\documentclass[11pt]{article}
\usepackage[a4paper,margin=24mm,top=22mm,bottom=23mm]{geometry}
\usepackage[T1]{fontenc}
\usepackage{amsmath,amsthm}
\usepackage{newtxtext,newtxmath}
\usepackage{microtype}
\PassOptionsToPackage{hyphens}{url}
\usepackage[hidelinks]{hyperref}
\urlstyle{same}
\hypersetup{pdfauthor={Roelof van Dijk},
  pdftitle={A Frobenius determinant proof of the non-existence of Landau--Siegel zeros}}
\setlength{\parindent}{1.1em}
\setlength{\parskip}{3pt}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}
\newcommand{\Z}{\mathbb Z}
\newcommand{\Q}{\mathbb Q}
\newcommand{\F}{\mathbb F}
\newcommand{\Nm}{\operatorname{N}}
\begin{document}
\begin{center}
{\Large\bfseries A Frobenius determinant proof}\\[3pt]
{\Large\bfseries of the non-existence of Landau--Siegel zeros}\\[9pt]
Roelof van Dijk
\end{center}

\begin{abstract}
We give a variant of the OpenAI proof~\cite{OpenAI} of the non-existence
of Landau--Siegel zeros, following the simplifications of
Kerr~\cite{Kerr}, Green~\cite{Green}
and Barreto~\cite{Barreto}.
The Frobenius congruence proves both divisibility and nonvanishing of
the determinant. Reduction modulo the largest inert prime below the
cutoff reduces nonvanishing to a polynomial root count, replacing
resultants and Wronskians.
The argument was discovered in discussions with GPT-6 Astra and Opus 5.5
and the new arguments should be attributed to these models.
\end{abstract}

\begin{theorem}\label{elementary}
Let $d$ be a nonsquare integer, and put
\[
 I_d(x)=\sum_{\substack{p\le x\\p\ \text{odd}\\(d/p)=-1}}\frac{\log p}{p}.
\]
There is an absolute effective constant $C$ such that, for every $x\ge2$,
\begin{equation}\label{bound}
 I_d(x)\le\frac12\log x+\frac14\log|d|+\log\log(3x)+C.
\end{equation}
\end{theorem}

\begin{corollary}\label{zeros}
Let $\chi$ be a primitive nonprincipal real Dirichlet character of
conductor $q$. Every real zero $\beta\in(0,1)$ of $L(s,\chi)$ satisfies
\[
 (1-\beta)\log q
 \ge1-O\!\left(\frac{\log\log(3q)}{\log q}\right).
\]
In particular, $(1-\beta)\log q\ge c$ with an absolute effective $c>0$.
\end{corollary}

The choice of the largest inert prime makes the elementary bound uniform
in $x$. If $R$ is the largest odd inert prime at most $x$, then
$I_d(x)=I_d(R)$: estimating the sum at $R$ therefore suffices for every
cutoff, without information on the spacing of inert primes.
Reduction modulo $R$ proves nonvanishing; reduction at all inert primes
provides divisibility. The proof compares that divisibility with the
size of the same determinant.

\begin{proof}[Proof of Theorem~\ref{elementary}]
We use Chebyshev's bound, Mertens' estimate, and their elementary consequence
\[
 \sum_{p\le y}\log p\ll y,\qquad
 \sum_{p\le y}\frac{\log p}{p}=\log y+O(1),\qquad
 \sum_{p\ge y}\frac{\log p}{p^2}\ll y^{-1}.
\]
The last estimate follows from the first by partial summation.
All implied constants below are absolute and effective.
If there is no odd inert prime at most $x$, there is nothing to prove.
Otherwise let $R$ be the largest such prime. Then $I_d(x)=I_d(R)$.
If $R<e^3$ or $R\le\sqrt{|d|}$, Mertens' estimate proves
\eqref{bound}. Hence assume $R\ge e^3$ and $R>\sqrt{|d|}$, and put
$L=\log R$, $S=\lceil R/L\rceil$ and $N=RS$.

Take all $\alpha=u+v\sqrt d$ with integers
\[
 0\le u<\sqrt R\,|d|^{1/4},\qquad
 0\le v<\sqrt R\,|d|^{-1/4}.
\]
The side lengths multiply to $R$, so their ceilings give at least $R$
integer points. They are balanced so that $u$ and $v\sqrt{|d|}$ have
the same upper bound $\sqrt R\,|d|^{1/4}$.
Thus both $\alpha$ and its quadratic conjugate
$\bar\alpha$ have modulus less than $H=2\sqrt R\,|d|^{1/4}\le2R$.
For every odd inert prime $p$, the binomial theorem, Fermat's little
theorem and Euler's criterion give
\begin{equation}\label{frobenius}
 \alpha^p=(u+v\sqrt d)^p
 \equiv u+v\,d^{(p-1)/2}\sqrt d
 \equiv u-v\sqrt d=\bar\alpha
 \pmod{p\Z[\sqrt d]}.
\end{equation}
Form the matrix
\[
 \mathcal A_{(r,s),(\alpha,j)}
   =(\partial_z^j z^rw^s)(\alpha,\bar\alpha),
 \qquad 0\le r<R,\quad 0\le s<S,\quad 0\le j<S.
\]
Its $N$ rows are the monomials $z^rw^s$; its columns are the chosen
points and derivative orders.

\emph{Nonvanishing.}
Modulo $R$, the ring $\Z[\sqrt d]$ is the field $\F_{R^2}$,
and \eqref{frobenius} gives $\bar\alpha=\alpha^R$.
The chosen points remain distinct because their integer coordinates
lie in $[0,R)$ (as $R>\sqrt{|d|}$), and $1,\sqrt d$ form an $\F_R$-basis.
A nontrivial row dependence over $\F_{R^2}$ would therefore give
\[
 G(z)=\sum_{r<R,\,s<S}c_{rs}z^{r+Rs}\in\F_{R^2}[z],
 \qquad G\ne0,\qquad \deg G<RS.
\]
The exponents are distinct, and $(z^R)'=0$ shows that the derivatives
of orders $0,\ldots,S-1$ of $G$ vanish at every chosen point.
Since $S<R$, the factorials of these orders are invertible.
Thus $G$ has at least $R$ distinct roots, each of multiplicity at least
$S$, contradicting $\deg G<RS$.
Choose an $N\times N$ minor $\Delta\in\Z[\sqrt d]$ nonzero modulo $R$.
It uses every row, and
\[
 \mathcal N=|\Nm_{\Q(\sqrt d)/\Q}\Delta|\in\Z_{>0}.
\]

\emph{Divisibility.}
For an odd inert prime $p$, put $h=\min(\lfloor r/p\rfloor,S-1-s)$
and replace each row by
\[
 z^rw^s\longmapsto z^{r-ph}(z^p-w)^h w^s.
\]
The condition $s+h<S$ keeps every term among the original rows.
All added terms have smaller $z$-degree, so this unitriangular change
preserves $\Delta$.
By \eqref{frobenius}, an undifferentiated factor $z^p-w$ is divisible
by $p$ at every point. Every positive $z$-derivative of this factor
has coefficients divisible by $p$.
In each term of Leibniz's rule, each of the $h$ copies of $z^p-w$
therefore contributes a factor $p$, whether differentiated or not.
Thus every derivative of the new row is divisible by $p^h$.
Extracting these row factors gives $p^{E_p}\mid\Delta$, where
$E_p=\sum_{r<R,\,s<S}h$. Conjugation also gives
$p^{E_p}\mid\bar\Delta$, hence $p^{2E_p}\mid\mathcal N$. The exponent satisfies
\[
 E_p=\sum_{r<R,\,s<S}\min(\lfloor r/p\rfloor,S-1-s)
 \ge\frac{SR(R-1)}{2p}-SR-\frac{R^3}{3p^2}.
\]
Indeed, $\sum_{s<S}\min(a,S-1-s)\ge Sa-a^2$ for integers $a\ge0$;
now use $\lfloor r/p\rfloor\ge r/p-1$.
We sum over $L\le p\le R$. The cutoff makes the quadratic error
$R^3\sum_{p\ge L}(\log p)/p^2\ll R^3/L\le NR$.
Chebyshev bounds the floor error by $O(NR)$, and Mertens allows us
to replace $R-1$ by $R$ with the same error. Consequently,
\begin{equation}\label{lower}
 \log\mathcal N\ge NR
 \sum_{\substack{L\le p\le R\\p\ \text{odd}\\(d/p)=-1}}
 \frac{\log p}{p}-O(NR).
\end{equation}

\emph{Size.}
Under either embedding, an entry in row $(r,s)$ has modulus at most
$R^S H^{r+s}$. Expanding the two conjugate determinants gives
\[
 \log\mathcal N\le2\log(N!)+2NS\log R+N(R+S-2)\log H
          \le NR\log H+O(NR),
\]
since $S=\lceil R/\log R\rceil$ and $H\le2R$.
Compare with \eqref{lower} and restore the primes below $L$ by Mertens:
\[
 I_d(x)=I_d(R)\le\log H+\log L+O(1)
 =\frac12\log R+\frac14\log|d|+\log\log R+O(1).
\]
Since $R\le x$, this proves \eqref{bound}.
\end{proof}

\begin{proof}[Proof of Corollary~\ref{zeros}]
Choose squarefree $d\ne1$ with $|d|\le q$ and
$\chi(p)=(d/p)$ for odd primes $p$.
For $0<\eta\le1$, partial summation in \eqref{bound}, absorbing the
possible contribution of $p=2$ into the error term, gives
\[
 \sum_{\chi(p)=-1}\frac{\log p}{p^{1+\eta}}
 \le\frac1{2\eta}+\frac14\log q+O(\log(2/\eta)).
\]
Here the logarithmic error follows from Jensen's inequality after
$u=\log t$; the measure $\eta e^{-\eta u}\,du$ has mean $1/\eta$.
Higher prime powers contribute $O(1)$ uniformly in $\eta$, so the Euler
product for $\zeta$ gives
\[
 \sum_p\frac{\log p}{p^{1+\eta}}
 =-\frac{\zeta'}\zeta(1+\eta)+O(1)=\frac1\eta+O(1).
\]
Since $\chi(p)$ is $-1$, $0$ or $1$, the Euler product for $L$ gives
\begin{align*}
 \frac{L'}L(1+\eta,\chi)
 &=-\sum_p\frac{\chi(p)\log p}{p^{1+\eta}}+O(1)\\
 &=2\sum_{\chi(p)=-1}\frac{\log p}{p^{1+\eta}}
   -\sum_p\frac{\log p}{p^{1+\eta}}
   +\sum_{p\mid q}\frac{\log p}{p^{1+\eta}}+O(1).
\end{align*}
The terms $2\cdot(2\eta)^{-1}$ and $\eta^{-1}$ cancel.
Using $\sum_{p\mid q}(\log p)/p\ll\log\log(3q)$, we obtain
\[
 \frac{L'}L(1+\eta,\chi)
 \le\frac12\log q
       +O\bigl(\log(2/\eta)+\log\log(3q)\bigr).
\]
Finally, the Hadamard formula~\cite[Cor.~10.18]{MV}, retaining $\beta$, gives
\[
 \frac1{1-\beta+\eta}
 \le\frac{L'}L(1+\eta,\chi)+\frac12\log q+O(1)
 \le\log q+O\bigl(\log(2/\eta)+\log\log(3q)\bigr).
\]
Take $\eta=(\log q)^{-2}$ and invert.
For the finitely many smaller conductors, nonvanishing at $1$
gives a smaller effective positive $c$.
\end{proof}

Tao~\cite{Tao} asks for explicit bounds on a hypothetical tenth
class-number-one discriminant obtained from arguments excluding
Siegel zeros. The determinant above gives the following result.

\begin{corollary}[Tenth-discriminant bound]\label{tenth}
Let $K$ be an imaginary quadratic field and put
$D=|\operatorname{disc}K|$. Then
\begin{equation}\label{tenth-bound}
 h(K)=1\quad\Longrightarrow\quad D<8\cdot10^9.
\end{equation}
\end{corollary}
Here $D$ is the absolute field discriminant; the fields $\Q(i)$ and
$\Q(\sqrt{-2})$ have $D=4$ and $D=8$, respectively.

\begin{proof}
The class-number-one hypothesis supplies the required primes directly.
Every odd prime $p<D/4$ is inert: otherwise a prime ideal of norm $p$
would be principal, giving an algebraic integer of norm $p$, hence
$4p=u^2+Dv^2$ with $u,v\in\Z$. The inequality $4p<D$ forces $v=0$,
which would make $p$ a square.

Suppose $D\ge8\cdot10^9$. Bertrand's theorem gives an odd prime
$D/8<R<D/4$, so $R\ge10^9$ and every odd prime at most $R$ is inert.
Put $S=\lfloor R/27\rfloor$, $a=(R/8)^{1/4}$,
$V=\lceil a\rceil$ and $U=\lceil R/V\rceil$.
The cutoff $27$ balances the $1/c$ coefficient in the size bound for
$S=\lfloor R/c\rfloor$ against the small-prime loss.
We have $a\ge100$ and
\[
 U\le8a^3+1,\qquad V\le a+1,\qquad D\le64a^4.
\]
Moreover, $UV\ge R$ and $U,V\le R$.
Choose $R$ points from the centered $U\times V$ integer rectangle.
The coordinate ranges have lengths at most $R$, so the points remain
distinct modulo $R$. Both conjugates have
modulus at most
\[
 B=5.7a^3,\qquad \log B\le\tfrac34\log R+0.182.
\]
Indeed, the displayed bounds give
$4|u+v\sqrt{-D}|^2\le U^2+DV^2\le4(5.7a^3)^2$.
Use every derivative order $j<S$ at each point and apply the Frobenius
congruence~\eqref{frobenius} with $d=-D$. Since $S<R$, the same root
count proves that the resulting square determinant $\Delta$ is nonzero.
Write $\mathcal N=|\Nm\Delta|$.

The total derivative order over the columns equals the total $w$-degree over
the rows:
\[
 \sum_{\text{columns}}j
 =R\sum_{j=0}^{S-1}j=\frac{RS(S-1)}2
 =\sum_{r<R,\,s<S}s.
\]
Hence the exponent of $B$ in each determinant term is
\[
 \sum_{r<R,\,s<S}(r+s)-\sum_{\text{columns}}j
 =\frac{RS(R-1)}2.
\]
The size bound is therefore
\[
 \log\mathcal N\le2\log((RS)!)+RS(R-1)\log B
                         +RS(S-1)\log R.
\]
At each odd prime $p\le R$, the row change gives
$p^{2E_p}\mid\mathcal N$, with
\[
 E_p=\sum_{k=1}^{S-1}(S-k)\max(R-kp,0).
\]
For $p<27$ this is an arithmetic sum, since $pS\le R$.
For $p\ge27$, Legendre's formula for the prime
valuations of $\prod_{r<R}r!$ supplies the lower bound.
With $t=\log R$ and $\ell=\log\mathcal N/(R^2S)$,
Appendix~\ref{appendix:constants} derives the estimates
\begin{equation}\label{explicit-comparison}
 t-\frac{2103}{500}\le\ell
 \le\frac{85}{108}t+\frac{23}{125}.
\end{equation}
Comparing the bounds gives
\[
 \frac{23}{108}t\le\frac{439}{100}.
\]
Yet $R\ge10^9$ gives $t\ge207/10$, so
\[
 \frac{23}{108}t\ge\frac{529}{120}>\frac{439}{100},
\]
a contradiction.
\end{proof}

The bound is small enough to finish by a finite residue sieve.
For $D>8$, class number one also forces $D\equiv3\pmod8$.
Testing inertness at odd primes $p\le251$ with $4p<D$
leaves only $D=11,19,43,67,163$ in $8<D<8\cdot10^9$.
Together with the small discriminants and the converse from Minkowski's
bound, this gives
\[
 h(K)=1\quad\Longleftrightarrow\quad
 D\in\{3,4,7,8,11,19,43,67,163\}.
\]
{\small
\begin{thebibliography}{6}
\setlength{\itemsep}{0pt}
\bibitem{OpenAI} OpenAI, \emph{Uniform exclusion of Landau--Siegel zeros},
OpenAI Math Release, 1 October 2026.
\url{https://github.com/openai/math/blob/main/preprints/Uniform-exclusion-of-Landau-Siegel-zeros-October-1-2026/paper.pdf}.
\bibitem{Kerr} B.~Kerr, \emph{A simpler proof that there are no Landau--Siegel
zeros}, arXiv:2610.21099v1, 2026.
\url{https://arxiv.org/abs/2610.21099v1}.
\bibitem{Green} B.~Green, \emph{On Landau--Siegel zeros},
hexagon:2610.00311v2, 2026.
\url{https://hexagonmath.org/pdf/2610.00311v2}.
\bibitem{Barreto} K.~Barreto, \emph{An elementary proof of the non-existence
of Landau--Siegel zeros}, manuscript.
\url{https://x.com/AcerFur/status/2112239827077026128}.
\bibitem{MV} H.~L. Montgomery and R.~C. Vaughan,
\emph{Multiplicative Number Theory I: Classical Theory},
Cambridge University Press, 2006.
\bibitem{Tao} T.~Tao, \emph{An experimental AI-math challenge: using tenth
discriminant bounds to explore arguments ruling out Siegel zeroes},
What's New, 20 October 2026.
\url{https://terrytao.wordpress.com/2026/10/20/an-experimental-ai-math-challenge-using-tenth-discriminant-bounds-to-explore-arguments-ruling-out-siegel-zeroes/}.
\end{thebibliography}
}

\clearpage
\appendix
\section{Full derivation of Corollary~\ref{tenth}}\label{appendix:constants}
We derive \eqref{explicit-comparison}. Throughout this appendix,
$R\ge10^9$, $S=\lfloor R/27\rfloor$, $t=\log R$ and
$\ell=\log\mathcal N/(R^2S)$, where $\mathcal N$ is the determinant
norm in the proof of Corollary~\ref{tenth}.

\emph{Upper bound.}
The inequality $\log u\le u-1$ gives $t\le R/10^9+20$, hence
$t/R\le21/10^9$.
Since $S\le R/27$ and $\log((RS)!)\le RS\log(RS)\le2RSt$,
the determinant size bound gives
\[
 \ell\le\log B+\frac{t}{27}+\frac{4t}{R}
 \le\frac{85}{108}t+0.182+\frac{84}{10^9}
 <\frac{85}{108}t+\frac{23}{125}.
\]

\emph{Lower bound.}
For $p\le23$, we have $pS\le R$, so summing the formula for $E_p$ gives
\[
 E_p=S(S-1)\left(\frac R2-\frac{p(S+1)}6\right).
\]
For $p\ge29$, put $A_p=\sum_{r<R}\lfloor r/p\rfloor$.
The loss from truncating the $w$-degree satisfies
\[
 0\le SA_p-E_p
 \le\frac12\sum_{r<R}
       \left(\lfloor r/p\rfloor^2+\lfloor r/p\rfloor\right)
 \le\frac{R^3}{6p^2}+\frac{R^2}{4p}.
\]
To sum $A_p\log p$, use Legendre's identity
\[
 F:=\sum_{r<R}\log(r!)
 =\sum_{p\le R}\log p\sum_{r<R}\sum_{\nu\ge1}
       \left\lfloor\frac r{p^\nu}\right\rfloor.
\]
Integral comparison, using $\log(r!)\ge r\log r-r$ for $r\ge1$, gives
\[
 F\ge\frac{R^2}{2}t-\frac{3R^2}{4}-Rt.
\]
We use the following prime sums:
\begin{align*}
 C_0&:=\sum_{p\le23}\frac{\log p}{p-1}\le2.906,
 & A&:=\sum_{3\le p\le23}\log p\in[18.529,20],\\
 C_1&:=\sum_{29\le p\le R}\frac{\log p}{p(p-1)}\le0.037209,
 & W&:=\sum_{3\le p\le23}p\log p\le260.911,\\
 C_2&:=\sum_{29\le p\le R}\frac{\log p}{p^2}\le0.036547.
\end{align*}
For $C_1$ and $C_2$, sum over primes up to $1000$ and bound the tails
by partial summation from the elementary Chebyshev estimate
$\vartheta(y):=\sum_{p\le y}\log p\le2(\log2)y$:
\[
 \sum_{p>1000}\frac{\log p}{p^2}\le\frac{4\log2}{1000},
 \qquad
 \sum_{p>1000}\frac{\log p}{p(p-1)}\le\frac{4\log2}{999}.
\]
All the finite numerical bounds above follow from rational bounds
for the logarithms. Write $p=2^b y$ with $1\le y<2$, and use
\[
 0\le\log y-2\sum_{j=0}^{11}\frac{z^{2j+1}}{2j+1}
 \le\frac{2z^{25}}{25(1-z^2)},\qquad z=\frac{y-1}{y+1},
\]
with the same formula at $y=2$ for $\log2$.

In Legendre's identity, all powers of primes $p\le23$ contribute
at most $R^2C_0/2$, and powers $p^\nu$ with $p\ge29$, $\nu\ge2$
contribute at most $R^2C_1/2$. Thus
\[
 \sum_{29\le p\le R}A_p\log p
 \ge F-\frac{R^2}{2}(C_0+C_1).
\]
Also, $\sum_{p\le R}\lfloor R/p\rfloor\log p\le\log(R!)$
and the same Chebyshev estimate give
\[
 \sum_{p\le R}\frac{\log p}{p}
 \le\frac{\log(R!)+\vartheta(R)}{R}
 \le t+2\log2<t+20.
\]
Combining the small- and large-prime contributions to
$\log\mathcal N\ge2\sum_{3\le p\le R}E_p\log p$ now gives
\begin{align*}
 \ell\ge{}&t-\frac32-\frac{2t}{R}-C_0-C_1
       -\frac{R}{3S}C_2-\frac{t+20}{2S}\\
 &\hspace{20mm}+\frac{S-1}{R}A-\frac{S^2-1}{3R^2}W.
\end{align*}
Since $R/27-1<S\le R/27$, the rounding terms satisfy
\begin{align*}
 \varepsilon
 &:=\frac{2t}{R}+\left(\frac{R}{3S}-9\right)C_2
       +\frac{t+20}{2S}+\frac{2A}{R}\\
 &\le\frac{82}{10^9}
       +\frac{243\cdot0.036547}{10^9-27}
       +\frac{1107}{2(10^9-27)}<10^{-6}.
\end{align*}
Consequently,
\begin{align*}
 \ell&\ge t-\left(\frac32+C_0+C_1+9C_2-\frac{A}{27}
                         +\frac{W}{2187}+\varepsilon\right)\\
 &>t-\left(\frac32+2.906+0.037209+9\cdot0.036547
             -\frac{18.529}{27}+\frac{260.911}{2187}+10^{-6}\right)\\
 &>t-\frac{2103}{500}.
\end{align*}
\end{document}