragraph where the authors claim that for i not in I with νG(ri) > σ, setting φ_i** = φ_i + π, q_i** = q_i − πq_i*, and r_i** = r_i − πr_i* + π^2q_i* gives the Euclidean quotient and remainder of f by φ_i**. That claim is false in general. The stress-test example is correct: for K = Q with the 5-adic valuation and f = X^3 + X^2 + 25, reducing mod 5 gives X^2(X+1), so for the simple factor φ_i = X+1 we get q_i* = X−1, r_i* = 1, and then r_i** = 25X−5 has degree 1, equal to deg(φ_i**), so it cannot be the remainder. The actual remainder of f by X+6 is −155.\n\nThere is also a separate issue in the final contradiction argument. The authors write that ω(g(α)) = miσ/li < σ, but if the chosen i has mi = 0 (which can happen for a simple factor, or even for a repeated factor when φ_i does not divide g), then miσ/li = 0, which is not less than σ. The conclusion still follows if one splits off the mi = 0 case: then φ_i does not divide g, so by Lemma 2.2(iv) ω(g(α)) = 0, and since ν(b) ≥ σ, ω(θ) ≤ −σ < 0. So the theorem itself appears true, but the written proof is not complete.\n\nOther soft spots are minor: Lemma 2.3's phrase \"r_i is zero\" should read \"r_i maps to zero in kν[X]\", and the introduction has garbled sentences. The citation pattern is fine; the self-citation [1] is not problematic because Lemma 2.2 is proved in full. Separability is explicitly assumed, which is appropriate.\n\nMy recommendation: this paper deserves serious peer review, but the referee should ask for a corrected proof of Theorem 2.5(ii) that either fixes the lift-adjustment step or removes it by handling mi = 0 separately. The new results are substantial enough that a revision would be worth engaging with.","headline":"A genuinely useful generalization of Dedekind's criterion to arbitrary-rank valued fields, but the proof of the main sufficiency direction has a fixable gap in the lift-adjustment step.","tokens_in":12327,"tokens_out":5915,"would_cite":true,"duration_ms":56455,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["12J10","13A18","13B22"],"pacs":[],"model":"deepseek-v4-flash","headline":"Over arbitrary-rank valued fields, integral closedness of a simple extension ring reduces to a finite remainder check.","keywords":["Dedekind's criterion","valued field","extensions of a valuation","integral closure","Gaussian valuation","Henselization","ramification index","Eisenstein polynomial"],"falsifier":"For $K=F(X,Y)$ with the lexicographic valuation on $\\mathbb{Z}^2$ (where $\\min(\\Gamma_\\nu^+)=(0,1)$) and $f(Z)=Z^3+YZ+X$, the theorem predicts $R_\\nu[\\alpha]$ is not integrally closed because the remainder of $f$ modulo $Z$ is $r=X$ and $\\nu_G(r)=(1,0)\\neq(0,1)$. Directly computing the integral closure of $R_\\nu[Z]/(f)$ in this case, and either exhibiting an element of $S\\setminus R_\\nu[\\alpha]$ or finding $S=R_\\nu[\\alpha]$, would settle the criterion on a concrete instance.","tokens_in":11182,"feed_emoji":"🧮","tokens_out":8933,"duration_ms":82310,"temperature":0.7,"pith_summary":"This paper gives necessary and sufficient conditions, for an arbitrary-rank valued field $(K,\\nu)$, under which the subring $R_\\nu[\\alpha]$ of a finite separable extension $K(\\alpha)$ equals the full integral closure of the valuation ring. The answer is read from the reduction of $f$ modulo the maximal ideal: if $\\bar f = \\prod \\bar\\varphi_i^{\\ell_i}$, then $R_\\nu[\\alpha]$ is integrally closed exactly when, for every repeated factor, the remainder of $f$ upon Euclidean division by a lift $\\varphi_i$ has Gaussian valuation equal to the smallest positive element of the value group. If no smallest positive value exists, integral closedness holds precisely in the square-free case. The same condition is rephrased in terms of the valuations on $K(\\alpha)$ extending $\\nu$, and it yields the ramification indices and residue degrees of all those extensions.","feed_headline":"A remainder test decides when R[α] is integrally closed","feed_subtitle":"Check one Gaussian valuation per repeated factor: it must equal the smallest positive value, with no Henselian assumption.","key_machinery":"The load-bearing object is the pair consisting of the Gaussian valuation $\\nu_G$ and the remainder $r_i$: $\\nu_G$ takes the minimum of the $\\nu$-values of the coefficients of a polynomial, and $r_i$ is the remainder of $f$ modulo the lifted irreducible factor $\\varphi_i$. The criterion's condition $\\nu_G(r_i)=\\min(\\Gamma_\\nu^+)$ is a finite, checkable measurement of how far $f$ is from being a product of powers of the $\\varphi_i$ modulo the maximal ideal. The other central mechanism is the Henselization correspondence of Lemma 2.1: valuations on $K(\\alpha)$ extending $\\nu$ biject with irreducible factors of $f$ over the Henselization $K^h$, allowing each extension to be identified by the unique residue factor $\\varphi_i$ that becomes positive at $\\varphi_i(\\alpha)$.","core_discovery":"Under the paper's standing assumptions, Theorem 2.5 states: if $I = \\{\\,i \\mid \\ell_i \\ge 2\\,\\}$ is nonempty, then $R_\\nu[\\alpha]$ is integrally closed if and only if $\\nu_G(r_i) = \\min(\\Gamma_\\nu^+)$ for every $i \\in I$, where $r_i$ is the remainder on Euclidean division of $f$ by the monic lift $\\varphi_i$ of the $i$-th irreducible residue factor, and $\\nu_G$ is the Gaussian extension of $\\nu$ to $K[X]$. Theorem 2.9 gives an equivalent valuation-extension form: integral closedness holds exactly when $\\nu$ has $s$ distinct extensions to $K(\\alpha)$, one for each residue factor, and for each repeated factor the value $\\ell_i\\,\\omega_i(\\varphi_i(\\alpha))$ is the minimum positive element of $\\Gamma_\\nu$. When this happens, Corollary 2.10 computes $e(\\omega_i/\\nu)=\\ell_i$ and $f(\\omega_i/\\nu)=\\deg(\\varphi_i)$, and the fundamental inequality becomes an equality.","pith_inferences":["The criterion is computational in practice: for an explicit valued field with decidable value-group order, one can test integral closedness by Euclidean division and Gaussian valuation alone, without constructing the integral closure.","The paper's proof depends on separability through the Henselization correspondence, so a natural stress test is to search for a non-separable $f$ where the remainder condition holds but integral closedness fails; such an example would mark the theorem's boundary.","The equality $e(\\omega_i/\\nu)=\\ell_i$, $f(\\omega_i/\\nu)=\\deg(\\varphi_i)$ suggests that integrally closed orders are exactly those that realize the full splitting of $\\nu$ predicted by the residue factorization, which could guide constructions of extensions with prescribed ramification data."],"forward_implications":["If $\\Gamma_\\nu^+$ has a least element $\\sigma$, the criterion reduces to checking finitely many remainders: $R_\\nu[\\alpha]$ is integrally closed if and only if every repeated-factor remainder has Gaussian value exactly $\\sigma$.","If $\\Gamma_\\nu^+$ has no least element, then integral closedness can occur only in the square-free case $\\ell_i=1$ for all $i$; any repeated residue factor forces $R_\\nu[\\alpha]$ to be non-closed.","When the criterion holds, $\\nu$ splits into exactly $s$ extensions, with ramification index $\\ell_i$ and residue degree $\\deg(\\varphi_i)$ attached to the $i$-th residue factor, so the fundamental inequality becomes an equality.","For a pure extension $X^n-a$ with $a \\in M_\\nu$ and $\\min(\\Gamma_\\nu^+)=\\sigma$, integral closedness of $R_\\nu[\\alpha]$ is equivalent to $\\nu(a)=\\sigma$; in the coprime case the integral closure is exhibited explicitly as $R_\\nu[\\theta^v/\\pi^u]$."],"supporting_citations":[{"why":"Supplies the bijection between valuations extending $\\nu$ and irreducible factors over the Henselization that underlies Lemma 2.1.","marker":"[3, 17.17]"},{"why":"The earlier generalized criterion that this paper re-proves and sharpens.","marker":"[5]"},{"why":"The previous proof that is computationally enhanced; its hidden separability assumption is flagged in the introduction.","marker":"[7, Theorem 1.1]"},{"why":"Gives the integral closure as the intersection of valuation rings, used repeatedly to test membership in $S$.","marker":"[4, Corollary 3.1.4]"},{"why":"The fundamental inequality whose equality is established in Corollary 2.10.","marker":"[4, Theorem 3.3.4]"}],"fun_headline_variants":["Remainder test decides integral closure in valued fields","Gaussian remainders reveal when R[α] is integrally closed","Integral closure test via remainders and Gaussian valuations","Valued-field criterion: check each repeated factor's remainder"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The argument rests on a bijection, supplied by Lemma 2.1, between valuations on $K(\\alpha)$ extending $\\nu$ and irreducible factors of $f$ over the Henselization $K^h$; this bijection is only guaranteed for separable $f$, so if $f$ fails to be separable the whole characterization loses its footing.","fun_headline_variants_meta":{"raw":{"variants":["Remainder test decides integral closure in valued fields","Gaussian remainders reveal when R[α] is integrally closed","Integral closure test via remainders and Gaussian valuations","Valued-field criterion: check each repeated factor's remainder"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000298,"raw_usage":{"total_tokens":1694,"prompt_tokens":884,"completion_tokens":810,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":500,"completion_tokens_details":{"reasoning_tokens":742}},"tokens_in":500,"tokens_out":810,"duration_ms":7511,"temperature":1.0,"reasoning_tokens":742,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-14T12:48:54.721490+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"For $K=F(X,Y)$ with the lexicographic valuation on $\\mathbb{Z}^2$ (where $\\min(\\Gamma_\\nu^+)=(0,1)$) and $f(Z)=Z^3+YZ+X$, the theorem predicts $R_\\nu[\\alpha]$ is not integrally closed because the remainder of $f$ modulo $Z$ is $r=X$ and $\\nu_G(r)=(1,0)\\neq(0,1)$. Directly computing the integral closure of $R_\\nu[Z]/(f)$ in this case, and either exhibiting an element of $S\\setminus R_\\nu[\\alpha]$ or finding $S=R_\\nu[\\alpha]$, would settle the criterion on a concrete instance.","supporting_citations":[{"cited_title":"Ershov, A Dedeking criterion for arbitrary valuation rings , Doklady Math 74 (2006), 650–652","cited_arxiv_id":null,"evidence_quote":"The earlier generalized criterion that this paper re-proves and sharpens."}],"review_version":1}