A simpler proof that there are no Landau--Siegel zeros

Abstract: This paper simplifies the recent OpenAI proof that there are no Landau--Siegel zeros.

• The paper presents a shorter proof establishing the Landau–Siegel zero exclusion theorem, demonstrating that there is an absolute constant c > 0 such that if β ∈(0,1) is a real zero of L(s, χ), then (1−β)log |q| ≥ c., where χ(n) = (q/n) is associated with a fundamental discriminant q.
• The simplified approach leverages the pigeonhole principle on a lattice of polynomials and Masser's theorem, eliminating the need for interpolation and determinant estimates.
• Critical to the proof is the application of an intersection between a prime bias analytic input and an algebraic construction that takes advantage of numerous primes with χ(p) = -1.

Main result and proof strategy

The paper presents a shorter proof of the nonexistence of Landau–Siegel zeros for primitive nonprincipal real Dirichlet characters. Its principal theorem asserts that there is an absolute constant $c>0$ such that, whenever $\chi(n)=(q/n)$ is associated with a fundamental discriminant $q$ and $\beta\in(0,1)$ is a real zero of $L(s,\chi)$,

$(1-\beta)\log |q|\ge c.$

Thus the result gives the standard effective-looking zero-free region at the Landau–Siegel scale: a real zero cannot approach $1$ more rapidly than a constant multiple of $1/\log |q|$. The paper’s stated simplification concerns the algebraic construction used to produce an auxiliary polynomial. In place of interpolation and determinant estimates, it applies the pigeonhole principle to a lattice of polynomials and then invokes a theorem of Masser to obtain the contradiction.

The proof has two inputs. The analytic input bounds a weighted prime sum over primes for which $\chi(p)=1$:

$$\sum_{\substack{p\le X\ \chi(p)=1}}\frac{\log p}{p} \ll \log |q|+(1-\beta)(\log X)^2.$$

The algebraic input shows that, if the complementary set of primes with $\chi(n)=(q/n)$0 is sufficiently large in the weighted sense, then there exists a nonzero polynomial of controlled weighted degree that vanishes on a four-dimensional box of algebraic points. Masser’s theorem then forces the total degree of this polynomial to be too large, yielding a contradiction.

The analytic prime-bias input

The argument begins with a consequence of Davenport’s treatment of exceptional zeros. If $\chi(n)=(q/n)$1 is a real zero of $\chi(n)=(q/n)$2, then for every $\chi(n)=(q/n)$3,

$\chi(n)=(q/n)$4

The proof uses this estimate indirectly. By Mertens’ estimate, the unrestricted sum over primes is approximately $\chi(n)=(q/n)$5. Removing the primes dividing $\chi(n)=(q/n)$6 and subtracting the contribution from $\chi(n)=(q/n)$7 leaves a lower bound for the primes satisfying $\chi(n)=(q/n)$8:

$\chi(n)=(q/n)$9

Consequently, if $q$0 is sufficiently small, one can choose $q$1 large enough relative to $q$2 so that the $q$3 primes carry substantial weighted mass. This is the only analytic information about the exceptional zero used in the proof.

The argument is therefore a quantitative contradiction between two mechanisms. The analytic estimate supplies many primes at which Frobenius acts nontrivially on $q$4, while the algebraic construction converts those Frobenius congruences into excessive vanishing of an auxiliary polynomial.

The quartic field and Frobenius congruences

The algebraic setting is the biquadratic field

$q$5

with order

$q$6

The four embeddings of $q$7 are encoded through the automorphisms

$q$8

and

$q$9

For

$\beta\in(0,1)$0

the Frobenius action modulo a prime $\beta\in(0,1)$1 satisfies

$\beta\in(0,1)$2

For primes with $\beta\in(0,1)$3, this yields either

$\beta\in(0,1)$4

when $\beta\in(0,1)$5, or

$\beta\in(0,1)$6

when $\beta\in(0,1)$7.

These congruences motivate evaluating polynomials at the triples

$\beta\in(0,1)$8

The construction is designed so that factors of the form $\beta\in(0,1)$9 or $L(s,\chi)$0 vanish modulo primes with $L(s,\chi)$1. This is the central algebraic use of the character values.

The weighted polynomial lattice

For a degree parameter $L(s,\chi)$2, the paper considers the $L(s,\chi)$3-lattice $L(s,\chi)$4 generated by monomials

$L(s,\chi)$5

The asymmetric weight is essential. Powers of $L(s,\chi)$6 and $L(s,\chi)$7 are assigned weight $L(s,\chi)$8, whereas powers of $L(s,\chi)$9 have weight $(1-\beta)\log |q|\ge c.$0. This permits many factors involving $(1-\beta)\log |q|\ge c.$1 or $(1-\beta)\log |q|\ge c.$2 while keeping the weighted degree bounded.

For primes $(1-\beta)\log |q|\ge c.$3, the monomial basis can be replaced by the basis

$(1-\beta)\log |q|\ge c.$4

where $(1-\beta)\log |q|\ge c.$5 or $(1-\beta)\log |q|\ge c.$6. The change of basis is triangular with diagonal entries equal to one. If $(1-\beta)\log |q|\ge c.$7 is a prime with $(1-\beta)\log |q|\ge c.$8, the corresponding factor $(1-\beta)\log |q|\ge c.$9 vanishes at the selected evaluation points modulo $1$0. Repeated factors produce divisibility by powers of $1$1.

If

$1$2

then the number of possible residue vectors of evaluations modulo $1$3 is reduced by a factor of $1$4. The exponent $1$5 reflects the degree $1$6.

This is the point at which the proof departs from an interpolation-based approach. Rather than constructing a polynomial by solving a system of linear equations and estimating a determinant, it counts bounded-height polynomials and compares that count with the number of possible evaluation residue vectors.

Pigeonhole construction of the vanishing polynomial

The parameters are chosen as

$1$7

and

$1$8

The relevant prime set is

$1$9

The key sufficient condition is

$1/\log |q|$0

Under this condition, the pigeonhole argument produces a nonzero $1/\log |q|$1 satisfying

$1/\log |q|$2

|f|_B

\max_{i+8(j+k)\le D} |c_{i,j,k}|B^{i+j+k}.

$1/\log |q|$3

P=\prod_{p\in\mathcal S}p^{k_p}.

$1/\log |q|$4

k_p=\max{1,\lfloor D/p\rfloor}, \qquad H=\frac{P}{4r},

$1/\log |q|$5

|F(\theta_n,\sigma\theta_n,\sigma\tau\theta_n)| \le \frac P2.

$1/\log |q|$6

r=\frac{D^3}{384}+O(D^2),

$1/\log |q|$7

W=\frac{5D^4}{6144}+O(D^3),

$1/\log |q|$8

E_p=\frac{D^4}{1536p}+O(D^3).

$1/\log |q|$9

\sum_{p\in\mathcal S}E_p\log p

\frac{D^4}{1536} \sum_{p\in\mathcal S}\frac{\log p}{p} +O(D^3Q).

$\chi(p)=1$0

\sum_{j=1}^s m_j\gamma_j, \qquad 0\le m_j\le M,

$\chi(p)=1$1

(M/n)^{\mu(\Gamma)},

$\chi(p)=1$2

\gamma_1=(1,1,1),

$\chi(p)=1$3

\gamma_2=(\sqrt q,-\sqrt q,-\sqrt q),

$\chi(p)=1$4

\gamma_3=(\sqrt2,\sqrt2,-\sqrt2),

$\chi(p)=1$5

\gamma_4=(\sqrt{2q},-\sqrt{2q},\sqrt{2q}).

$\chi(p)=1$6

(\theta_n,\sigma\theta_n,\sigma\tau\theta_n)

n_1\gamma_1+n_2\gamma_2+n_3\gamma_3+n_4\gamma_4.

$\chi(p)=1$7

4\sqrt{2q}\ne 0.

$\chi(p)=1$8

(\sqrt{2q},\sqrt2,-\sqrt q,-1).

$\chi(p)=1$9

\mu(\Gamma)\ge \frac43.

$$\sum_{\substack{p\le X\ \chi(p)=1}}\frac{\log p}{p} \ll \log |q|+(1-\beta)(\log X)^2.$$0

\deg F \ge \left(\frac{N-1}{3}\right)^{4/3}.

$$\sum_{\substack{p\le X\ \chi(p)=1}}\frac{\log p}{p} \ll \log |q|+(1-\beta)(\log X)^2.$$1

D= \left\lfloor \frac12\left(\frac{N-1}{3}\right)^{4/3} \right\rfloor,

$$\sum_{\substack{p\le X\ \chi(p)=1}}\frac{\log p}{p} \ll \log |q|+(1-\beta)(\log X)^2.$$2

\deg F>D,

$$\sum_{\substack{p\le X\ \chi(p)=1}}\frac{\log p}{p} \ll \log |q|+(1-\beta)(\log X)^2.$$3

which contradicts the construction.

This contradiction is obtained whenever $$\sum_{\substack{p\le X\ \chi(p)=1}}\frac{\log p}{p} \ll \log |q|+(1-\beta)(\log X)^2.$$4 is sufficiently small. The proof first takes $$\sum_{\substack{p\le X\ \chi(p)=1}}\frac{\log p}{p} \ll \log |q|+(1-\beta)(\log X)^2.$$5 sufficiently large, then chooses the permissible smallness threshold for $$\sum_{\substack{p\le X\ \chi(p)=1}}\frac{\log p}{p} \ll \log |q|+(1-\beta)(\log X)^2.$$6, and finally assumes $$\sum_{\substack{p\le X\ \chi(p)=1}}\frac{\log p}{p} \ll \log |q|+(1-\beta)(\log X)^2.$$7 sufficiently large. The finitely many remaining discriminants can be absorbed into the absolute constant in the theorem.

Limitations and open questions

The paper’s simplification depends on several structural inputs that are not reproved in full. The analytic inequality for the prime bias is imported from Davenport, while the polynomial-degree lower bound is a specialized consequence of Masser’s theorem. The proof therefore simplifies the auxiliary-polynomial step rather than replacing the established analytic and transcendence-theoretic components.

The exposition also relies on asymptotic estimates stated uniformly over $$\sum_{\substack{p\le X\ \chi(p)=1}}\frac{\log p}{p} \ll \log |q|+(1-\beta)(\log X)^2.$$8 and on the assertion that the displayed numerical margins dominate all error terms for sufficiently large $$\sum_{\substack{p\le X\ \chi(p)=1}}\frac{\log p}{p} \ll \log |q|+(1-\beta)(\log X)^2.$$9. The argument is asymptotic and does not provide an explicit numerical value of the absolute constant $\chi(n)=(q/n)$00. In addition, the reduction from the large-$\chi(n)=(q/n)$01 case to all fundamental discriminants is stated rather than quantified.

A specific question left open by the proof is whether the pigeonhole construction can be optimized to yield explicit constants, smaller extension fields, or weaker weighted-degree losses than the factor $\chi(n)=(q/n)$02 assigned to $\chi(n)=(q/n)$03 and $\chi(n)=(q/n)$04. The paper establishes that these choices suffice, but does not claim that they are optimal.

Conclusion

The paper gives a concise algebraic proof of the Landau–Siegel zero exclusion theorem. Its central contribution is methodological: a pigeonhole argument on a weighted polynomial lattice replaces interpolation and determinant estimates. Frobenius congruences at primes with $\chi(n)=(q/n)$05 create strong modular restrictions on polynomial evaluations; the analytic prime-bias estimate ensures that enough such primes are available; and Masser’s theorem shows that a nonzero polynomial vanishing on the resulting four-dimensional box would require degree exceeding the construction’s upper bound. The resulting contradiction establishes $\chi(n)=(q/n)$06 for an absolute constant $\chi(n)=(q/n)$07 (2610.21099).