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

What carries the argument

Three ingredients carry the argument. The polynomial lattice $L_D$ is the $\mathcal{O}$-span of monomials $x_1^i x_2^j x_3^k$ with $i+8(j+k)\le D$; the factor $8$ keeps the degree bound compatible with the grid side $N=|q|^\gamma$. For each prime $p\in S$, the basis $G_{i,j,k}=x_1^v(x_1^p-x_\ell)^e x_2^j x_3^k$ with $i=pe+v$, $0\le v<p$, and $\ell\in\{2,3\}$ chosen by $(2/p)$, turns the Frobenius congruences into vanishing modulo $p^{\lfloor i/p\rfloor}$ at $(\theta_n,\sigma\theta_n,\sigma\tau\theta_n)$. The exponent sum $E_p=\sum_{i+8(j+k)\le D}\lfloor i/p\rfloor$ measures how many residue classes each prime kills, and the counting lemma converts this into a bound on the image of $L_D$ modulo $P$. The pigeonhole principle, applied to the height box $\{f\in L_D:\|f\|_B\le H\}$, supplies two polynomials with the same evaluations modulo $P$, and a norm argument upgrades their difference to an exact vanishing polynomial. The final step is the lattice degree bound: a nonzero polynomial vanishing on the full grid $n_1\gamma_1+\cdots+n_4\gamma_4$ with $0\le n_j\le N-1$ must have degree at least $((N-1)/3)\,\mu(\Gamma)$, and Lemma 7 gives $\mu(\Gamma)\ge 4/3$, so the required degree is greater than $D$.