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Two bounds for the determinant

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

The determinant \(\Delta\) of A weighted determinant in a biquadratic field is a nonzero element of \(\mathcal R=\mathbf Z[a,b]\), so its norm is a nonzero integer. This lesson bounds that integer from above by Hadamard's inequality and from below by its divisibility at the primes \(p>H\) with \(\chi(p)=-1\), and compares the two bounds to prove Theorem A. The argument follows OpenAI [OpenAI-LS, Sections 5 and 6].

The notation is that of the previous lesson: \(\chi\), \(q\), \(d\ne2\), \(K=\mathbf Q(a,b)\), \(\sigma\), \(\tau\), \(\operatorname{Nm}\), integers \(1\le H\le N\), \(U=N^{4/3}\), \(M=N^4\), the retained set \(\mathcal P\), the determinant \(\Delta\), and the sums \(S_1\), \(S_2\). As in Real zeros and the prime bias, \(\ell=\log q\).

1. The Archimedean bound

Lemma 4.1. For every \(n\in E_N\) and every \(g\in\{1,\sigma,\tau,\sigma\tau\}\), \[ \begin{aligned} |g(\theta_n)|&\le(N-1)(1+\sqrt{|d|})(1+\sqrt2)\\ &<8N\sqrt q . \end{aligned} \]

Proof. \(g(\theta_n)=n_1\pm n_2a\pm n_3b\pm n_4ab\) with \(0\le n_i\le N-1\), \(|a|=\sqrt{|d|}\) and \(|b|=\sqrt2\). The triangle inequality gives the first bound, as \[ \begin{aligned} &1+\sqrt{|d|}+\sqrt2+\sqrt{2|d|}\\ &\quad=(1+\sqrt{|d|})(1+\sqrt2). \end{aligned} \] Since \(1\le|d|\le q\), \(1+\sqrt{|d|}\le2\sqrt q\), and \(2(1+\sqrt2)<8\). \(\square\)

Proposition 4.2. \[ \begin{aligned} &\frac14\log|\operatorname{Nm}(\Delta)|\\ &\quad\le\frac M2\log M+(S_1+S_2)\log N\\ &\qquad+(S_1+S_2)\Bigl(\frac12\ell+\log8\Bigr). \end{aligned} \]

Proof. Let \(g\in\{1,\sigma,\tau,\sigma\tau\}\). As \(g\) is a ring homomorphism and a determinant is a polynomial in the entries, \(g(\Delta)\) is the determinant of the rows \(g(R_\alpha)\), \(\alpha\in\mathcal P\). The entries of \(g(R_\alpha)\) are \[ g(\theta_n)^{\alpha_1}\,g\sigma(\theta_n)^{\alpha_2}\,g\sigma\tau(\theta_n)^{\alpha_3}, \] and \(g\sigma,g\sigma\tau\in\{1,\sigma,\tau,\sigma\tau\}\). By Lemma 4.1 every entry has absolute value at most \((8N\sqrt q)^{\alpha_1+\alpha_2+\alpha_3}\), so the row has Euclidean norm at most \(\sqrt M(8N\sqrt q)^{\alpha_1+\alpha_2+\alpha_3}\). Hadamard's inequality, Lemma 1.3 of The real-place integral formula and two bounds, applied to the transposed matrix, gives \[ |g(\Delta)|\le M^{M/2}(8N\sqrt q)^{S_1+S_2}. \] Multiply over the four values of \(g\), take logarithms and divide by \(4\). \(\square\)

2. The Frobenius congruence

Lemma 4.3. Let \(p\) be an odd prime with \(p\nmid q\), and let \(g_p\) be the automorphism of \(K\) with \[ g_p(a)=\Bigl(\frac dp\Bigr)a,\qquad g_p(b)=\Bigl(\frac2p\Bigr)b . \] Then \(\theta^p\equiv g_p(\theta)\pmod{p\mathcal R}\) for every \(\theta\in\mathcal R\). If \(\chi(p)=-1\), then \(g_p=\sigma\) when \((2/p)=1\), and \(g_p=\sigma\tau\) when \((2/p)=-1\).

Proof. Since \(p\nmid2d\), Euler's criterion Dirichlet characters, Section 4 gives \(d^{(p-1)/2}\equiv(d/p)\) and \(2^{(p-1)/2}\equiv(2/p)\) modulo \(p\). Hence, in \(\mathcal R\), \[ \begin{aligned} a^p&=a\,d^{(p-1)/2}\equiv\Bigl(\frac dp\Bigr)a,\\ b^p&=b\,2^{(p-1)/2}\equiv\Bigl(\frac2p\Bigr)b \end{aligned} \] modulo \(p\mathcal R\). The ring \(\mathcal R/p\mathcal R\) is commutative of characteristic \(p\), so \(x\mapsto x^p\) is a ring homomorphism of it: the binomial coefficients \(\binom pk\), \(0<k<p\), are divisible by \(p\). It fixes the classes of integers, by Fermat's little theorem. For \(\theta=z_1+z_2a+z_3b+z_4ab\) it follows that \[ \theta^p\equiv z_1+z_2a^p+z_3b^p+z_4a^pb^p\equiv g_p(\theta). \] If \(\chi(p)=-1\), then \((d/p)=\chi(p)=-1\) by Section 1 of the previous lesson. \(\square\)

Call a prime \(p\) admissible if \(p>H\), \(p\nmid2q\) and \(\chi(p)=-1\).

3. Divisibility

Proposition 4.4 (OpenAI). For every admissible prime \(p\), \[ \Delta\in p^{E_p}\mathcal R,\qquad E_p=\sum_{\alpha\in\mathcal P}\Bigl\lfloor\frac{\alpha_1}p\Bigr\rfloor . \]

Proof. Let \(j=2\) if \(g_p=\sigma\) and \(j=3\) if \(g_p=\sigma\tau\), and let \(e_1,e_2,e_3\) be the unit vectors of \(\mathbf Z^3\). For \(\alpha\in\mathcal P\) write \(\alpha_1=kp+r\) with \(0\le r<p\), and let \(\widetilde R_\alpha\) be the vector whose entry at \(n\in E_N\) is \[ \theta_n^r\bigl(\theta_n^p-g_p(\theta_n)\bigr)^k\,\sigma(\theta_n)^{\alpha_2}\,\sigma\tau(\theta_n)^{\alpha_3}. \] Expanding the power by the binomial theorem, and using that \(g_p(\theta_n)\) is the conjugate raised to the exponent \(\alpha_j\), gives \[ \begin{aligned} \widetilde R_\alpha={}&R_\alpha\\ &+\sum_{m=1}^k(-1)^m\binom kmR_{\alpha-pme_1+me_j}. \end{aligned} \] The index \(\alpha-pme_1+me_j\) has weight \(w(\alpha)-m(p-H)<w(\alpha)\), because \(p>H\). So it precedes \(\alpha\), and by part 1 of Lemma 3.4 its row is a \(K\)-linear combination of retained rows preceding \(\alpha\). Thus the matrix with rows \(\widetilde R_\alpha\), \(\alpha\in\mathcal P\) in increasing order, is \(L\) times the matrix with rows \(R_\alpha\), for a lower triangular matrix \(L\) over \(K\) with diagonal entries \(1\), and \[ \det\bigl(\widetilde R_\alpha\bigr)_{\alpha\in\mathcal P}=\Delta . \] On the other hand \(\theta_n^p-g_p(\theta_n)\in p\mathcal R\) by Lemma 4.3, so every entry of \(\widetilde R_\alpha\) lies in \(p^k\mathcal R\), with \(k=\lfloor\alpha_1/p\rfloor\). Dividing each row by its power of \(p\) leaves a matrix over \(\mathcal R\), whose determinant lies in \(\mathcal R\). Hence \(\Delta\in p^{E_p}\mathcal R\). \(\square\)

The entries of \(L\) need not lie in \(\mathcal R\); the divisibility comes from the entries of the rows \(\widetilde R_\alpha\) alone.

Corollary 4.5. Let \(\mathcal A_U\) be the set of admissible primes \(p\le U\). Then \[ \begin{aligned} &\frac14\log|\operatorname{Nm}(\Delta)|\ge\sum_{p\in\mathcal A_U}E_p\log p\\ &\quad\ge S_1\sum_{p\in\mathcal A_U}\frac{\log p}p-1.39\,MU . \end{aligned} \]

Proof. Write \(\Delta=p^{E_p}\rho_p\) with \(\rho_p\in\mathcal R\). Then \(\operatorname{Nm}(\Delta)=p^{4E_p}\operatorname{Nm}(\rho_p)\) with \(\operatorname{Nm}(\rho_p)\in\mathbf Z\), by Lemma 3.2. So \(p^{4E_p}\) divides the nonzero integer \(\operatorname{Nm}(\Delta)\) for every admissible \(p\), and so does the product of these coprime prime powers over \(p\le U\). This gives the first inequality. For the second, \(\lfloor x\rfloor\ge x-1\) gives \(E_p\ge S_1/p-M\), and Chebyshev's bound \(\sum_{p\le U}\log p<(2\log2)U\), Theorem 2.1 of Counting primes by elementary means: Chebyshev and Mertens, gives the rest, as \(2\log2<1.39\). \(\square\)

4. Proof of Theorem A

Suppose Theorem A is false. By Corollary 1.6 of Real zeros and the prime bias there are primitive nonprincipal real characters of conductors \(q\to\infty\) with real zeros \(\beta\) such that \(\delta=(1-\beta)\ell\to0\). Fix \[ \begin{aligned} H&=\frac{12C_0}{c_0}=12\cdot192\cdot4\cdot97^2\\ &=86{,}713{,}344, \end{aligned} \] with \(c_0,C_0\) from Lemma 3.5, and for each character of the sequence put \(N=\lceil q^{55}\rceil\). For all large \(q\) in the sequence, \(q\ne8\), \(N\ge H\), \(H^{2/3}N^{4/3}\ge2\cdot97^2\) and \(U\ge3\). Then all results of the last three lessons apply, and Lemma 3.5 gives \[ \rho:=\frac{S_2}{S_1}\le\frac1{12} . \]

The primes of \(\mathcal A_U\) are those counted in Corollary 1.4 with \(X=U\). Corollaries 1.4 and 4.5 give \[ \begin{aligned} &\frac14\log|\operatorname{Nm}(\Delta)|\\ &\quad\ge S_1\log U-4S_1\ell-\frac{3S_1\delta(\log U)^2}{\ell}\\ &\qquad-S_1(\log H+2c_M)-1.39\,MU . \end{aligned} \] Proposition 4.2, with \(\log N=\frac34\log U\) and \(\log M=3\log U\), gives \[ \begin{aligned} &\frac14\log|\operatorname{Nm}(\Delta)|\le\frac32M\log U\\ &\quad+S_1(1+\rho)\Bigl(\frac34\log U+\frac12\ell+\log8\Bigr). \end{aligned} \] Combine the two bounds and divide by \(S_1\log U>0\): \[ \begin{aligned} 1\le{}&\frac34(1+\rho)+\Bigl(4+\frac{1+\rho}2\Bigr)\frac\ell{\log U}\\ &+\frac{3\delta\log U}{\ell}\\ &+\frac{(1+\rho)\log8+\log H+2c_M}{\log U}\\ &+\frac{1.39\,MU}{S_1\log U}+\frac{3M}{2S_1} . \end{aligned} \] The terms on the right are bounded as follows.

  1. \(\frac34(1+\rho)\le\frac34\cdot\frac{13}{12}=\frac{13}{16}\).
  2. Since \(N\ge q^{55}\), \(\log U=\frac43\log N\ge\frac{220}3\ell\), so the second term is at most \(\frac{109}{24}\cdot\frac3{220}=\frac{109}{1760}<\frac1{16}\).
  3. Since \(N\le2q^{55}\) and \(\ell>1\), \(\log U\le\frac43(55\ell+\log2)\le75\ell\), so the third term is at most \(225\delta\), which tends to \(0\).
  4. The fourth term tends to \(0\), because \(H\) is fixed and \(\log U\to\infty\).
  5. By Lemma 3.5, \(S_1\ge c_0MH^{2/3}U\), so the last two terms are at most \(\frac{1.39}{c_0H^{2/3}\log U}\) and \(\frac3{2c_0H^{2/3}U}\), which tend to \(0\).

Letting \(q\to\infty\) along the sequence gives \[ 1\le\frac{13}{16}+\frac{109}{1760}<\frac{13}{16}+\frac1{16}=\frac78, \] a contradiction. This proves Theorem A. \(\square\)

Banks [Banks] arranges the proof so that it gives an explicit constant: \(c=3.0\cdot10^{-3}\) for \(q\ge10^{10}\).

5. A zero-free region without exception

Corollary B. There is an absolute constant \(c_B>0\) such that, for every modulus \(q\ge1\), no Dirichlet \(L\)-function \(L(s,\chi)\) with \(\chi\) modulo \(q\) vanishes in the region \[ \operatorname{Re}s\ge1-\frac{c_B}{\log(q(|\operatorname{Im}s|+2))} . \]

Proof. Let \(c\) be the constant of Theorem 5.1 of Zero-free regions and the exceptional zero, let \(c_A\) be the constant of Theorem A, and put \(c_B=\min(c,c_A)\). By that theorem, the only possible zero of \(\prod_{\chi\bmod q}L(s,\chi)\) in the region with constant \(c\) is a simple real zero \(\beta_1<1\) of one real nonprincipal character \(\chi_1\) modulo \(q\); the proof there takes \(c<\log2\), so the region lies in \(\operatorname{Re}s>0\) and \(\beta_1>0\). The character \(\chi_1\) is induced by a primitive real nonprincipal character \(\chi^*\) of conductor \(q^*\mid q\), \(q^*\ge3\). By (3.3) of The functional equation of Dirichlet L-functions, \(L(s,\chi_1)\) is \(L(s,\chi^*)\) times finite Euler factors whose zeros lie on \(\operatorname{Re}s=0\). So \(L(\beta_1,\chi^*)=0\), and Theorem A gives \[ 1-\beta_1\ge\frac{c_A}{\log q^*}\ge\frac{c_A}{\log q}>\frac{c_B}{\log(2q)} . \] Thus \(\beta_1\) lies outside the region with constant \(c_B\), which is contained in the region with constant \(c\). \(\square\)

6. Why a real zero is needed

The two bounds hold for every \(d\ne1,2\) and all \(1\le H\le N\); only the distribution of the primes forced by a real zero close to \(1\) makes them incompatible. Fix \(d\ne1,2\) and \(H\), and let \(N\to\infty\), so that \(\ell/\log U\to0\).

For odd \(p\nmid q\), the pair \(\bigl((d/p),(2/p)\bigr)=\bigl(\chi(p),\chi_8(p)\bigr)\), with \(\chi_8\) as in (4.3) of Dirichlet characters, depends only on \(p\) modulo \(8q\). The map \(u\mapsto(\chi(u),\chi_8(u))\) is a homomorphism from the units modulo \(8q\) onto \(\{\pm1\}^2\): its image is not contained in a subgroup of order \(2\), because \(\chi\) and \(\chi_8\) are nonprincipal and different, having the conductors \(q\ne8\) and \(8\). So each sign pair occurs for a quarter of the residue classes. By Theorem 4.2 of Dirichlet's theorem on primes in arithmetic progressions, the primes up to \(U\) of each sign pair carry the mass \(\frac14\log U+O_{q}(1)\). Proposition 4.4 applies to the two pairs with \((d/p)=-1\), and by Exercise 4.2 its conclusion also holds when \((d/p)=(2/p)=1\). If the argument of Corollary 4.5 uses the primes of all three pairs, its lower bound has leading term \(\frac34S_1\log U\), while the upper bound of Proposition 4.2 has leading term \(\frac34(S_1+S_2)\log U\). They differ by \(\frac34S_2\log U\ge0\): no contradiction, as must be the case.

At the primes with \((d/p)=1\) and \((2/p)=-1\), \(\theta^p\equiv\tau(\theta)\), and \(\tau(\theta_n)\) is not one of the three conjugates in the rows. The divisibility can then fail. For \(d=-1\), \(N=2\), \(H=3\) and the tie-breaking rule of Section 3 of the previous lesson, a computation with exact arithmetic in \(K\) gives \(E_5=3\), while \(5\) does not divide \(\operatorname{Nm}(\Delta)\). The condition \(p>H\) cannot be dropped either: for \(d=5\), \(N=2\), \(H=3\) and \(p=3\), where \((5/3)=-1\), the same computation gives \(E_3=9\), while \(3^{20}\) is the exact power of \(3\) dividing \(\operatorname{Nm}(\Delta)\), less than \(3^{36}\).

7. Exercises

4.1 (easy). Let \(d=5\) and \(p=3\). Compute \((1+a)^3\) in \(\mathcal R\) and check that \((1+a)^3\equiv1-a\pmod{3\mathcal R}\), as Lemma 4.3 predicts.

4.2 (medium). Let \(p\) be an odd prime with \(p\nmid q\) and \((d/p)=(2/p)=1\). Show that \(\Delta\in p^{E_p}\mathcal R\) for every \(H\ge1\), without the condition \(p>H\).

4.3 (medium). Replace \(K\) by \(\mathbf Q(\sqrt d)\): take \(n\in\{0,\dots,N-1\}^2\), \(\theta_n=n_1+n_2a\), and rows \(\bigl(\theta_n^{\alpha_1}\sigma(\theta_n)^{\alpha_2}\bigr)_n\). With \(M=N^2\) rows chosen in the same way, the natural scale is \(U=N\). Explain why the comparison of the two bounds then gives no contradiction, and what the adjunction of \(\sqrt2\) changes.

4.4 (easy). Check that \(12\cdot192\cdot4\cdot97^2=86{,}713{,}344\), that \(\frac{109}{24}\cdot\frac3{220}=\frac{109}{1760}<\frac1{16}\), and that \(\frac43(55+\log2)\le75\).

8. Solutions

4.1. In \(\mathcal R\), \[ \begin{aligned} (1+a)^3&=1+3a+3a^2+a^3\\ &=1+3a+15+5a=16+8a\\ &\equiv1+2a\equiv1-a\pmod3 . \end{aligned} \] Here \((5/3)=-1\) and \((2/3)=-1\), so \(g_3=\sigma\tau\), and \(\sigma\tau(1+a)=1-a\).

4.2. Here \(g_p\) is the identity, so \(\theta^p\equiv\theta\pmod{p\mathcal R}\). Replace the first factor \(\theta_n^{\alpha_1}\) by \(\theta_n^r(\theta_n^p-\theta_n)^k\). The expansion gives \(R_\alpha+\sum_{m\ge1}(-1)^m\binom kmR_{\alpha-(p-1)me_1}\), and these indices have weight \(w(\alpha)-(p-1)m<w(\alpha)\) for every \(H\). The rest of the proof of Proposition 4.4 is unchanged.

4.3. The upper bound has main term \((S_1+S_2)\log N\) and the lower bound \(S_1\log U\), and \(\log N=\log U\): the two main terms agree up to \(S_2\log U\ge0\), even if all primes up to \(U\) were admissible. With \(\sqrt2\) adjoined, there are four coordinates \(n_1,\dots,n_4\) but only three conjugates in the rows. Then \(M=N^4=U^3\), so \(\log N=\frac34\log U\), and this gap of \(\frac14\log U\) per unit of degree is what Theorem A uses.

4.4. \(12\cdot192=2304\), \(4\cdot97^2=37636\), and \(2304\cdot37636=86{,}713{,}344\). Next \(\frac{109\cdot3}{24\cdot220}=\frac{327}{5280}=\frac{109}{1760}\), and \(109\cdot16=1744<1760\). Finally \(\frac43(55+0.7)=74.27\le75\).

References