REVIEW 3 major objections 4 minor 7 references
A New Class of Irreducible Polynomials
T0 review · 3 major / 4 minor · reviewed 2026-08-14 · deepseek-v4-flash
Pith's one-line read The paper proves that a primitive integer polynomial is irreducible if all its zeros lie outside a disc of radius $d$ and its constant term is a prime power times $d$, provided $p^k$ divides a consecutive block of coefficients whose…
desk verdict Plausible extension of Murty–Girstmair to prime powers, but the proof of the key lemma has a false divisibility step, so the main theorems are unproven as written. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The load-bearing device is Lemma 3, a divisibility lemma about the coefficient sequence of a product. It says: if $p^k$ divides the first $j$ coefficients of $f_1 f_2$, $p^{k+1}$ does not divide the constant term, $\gcd(k,j)=1$, and $p$ divides both factor constants, then $p$ divides the next coefficient $a_j$. The proof works by recursively comparing highest powers of $p$ in the convolution formulas $a_t = \sum_i b_i c_{t-i}$, using $\gcd(k,j)=1$ to keep the block of preserved divisibility aligned. The zero-location hypothesis acts as a companion: it guarantees that in any factorization both factor constants exceed $d$ in absolute value, so the prime-power shape of $a_0$ cannot be split without violating Lemma 3 or the size bound.
What would settle it
Take $p=2$, $k=5$, $j=4$, $f_1=2+4x+x^2$, and $f_2=16+16x+8x^2+8x^3$. The product has its first four coefficients divisible by $2^5$, its constant term equals $2^5$, and both factor constants are even, so Lemma 3's hypotheses hold. Yet the proof's intermediate claim that $2^5$ divides $a_3 - b_0 c_3 - b_3 c_0$ is false: $a_3 - b_0 c_3 - b_3 c_0 = 64-16-0 = 48$, and $32$ does not divide $48$. Checking whether a different argument can still force $p \mid a_j$ in this instance would resolve whether the lemma as stated is sound.
Extended reading notes
Core claim
The central claim is Theorem 1 (and its leading-end analogue Theorem 2): a primitive $f \in \mathbb{Z}[x]$ with every zero $\theta$ satisfying $|\theta|>d$ is irreducible in $\mathbb{Z}[x]$ if $a_0 = \pm p^k d$ with $p \nmid d$, and there is $j$ with $\gcd(k,j)=1$ such that $p^k \mid a_0, \dots, a_{j-1}$ and (for $k>1$) $p \nmid a_j$. Theorem 2 is the analogue at the leading coefficient end, with the additional hypothesis $|a_0/q| \le |a_n|$ where $q$ is the smallest prime divisor of $a_0$. The proof supposes a factorization $f_1 f_2$ and uses the zero-location hypothesis to get $|b_0|>d$ and $|c_0|>d$ from the factor constants. Lemma 3 is then invoked to force $p \mid a_j$ whenever both factor constants are divisible by $p$, contradicting the hypothesis; the remaining cases are excluded by the size comparison $|b_0|, |c_0| > d$. The same lemma also yields a Newton-polygon-free proof of the classical prime-power Eisenstein-type theorem (Theorem B).
Load-bearing premise
The whole proof rests on Lemma 3: that whenever $p^k$ divides the first $j$ coefficients of a product (with $p^{k+1}$ not dividing the constant term and $\gcd(k,j)=1$), any factorization whose constant terms are both divisible by $p$ must have $p$ dividing the next coefficient; if this lemma fails, the arguments for both main theorems collapse.
Editorial extensions
If this is right
- Theorems 1 and 2 provide explicit infinite families of irreducible polynomials, such as $X_{j,k}$ and $Y_{j,k,m}$ in the examples, all with zeros outside the unit disc.
- The criterion extends the earlier constant-coefficient theorem from prime values of $|f(0)|/d$ to prime powers, under mild coefficient conditions.
- Lemma 3 gives a proof of the classical prime-power Eisenstein-type theorem (Theorem B) that does not rely on Newton polygons.
- The conditions are purely coefficient-based, so irreducibility can be checked without computing roots or factoring the polynomial.
- Both the constant-coefficient and leading-coefficient versions are covered, and the second theorem requires only an additional divisibility-size comparison.
Reading between the lines
- A natural next step is to ask whether the prime power $p^k$ in Lemma 3 can be replaced by a product of powers of distinct primes, with the block length $j$ coprime to each exponent; the p-adic valuation chase would need a multivalued version.
- Example 3 already shows that the root-location hypothesis can be certified by a coefficient-dominance inequality; combining that observation with Theorems 1 and 2 would yield a fully coefficient-based irreducibility test, which the paper does not spell out.
- The role of the condition $\gcd(k,j)=1$ suggests the criterion might extend to Eisenstein-type polynomials with higher ramification, where the $j$-block corresponds to a jump in a Newton polygon.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The manuscript proposes two irreducibility criteria for primitive integer polynomials whose zeros lie outside the closed disc |z|≤d. Theorem 1 assumes the constant term is ±p^k d, with p^k dividing an initial block a_0,...,a_{j-1}, gcd(k,j)=1, and p∤a_j for k>1; Theorem 2 is the analogous criterion at the leading coefficient. The proofs are built on an auxiliary divisibility lemma, Lemma 3, which is supposed to force p|a_j whenever a product polynomial has p^k dividing its first j coefficients and both factor constants are divisible by p. Several examples are given, including a family X_{j,k} claimed to satisfy Theorem 1.
Significance. If the theorems are correct, they give a natural prime-power extension of the Murty–Girstmair criterion and of Eisenstein–Dumas style arguments, and the paper is self-contained with no fitted parameters or circular dependencies. The intended statements are plausible and the elementary machinery could be of independent interest. However, the current text does not provide valid proofs of the main results: Lemma 3, which is load-bearing for both theorems, contains a false divisibility step, and the proofs of Theorems 1 and 2 have additional structural gaps. The paper therefore needs substantial revision before its claims can be accepted.
major comments (3)
- [Proof of Lemma 3, Case I, Subcase I, even-j paragraph after Eq. (4)] The displayed assertion that p^k divides (a_{j-1} - b_0c_{j-1} - ... - b_{j-1}c_0) = b_{(j-2)/2}c_{j/2} + b_{j/2}c_{(j-2)/2} does not follow from Eq. (4). Equation (4) only controls indices up to κ=(j-2)/2, so it cannot justify subtracting b_0c_{j-1} and b_{j-1}c_0; one would need p^k|b_0c_{j-1} and p^k|b_{j-1}c_0. The step is demonstrably false in a valid instance: take p=2, k=5, j=4, f1=2+4x+x^2, f2=16+16x+8x^2+8x^3. Then a0=32, a1=a2=96, a3=64, a4=40, so the hypotheses of Lemma 3 hold, but a3 - b_0c_3 - b_3c_0 = 64 - 16 - 0 = 48, which is not divisible by 32. Since Theorem 1 and Theorem 2 both invoke Lemma 3 in the p|b_0, p|c_0 case, the main proofs are incomplete as written.
- [Proof of Theorem 1, Eq. (13)] The equality a_m = b_m c_{n-m} is false in general: the coefficient of x^m in f1f2 is the sum Σ_i b_i c_{m-i}, not a single product of the leading coefficient of f1 with a coefficient of f2. The contradiction in the p∤c_0 case therefore does not follow from Eq. (13). A correct argument is available directly from a_0=b_0c_0: if p^k|b_0, then c_0 divides d, so |c_0|≤d, contradicting |c_0|>d from the zero-location assumption. The proof should be rewritten with this substitution; as it stands, the displayed reasoning is invalid.
- [Proof of Theorem 2, second paragraph] The case analysis in the proof of Theorem 2 omits the case p|b_m with p∤c_{n-m}. The proof treats p∤b_m and then jumps to the case p|b_m, p|c_{n-m}. Since a_n=b_m c_{n-m}=±p^k d, the mixed case p|b_m, p∤c_{n-m} is possible and must be handled separately. A symmetric version of the first paragraph, using |c_0|≥q instead of |b_0|≥q, should close the gap, but the argument is absent from the manuscript.
minor comments (4)
- [Section 3, Example 1] The polynomial X_{j,k} in Eq. (14) has a_0=p^{k+1}=p^k·p, so the parameter d in Theorem 1 is d=p, not d=1. The proof in Eqs. (15)-(16) only establishes |ζ|>1 for each zero, which is weaker than the required |ζ|>p. Thus the example does not instantiate Theorem 1 as written.
- [Section 3, Example 2] The statement that Y_{j,k,m} satisfies the hypotheses of Theorem 2 is asserted without verifying the zero-location condition |θ|>d for its zeros; an explicit verification, or a reference to a general lemma, is needed.
- [Proof of Lemma 3, definition of ℓ] The sentence introducing ℓ with ℓ≤k−ℓ should state explicitly that the factors f1 and f2 may be swapped so that v_p(b_0)≤v_p(c_0); otherwise the inequality ℓ≤k−ℓ is not guaranteed by the hypotheses alone.
- [Proof of Theorem 1, reference to Eq. (4)] The phrase 'in view of (4)' at the start of the proof of Theorem 1 is confusing because Eq. (4) belongs to the proof of Lemma 3 and concerns p-adic valuations of intermediate coefficients; the intended reference appears to be to the factorization a_0=b_0c_0.
Circularity Check
No significant circularity: the paper's irreducibility proofs rest on in-paper elementary divisibility lemmas and standard external criteria, not on fitted parameters or self-cited load-bearing results.
full rationale
The paper derives its main theorems from hypotheses on coefficient divisibility and zero locations. Theorem A is proved directly by comparing |f1(0)| = |bm| * product |theta| with d using zero-location hypotheses and the factorization a0 = b0 c0 = ±pd. Theorems 1 and 2 are proved by assuming a nontrivial factorization, using the zero-location hypothesis to force |b0| > d and |c0| > d, and then applying Lemma 3, which is an elementary divisibility lemma proved within the paper using valuations and the coefficient convolution formula. There are no fitted parameters, no empirical predictions, and no self-citations; the references are to classical criteria (Schönemann, Eisenstein, Dumas, Murty, Girstmair, Perron) and are not load-bearing in a circular way. The alleged flaw in Lemma 3's proof, if real, is a mathematical correctness issue rather than a circularity issue: the lemma is not assumed as an input equivalent to the theorem, and the main results would be repairable by an independent proof of the lemma. Therefore the derivation chain is not circular, and the appropriate score is 0.
Assumptions & free parameters
assumptions (3)
- standard math Every nonconstant complex polynomial has roots, and for f1 in Z[x], |f1(0)| equals the absolute leading coefficient times the product of root magnitudes.
- standard math If a prime p divides a product of integers, it divides one factor, and p-adic valuations add across products.
- standard math Triangle inequality for complex numbers, used to show coefficient dominance implies all roots lie outside a disc.
Cite this review
Pith. "Pith review of A New Class of Irreducible Polynomials." pith.science (2026). https://pith.science/paper/KPSSB3GS
@misc{pith2026190805587,
author = {Pith},
title = {Pith review of: A New Class of Irreducible Polynomials},
year = {2026},
howpublished = {\url{https://pith.science/paper/KPSSB3GS}},
note = {Machine review of arXiv:1908.05587}
}
read the original abstract
In this article, we propose a few sufficient conditions on polynomials having integer coefficients all of whose zeros lie outside a closed disc centered at the origin in the complex plane and deduce the irreducibility over the ring of integers.
Reference graph
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work page 1907
Reviewed August 14, 2026 · model on record in the stance chip above.
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