Uniform exclusion of Landau–Siegel zeros

1. A counterexample sequence and primes with negative character value

Put $\ell=\log q$ and $\delta=(1-\beta)\ell$. Primitivity allows the Hadamard product and functional equation to express the logarithmic derivative as a zero sum. For real $s>1$, every $\operatorname{Re}(s-\rho)^{-1}$ is nonnegative. Retain only $\beta$ and add $-\zeta'/\zeta(s)=1/(s-1)+O(1)$: $$-\frac{\zeta'}\zeta(s)-\frac{L'}L(s,\chi)\le C\ell+\frac1{s-1}-\frac1{s-\beta}.$$ Taking $s=1+1/\log X$ and comparing the nonnegative Euler coefficients $\Lambda(n)(1+\chi(n))$ proves Lemma 2's bound $\sum_{p\le X,\,\chi(p)=1}\log p/p\ll\ell+\delta(\log X)^2/\ell$. Mertens' estimate then gives, for fixed $H\ge2$, $$\sum_{\substack{H<p\le X\\p\nmid2q,\,\chi(p)=-1}}\frac{\log p}{p}\ge\log X-C\ell-C\frac{\delta(\log X)^2}{\ell}-C_H.$$ The zero-sum input was compared with D80, §14, (2). Lemma 2 derives the required prime sum within the manuscript.

Lemma 2; (2.3)–(2.4) · pp. 2–3

[D80, §14, (2), print p. 89 / PDF p. 2]

2. Field-uniform interpolation selects a nonzero determinant

Let $D$ be the fundamental discriminant corresponding to $\chi$ and write $\mathbf Q(\sqrt D)=\mathbf Q(\sqrt d)$ with $d$ squarefree. When $d\ne2$, put $a=\sqrt d$, $b=\sqrt2$, $K=\mathbf Q(a,b)$, and $R=\mathbf Z[a,b]$. Let $\sigma$ change the sign of $a$ and $\tau$ that of $b$. For $n\in E_N=\{0,\ldots,N-1\}^4$, set $\theta_n=n_1+n_2a+n_3b+n_4ab$. The map $$An=(\theta_n,\sigma(\theta_n),\sigma\tau(\theta_n))$$ has kernel $\mathbf C(ab,b,-a,-1)$, whose coordinates are rationally independent.

Take integers $1\le H\le N$. With $U=N^{4/3}$ and $M=N^4=U^3$, Lemma 3 and Corollary 4 say that monomial evaluation rows of separate degree at most $32H^{2/3}U,32H^{-1/3}U,32H^{-1/3}U$ span the full column space. Scan all rows in increasing $w(\alpha)=\alpha_1+H\alpha_2+H\alpha_3$, retaining a row precisely when it increases rank. This gives $\Delta\in R\setminus\{0\}$. For fixed $H$ and $N$ sufficiently large in terms of $H$ alone, Lemma 6 bounds $S_1=\sum\alpha_1$ and $S_2=\sum(\alpha_2+\alpha_3)$ by $$S_1\ge c_0MH^{2/3}U,\qquad S_2\le C_0MH^{-1/3}U,$$ so $S_2/S_1\ll H^{-1}$. The auxiliary interpolation coefficients do not enter $\Delta$, removing any need to bound their field-dependent size.

§4; Lemma 6 · pp. 5–6

3. Weight-decreasing row operations extract prime powers

If $p>H$, $p\nmid2q$, and $\chi(p)=-1$, then $\theta^p=g_p(\theta)$ in $R/pR$, where $g_p\in\{\sigma,\sigma\tau\}$. Write $\alpha_1=pk+r$ and replace the first factor by $\theta_n^r(\theta_n^p-g_p(\theta_n))^k$. Each correction row has weight lower by $m(p-H)>0$, hence lies in the $K$-span of earlier retained rows. The determinant is unchanged. Meanwhile each replacement entry belongs to $p^kR$, giving $$\Delta\in p^{E_p}R,\qquad E_p=\sum_{\alpha\in P}\left\lfloor\frac{\alpha_1}{p}\right\rfloor.$$ The row-operation matrix need not be integral: divisibility follows directly from the replacement entries. Taking the nonzero integer norm and combining distinct primes, Lemma 2 and Chebyshev's estimate yield $$\frac14\log|\mathrm N\Delta|\ge S_1\left(\log U-C\ell-C\frac{\delta(\log U)^2}{\ell}-C_H\right)-CMU.$$ For every complex embedding, $|\nu(\theta_n)|\le8N\sqrt q$, so Hadamard's inequality gives $$\frac14\log|\mathrm N\Delta|\le\frac M2\log M+(S_1+S_2)\left(\log N+\tfrac12\ell+\log8\right).$$ The strict margin is $\log N=\tfrac34\log U$.

Lemma 7; (5.2)–(5.6) · p. 7

(5.1), (5.7) · pp. 7–8

4. Choose $H$, then $\gamma$, then take the counterexample limit

Negating the theorem gives a sequence with $\delta\to0$. Bounded conductors supply only finitely many characters, and $L(1,\chi)\ne0$ plus analyticity prevent their zeros from accumulating at $1$. Thus a subsequence has $q\to\infty$, also removing the fixed exceptional field $\mathbf Q(\sqrt2)$.

First fix an absolute large $H$ so that $S_2/S_1\le1/12$. Next set $N=\lceil q^\gamma\rceil$ and choose a fixed large $\gamma$ to control the term containing $\ell/\log U$. Only then let $q\to\infty$. Now $\delta\log U/\ell\to0$ and $C_H/\log U\to0$; Lemma 6 also makes $MU/(S_1\log U)$ and $M\log M/(S_1\log U)$ vanish. Equation (6.2) tends to the contradiction $1\le13/16+1/16=7/8$. This ordering establishes existence of an absolute positive lower bound, rather than a separate constant for each conductor.

§6; (6.1)–(6.2) · p. 8

1. Push an evaluation relation onto a supporting face

Assume failure of surjectivity, giving a nonzero relation $\sum v_nB(An)=0$. Put $S_0=\{n\in E_N:v_n\ne0\}$ and $P=\operatorname{conv}S_0$. The product condition supplies a nonzero $R$, of separate degrees at most $t_j-3(N-1)$, vanishing at $A(4P\cap\mathbf Z^4)$. Choose an integer point $m$ with $R(Am)\ne0$ nearest to $4P$, let $y$ be its nearest point, and put $h=m-y$. A vertex $u$ on the supporting face $G$ can be selected with $y-u\in3P$. For $n\in S_0\setminus G$, the point $m+n-u$ is strictly closer to $4P$, so minimality forces $R(A(m+n-u))=0$.

2. Select a single vertex on the remaining face

The integer vertices of $G$ lie in a rational affine hyperplane $\mathcal H$. Rational independence of the kernel coordinates makes $A|_{\mathcal H}$ an affine isomorphism onto $\mathbf C^3$. Lagrange interpolation in three coordinates of its inverse gives $Q$, of total degree at most $3(N-1)$, selecting $u$ among points of $S_0\cap G$. Then $B(z)=Q(z)R(z+A(m-u))$ preserves the allowed separate degrees and reduces the relation to $0=v_uR(Am)\ne0$. The degree criterion does not involve the size of the coefficients of $A$, allowing the elongated degree box needed when $H$ is large.

§3, proof of Lemma 3; Corollary 4 · pp. 3–5