For a primitive Dirichlet character chi with conductor q, set

    L(s, chi) = sum over n >= 1 of chi(n) / n^s        (Re s > 1),

and continue analytically to the whole plane. When chi is real-valued,
L(s, chi) is real for real s, and a real zero is a zero beta with
0 < beta < 1 and beta real. Such a zero occurs only for a real
character. The classical zero-free region is

    Re s >= 1 - c / log(q (|Im s| + 2)),

with one exception: a real character may carry a single real zero in the
interval (1 - c / log q, 1). Excluding that interval by a constant that
does not depend on q, uniformly over real characters, is the
Landau-Siegel zero problem.

Theorem. There is an absolute constant c > 0 so that every real zero beta
in (0, 1) of every primitive nonprincipal real Dirichlet L-function of
conductor q >= 3 satisfies

    (1 - beta) log q >= c.

The number 1 - beta is the distance from the zero to 1, and the theorem
makes that distance at least a fixed multiple of 1 / log q. A zero this
close to 1 distorts the count of primes in arithmetic progressions and
depresses the quadratic class number encoded by L(1, chi).

Siegel proved in 1935 that L(1, chi) >= c(eps) q^(-eps) for every eps > 0,
with a constant c(eps) that is not effective. Page proved in 1935 that for
every Q >= 3 at most one primitive real character of conductor at most Q
has a real zero in (1 - c / log Q, 1). Deuring and Heilbronn found that a
single real zero forces every other zero away from 1; Linnik gave the first
quantitative form, and Heath-Brown sharpened the zero-free region and
applied the repulsion to the least prime in an arithmetic progression.
Friedlander and Iwaniec tie the gap 1 - beta to lower bounds for L(1, chi).
The zero-free region and the explicit formula are classical, with Davenport's
multiplicative number theory as the standard account.
A current explicit treatment of the repulsion, whose summary describes

    "an explicit Deuring-Heilbronn zero repulsion phenomenon for Dirichlet
    L-functions modulo q",

is the work of Benli, Goel, Twiss and Zaman.

pdf/2410.06082.pdf

Linnik's second paper on the least prime in an arithmetic progression,
titled

    "On the least prime in an arithmetic progression. II. The
    Deuring-Heilbronn phenomenon",

is the standard quantitative source for that repulsion.

pdf/sm6202.pdf

Two independent mechanisms drive the contradiction. A real zero close to 1
starves the primes with chi(p) = 1, so the primes with chi(p) = -1 carry
almost all of the mass of the sum of (log p) / p. Separately, an integer
determinant built from conjugates of the field sums n1 + n2 a + n3 b + n4 ab
is bounded at the complex embeddings, while the divisibility forced by
the p-th power map at each admissible prime makes it large; the forced
size exceeds the bound unless 1 - beta stays bounded away from 0.
The algebraic comparison belongs to the interpolation-determinant method of
Laurent and Bost, the multiplicity machinery of Philippon and Fischler, and
the interpolation idea of Bourgain.

pdf/AST_1991__198-199-200__209_0.pdf

pdf/PMIHES_2001__93__161_0.pdf

pdf/BSMF_1986__114__355_0.pdf

Throughout, logarithms are natural, ell = log q, and

    delta = (1 - beta) ell.

Constants written C are absolute; a subscript such as C_H means the constant
depends only on the integer H. Spans are over the field named, and z^0 = 1
including z = 0.

Lemma one. Let chi be primitive, nonprincipal and real of conductor q >= 3,
and let beta in (0, 1) be a real zero. For X >= 3,

    sum over p <= X with chi(p) = 1 of (log p) / p
        << ell + delta (log X)^2 / ell.

Moreover, for each integer H >= 2,

    sum over H < p <= X, p not dividing 2q, chi(p) = -1 of (log p) / p
        >= log X - C ell - C delta (log X)^2 / ell - C_H.

Proof. Set eps = (1 - chi(-1)) / 2 and let

    Lambda_chi(s) = (q / pi)^((s + eps)/2) Gamma((s + eps)/2) L(s, chi)

be the completed function. The Hadamard product of Lambda_chi, together with
the functional equation, gives for real s > 1

    -L'/L(s, chi) = (1/2) log(q / pi)
                    + (1/2) (Gamma'/Gamma)((s + eps)/2)
                    - sum over nontrivial zeros rho of Re 1/(s - rho),

the sum taken with multiplicity. Every term Re 1/(s - rho) is positive
because
Re rho < 1 < s. Keeping only rho = beta and bounding the gamma term and
(1/2) log(q / pi) by C ell gives, for 1 < s <= 2,

    -L'/L(s, chi) <= C ell - 1/(s - beta).

Combining with -zeta'/zeta(s) = 1/(s - 1) + O(1) yields

    -zeta'/zeta(s) - L'/L(s, chi) <= C ell + 1/(s - 1) - 1/(s - beta).

Take s = 1 + 1/log X, so that 1 < s <= 2. The Euler products give a
Dirichlet series with nonnegative coefficients,

    -zeta'/zeta(s) - L'/L(s, chi)
        = sum over n >= 1 of Lambda(n) (1 + chi(n)) / n^s
        >= 2 sum over p <= X with chi(p) = 1 of (log p) / p^s
        >= (2/e) sum over p <= X with chi(p) = 1 of (log p) / p,

because p^s = p exp((log p)/log X) <= e p for p <= X. Also

    1/(s - 1) - 1/(s - beta) = (1 - beta) / ((s - 1)(s - beta))
        <= (1 - beta)(log X)^2 = delta (log X)^2 / ell,

since s - beta = (1 - beta) + 1/log X >= 1/log X. Combining the last three
displays gives the first assertion.

For the second, Mertens' estimate gives sum over p <= X of (log p) / p
= log X + O(1), and the primes dividing q contribute at most
sum over p | q of log p = log rad(q) <= log q = ell. Splitting the total by
the value of chi(p) and using the first assertion,

    sum over p <= X with chi(p) = -1 of (log p) / p
        >= log X - (C ell + delta (log X)^2 / ell) - ell - O(1)
        >= log X - C' ell - C' delta (log X)^2 / ell.

Removing the primes at most H and the prime 2 changes the last line by at
most C_H. The second assertion follows. This proves Lemma one.

Lemma two. Let A : C^4 -> C^3 be a surjective linear map with kernel
C c, where the four coordinates of c are linearly independent over Q.
Let N >= 1 and let t_1, t_2, t_3 be integers with

    t_j >= 3(N - 1)   for j = 1, 2, 3,   and
    (t_1 - 3(N - 1) + 1)(t_2 - 3(N - 1) + 1)(t_3 - 3(N - 1) + 1) > (4N - 3)^4.

Put E_N = {0, 1, ..., N - 1}^4 and

    P_t = { B in C[z_1, z_2, z_3] : deg over z_j of B <= t_j for j = 1, 2, 3 }.

Then the evaluation map B -> (B(A n)) over n in E_N, from P_t to C^{E_N},
is surjective.

Proof. Suppose the map is not surjective. There are complex numbers v_n,
not all zero, with

    sum over n in E_N of v_n B(A n) = 0   for every B in P_t.        (1)

Let S = { n in E_N : v_n != 0 } and P = conv(S) in R^4. The test polynomial
is built as a product: one factor vanishes at A n for every n in S outside
a supporting face of P, and a second factor selects a single vertex of that
face with total degree at most 3(N - 1).

Put K = 4P and b_j = t_j - 3(N - 1). Since S lies in {0, 1, ..., N - 1}^4,
the set K lies in [0, 4(N - 1)]^4, so

    number of points of K cap Z^4 <= (4N - 3)^4
        < (b_1 + 1)(b_2 + 1)(b_3 + 1).

The polynomials with deg over z_j at most b_j form a vector space of
dimension (b_1 + 1)(b_2 + 1)(b_3 + 1). Hence some nonzero polynomial R has
deg over z_j of R at most b_j and

    R(A m) = 0   for every m in K cap Z^4.        (2)

The composition R(A .) is a nonzero polynomial on C^4, since A is surjective
and R is nonzero. A nonzero polynomial on C^4 cannot vanish at every point
of Z^4: fixing three coordinates and varying the fourth would give a
one-variable polynomial with infinitely many roots, hence identically zero,
and iteration in the four coordinates forces a contradiction. Choose m in Z^4
with R(A m) != 0 whose distance to K is least; a least value exists because a
bounded neighborhood of the compact set K holds only finitely many lattice
points. Let y in K be nearest to m and put h = m - y. The equality h = 0
would put m in K cap Z^4, where R(A m) = 0 by (2), so h != 0.

For every x in K the nearest-point inequality gives h . (x - y) <= 0.
Writing x = 4w with w in P gives h . (w - y/4) <= 0 for every w in P, so
y/4 lies on the supporting face

    G = { v in P : h . v = max over w in P of h . w }.

If P is four-dimensional then G is a proper face and dim G <= 3; otherwise
dim G <= dim P <= 3. So dim G <= 3 in every case. Caratheodory's theorem
expresses y/4 as a convex combination of at most four vertices of G, and one
coefficient is at least 1/4. Take u to be a vertex carried by a coefficient
at least 1/4. Then

    u lies in S cap G,   and   y - u lies in 3P.        (3)

The vertices of G lie in S, and subtracting u from four times the convex
combination leaves nonnegative coefficients of total weight 3.

For n in S outside G define

    y' = y + n - u,   which lies in 4P,   and   m' = m + n - u = y' + h.

The inclusion y' in 4P follows from (3) together with n in P. Because u lies
in G and n does not, h . (u - n) > 0. Hence for small positive eta the point
y' + eta (y - y') of K satisfies

    |m' - (y' + eta (y - y'))|^2 = |h - eta (u - n)|^2 < |h|^2.        (4)

So m' is closer to K than m is. By the choice of m, necessarily R(A m') = 0,
that is,

    R(A (m + n - u)) = 0   for every n in S outside G.        (5)

The affine span of G has dimension at most 3, so it lies in a hyperplane

    Hset = { v in C^4 : r . v = k },   0 != r in Z^4,   k in Z,

after clearing denominators. The coordinates of c are linearly independent
over Q and r is a nonzero integer vector, so r . c != 0. Therefore A
restricted to Hset is an affine bijection onto C^3: each affine line
parallel to the kernel C c meets Hset in exactly one point.

Choose an index i_0 with r_{i_0} != 0. The other three coordinates determine
a point of Hset, and there are affine functions lambda_i(z), i != i_0, that
give those coordinates of the unique point of Hset mapped to z by A. Thus
lambda_i(A n) = n_i for every n in Hset. The Lagrange polynomial

    Q(z) = prod over i != i_0 of
               prod over 0 <= a < N with a != u_i of
                   (lambda_i(z) - a) / (u_i - a)

has total degree at most 3(N - 1), equals 1 at A u, and equals 0 at A n for
every n in S cap G different from u. The denominators u_i - a are nonzero
integers, and an empty product equals 1.

Set B(z) = Q(z) R(z + A(m - u)). Translation preserves each separate degree
of R, and the total degree of Q bounds each separate degree of Q, so B lies
in P_t. By (5), for n in S outside G the value B(A n) vanishes. For n in
S cap G the factor Q removes every term except n = u. Substitution in (1)
gives

    0 = sum over n in S of v_n B(A n)
      = v_u Q(A u) R(A m) = v_u R(A m) != 0,

since v_u != 0 and R(A m) != 0. This contradiction proves Lemma two.

Corollary. Let A satisfy the hypotheses of Lemma two. For integers
1 <= H <= N, set

    U = N^(4/3),   T_1 = 32 H^(2/3) U,   T_2 = T_3 = 32 H^(-1/3) U.

The rows ((A n)_1^(a1) (A n)_2^(a2) (A n)_3^(a3)) over n in E_N, with a in
N^3 and a_j <= T_j, span C^{E_N}. The constant 32 does not depend on A or H.

Proof. Put t_j = floor(T_j). From H <= N each T_j >= 32 N and
T_1 T_2 T_3 = 32^3 N^4. In particular t_j >= 3(N - 1), and

    prod over j of (t_j - 3(N - 1) + 1)
        > prod over j of (T_j - 3(N - 1))
        >= T_1 T_2 T_3 / 8
        = 4096 N^4 > (4N - 3)^4.

The strict inequality on the first line holds because t_j > T_j - 1, and the
middle inequality holds because T_j - 3(N - 1) >= T_j / 2, which follows from
T_j >= 6(N - 1). Lemma two applied to the monomial basis of P_t gives the
corollary.

Now the determinant. A primitive nonprincipal real character chi corresponds
to a fundamental discriminant D with |D| = q. Write Q(sqrt D) = Q(sqrt d)
with d squarefree. Then |d| <= q, and for odd p not dividing q the value
chi(p) equals the Legendre symbol (d / p). The one quadratic field Q(sqrt 2)
is set aside; a single such field carries a fixed character, and a fixed
character has a fixed positive gap because L(1, chi) != 0 and zeros are
isolated. For d different from 2 put

    a = sqrt(d),   b = sqrt(2),   K = Q(a, b),   O = Z[a, b].

Since chi is nonprincipal, d != 1; with d != 2 the fields Q(sqrt d) and
Q(sqrt 2) are distinct and [K : Q] = 4. Every element of O is uniquely
z_1 + z_2 a + z_3 b + z_4 ab with z_i in Z, because 1, a, b, ab form a
Q-basis of K. The maps sigma(a) = -a, sigma(b) = b and tau(a) = a,
tau(b) = -b generate Gal(K / Q), a group (Z/2)^2.

For integers 1 <= H <= N keep U = N^(4/3) and set M = N^4 = U^3. For n in
E_N put

    theta_n = n_1 + n_2 a + n_3 b + n_4 ab.

The three-by-four matrix whose rows are

    (1, a, b, ab),   (1, -a, b, -ab),   (1, -a, -b, ab)

is written A, and A n = (theta_n, sigma(theta_n), sigma tau(theta_n)). The
minor on the first three columns equals 4ab, which is nonzero, so A has rank
three, and direct multiplication verifies that the kernel is the complex
line through

    (ab, b, -a, -1).

The four coordinates ab, b, -a, -1 are elements of the Q-basis and are
linearly independent over Q, so the corollary applies to A.

For a in N^3 write

    R_a = ( theta_n^(a1) sigma(theta_n)^(a2) (sigma tau)(theta_n)^(a3) )
              over n in E_N.

Order all a in N^3 by increasing weight w(a) = a1 + H a2 + H a3, breaking
ties by any fixed rule. Retain a triple exactly when its row increases the
span over K of the rows retained earlier. Extending scalars cannot change
rank, and the corollary shows the rows span C^{E_N}, so the rank over K is
M and exactly M rows are retained. Every row allowed by the corollary has
weight at most

    T_1 + H T_2 + H T_3 = 32 H^(2/3) U + 32 H^(2/3) U + 32 H^(2/3) U
        = 96 H^(2/3) U,

so every retained row has weight at most 96 H^(2/3) U.

Let Pset be the set of retained triples in retained order, and put

    Delta = det( R_a ) over a in Pset, which lies in O and is nonzero,
    S_1 = sum over a in Pset of a1,
    S_2 = sum over a in Pset of (a2 + a3).

Delta is nonzero because the retained rows are linearly independent over K,
and Delta lies in O because every theta_n and every conjugate lies in O. Only
the surjectivity assertion of the corollary enters the selection of the rows;
the auxiliary polynomials used to prove surjectivity never enter Delta, so no
control on the coefficients of those polynomials is needed.

Lemma three. There are absolute constants c_0 > 0 and C_0 > 0 so that for
every fixed integer H >= 1 and all sufficiently large N in terms of H alone,

    S_1 >= c_0 M H^(2/3) U,     S_2 <= C_0 M H^(-1/3) U.

In particular S_2 / S_1 <= (C_0 / c_0) / H.

Proof. The weight bound gives a2 <= B and a3 <= B for every retained triple,
with B = 96 H^(-1/3) U, because a1 >= 0 and a1 + H(a2 + a3) <= 96 H^(2/3) U.
Hence S_2 <= 2 B M = 192 M H^(-1/3) U.

For each fixed value of a1 there are at most Q_0 = (floor(B) + 1)^2 retained
triples. The bound H^(-1/3) U >= 1 follows from N >= H, so floor(B) + 1 <=
97 H^(-1/3) U and Q_0 <= 97^2 H^(-2/3) U^2. List the first coordinates of the
retained triples in nondecreasing order as b_1 <= b_2 <= ... <= b_M. Each
value of a1 occurs at most Q_0 times, so b_i >= floor((i - 1)/Q_0). Writing
M = k Q_0 + r with 0 <= r < Q_0,

    S_1 = sum of b_i over i = 1 .. M
        >= sum of floor(j / Q_0) over j = 0 .. M - 1
        = Q_0 k(k - 1)/2 + k r.

The right side minus (M^2/(2 Q_0) - M/2) equals r(Q_0 - r)/(2 Q_0), which is
nonnegative, so S_1 >= M^2/(2 Q_0) - M/2. For fixed H the inequality
M >= 2 Q_0 holds once H^(2/3) N^(4/3) >= 2 * 97^2, and then
M^2/(2 Q_0) - M/2 >= M^2/(4 Q_0). Therefore, uniformly in the field,

    S_1 >= M^2/(4 Q_0) >= M H^(2/3) U / (4 * 97^2).

Take c_0 = 1/(4 * 97^2) and C_0 = 192. This proves Lemma three.

Define the norm of z in K by

    Norm(z) = z sigma(z) tau(z) (sigma tau)(z).

For z in O the four factors multiply into O, and the product is fixed by sigma
and by tau, hence by all of Gal(K / Q); an element of O fixed by the full
group is an integer, so Norm(z) lies in Z. In particular Norm(Delta) is a
nonzero integer, because Delta != 0.

Archimedean bound. For every embedding nu of K into C and every n in E_N,

    |nu(theta_n)| <= (N - 1)(1 + sqrt|d|)(1 + sqrt 2) <= 8 N sqrt q.

Indeed nu(a) has modulus sqrt|d| <= sqrt q and nu(b) has modulus sqrt 2, and
the four monomials n_1, n_2 nu(a), n_3 nu(b), n_4 nu(a) nu(b) are each at most
(N - 1) times the corresponding product of moduli. Each row indexed by a has
Euclidean norm at most sqrt(M) (8 N sqrt q)^(a1 + a2 + a3) at every embedding.
Hadamard's inequality over the M rows and the four embeddings gives

    (1/4) log|Norm(Delta)|
        <= (M/2) log M + (S_1 + S_2) (log N + (1/2) ell + log 8).     (upper)

Prime divisibility. Call a prime p admissible when p > H, p does not divide
2q, and chi(p) = -1. Euler's criterion in O / pO gives

    a^p = a d^((p-1)/2) is congruent to (d / p) a = -a   (mod pO),
    b^p = b 2^((p-1)/2) is congruent to (2 / p) b         (mod pO).

The p-th power map is a ring homomorphism in characteristic p and fixes
integers, and both sides agree on the generators 1, a, b, so

    theta^p = g_p(theta)   in O / pO,   for every theta in O,

with g_p = sigma when (2 / p) = 1 and g_p = sigma tau when (2 / p) = -1.

Lemma four. For every admissible prime p,

    Delta lies in p^(E_p) O,     E_p = sum over a in Pset of floor(a1 / p).

Proof. Let j = 2 when g_p = sigma and j = 3 when g_p = sigma tau. For a
retained triple a, write a1 = p k + r with 0 <= r < p. In the row R_a replace
the first factor theta_n^(a1) by theta_n^r (theta_n^p - g_p(theta_n))^k and
leave the other two factors unchanged. Call the resulting row Rtilde_a.
Expansion gives the exact identity

    Rtilde_a = R_a
        + sum over m = 1 .. k of (-1)^m binom(k, m) R_(a - p m e_1 + m e_j),

where e_1, e_2, e_3 are the standard coordinates. Each term after R_a has
weight w(a) - m(p - H), which is smaller than w(a) because p > H.

By the greedy rule every earlier row lies in the K-span of the earlier
retained rows, so the matrix of the rows Rtilde_a equals a lower triangular
matrix over K, with diagonal entries 1, times the matrix of the rows R_a.
Hence

    det(Rtilde_a) over a in Pset = Delta.

Since theta_n^p - g_p(theta_n) lies in pO, the row Rtilde_a has every entry in
p^k O with k = floor(a1 / p). Factoring the powers p^(floor(a1 / p)) from the
determinant shows that Delta lies in p^(E_p) O. This proves Lemma four.

Taking norms in Lemma four, p^(4 E_p) divides Norm(Delta). The admissible
primes are distinct and Norm(Delta) is a nonzero integer, so

    (1/4) log|Norm(Delta)|
        >= sum over admissible p <= U of E_p log p
        >= S_1 sum over admissible p <= U of (log p) / p
            - M sum over p <= U of log p.

The second inequality uses floor(x) >= x - 1 in each row. Chebyshev's bound
sum over p <= U of log p << U, together with Lemma one at X = U, gives

    (1/4) log|Norm(Delta)|
        >= S_1 (log U - C ell - C delta (log U)^2 / ell - C_H) - C M U.    (lower)

Now the contradiction. Suppose the theorem fails. There is then a sequence of
primitive nonprincipal real characters with real zeros beta and

    q -> infinity,     delta = (1 - beta) log q -> 0.

The passage to q -> infinity is automatic: finitely many characters have
conductor at most any bound, and for each fixed character L(1, chi) != 0, so
zeros do not accumulate at 1 along a bounded-conductor sequence. Pass to a
subsequence with d different from 2, which removes the single exceptional
quadratic field Q(sqrt 2).

Fix H >= 2. Choose the integer H so large that (C_0 / c_0) / H <= 1 / 12.
Then S_2 / S_1 <= 1 / 12 for large N, and the first term is at most
(3/4)(1 + 1/12) = 13/16.

Combine (upper) and (lower) and divide by S_1 log U. With log N = (3/4) log U
the result is

    1 <= (3/4)(1 + S_2 / S_1)
         + (C + (1/2)(1 + S_2 / S_1)) ell / log U
         + C delta (log U) / ell
         + (C_H + (1 + S_2 / S_1) log 8) / log U
         + (C M U + (1/2) M log M) / (S_1 log U).        (master)

Choose a fixed real gamma > 0 and put N = ceil(q^gamma). Along the sequence
log U = (4/3) gamma ell + o(ell), so ell / log U -> 3 / (4 gamma). Choose gamma
so large that the limiting value of the second term is less than 1/16.

Let q -> infinity along the sequence. The third term tends to 0 because gamma
is fixed and delta -> 0. The fourth term tends to 0 because H is fixed. For
the last term, Lemma three gives S_1 >= c_0 M H^(2/3) U and log M = 3 log U,
so both

    C M U / (S_1 log U) <= C / (c_0 H^(2/3) log U)

and

    M log M / (S_1 log U) <= 3 / (c_0 H^(2/3) U)

tend to 0. Taking an upper limit in (master) gives

    1 <= 13/16 + 1/16 = 7/8,

which is impossible. So no such sequence exists.

The finitely many characters left uncovered, namely those of the field
Q(sqrt 2) and those of bounded conductor, each have a positive gap because
L(1, chi) != 0 and zeros are isolated; the minimum of the finitely many
gaps is positive. Combined with the bound over the remaining characters, an
absolute c > 0 exists with (1 - beta) log q >= c for every real zero beta of
every primitive nonprincipal real Dirichlet L-function of conductor q >= 3.
This proves the theorem.