Kummer's Criterion and Regular Primes in Lean

3.3. The first generalised Bernoulli number🔗

The case n = 1 is the one that appears in the analytic formula for \hminus; we record three forms used throughout.

Lemma3.3.1
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Lemma 3.3.2
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L∃∀N

Intermediate form for B_{1,\chi}. Let R be an integral domain and \chi : (\mathbb{Z}/N\mathbb{Z}) \to R a non-trivial Dirichlet character. Then B_{1,\chi} \;=\; \sum_{a \in \mathbb{Z}/N\mathbb{Z}} \chi(a)\,\frac{a}{N}.

Definition 3.1.1

Lean code for Lemma3.3.11 theorem
  • theoremdefined in BernoulliRegular/BernoulliGeneralized.lean
    complete
    theorem BernoulliRegular.BernoulliGen_one_of_ne_one.{u_1} {N : } {R : Type u_1}
      [CommRing R] [Algebra  R] [IsDomain R] [NeZero N]
      {χ : DirichletCharacter R N} ( : χ  1) :
      BernoulliRegular.BernoulliGen χ 1 =
         a, χ a * (algebraMap  R) (a.val / N)
    theorem BernoulliRegular.BernoulliGen_one_of_ne_one.{u_1}
      {N : } {R : Type u_1} [CommRing R]
      [Algebra  R] [IsDomain R] [NeZero N]
      {χ : DirichletCharacter R N}
      ( : χ  1) :
      BernoulliRegular.BernoulliGen χ 1 =
         a,
          χ a * (algebraMap  R) (a.val / N)
    Intermediate form (T006): for a non-trivial Dirichlet character `χ`,
    the `1/2` constant term in `B_1(a/N) = a/N - 1/2` does not contribute
    after summing, because `∑ χ(a) = 0`. 
Proof for Lemma 3.3.1
uses 0

Since B_1(X) = X - \tfrac12 and the exponent 1 - 1 = 0 in the definition makes the leading N^{n-1} factor equal to 1, B_{1,\chi} = \sum_a \chi(a) \cdot \frac{a}{N} - \sum_a \chi(a) \cdot \frac{1}{2}. The second sum equals \tfrac12 \cdot \sum_a \chi(a), which is zero for non-trivial \chi by the character-sum identity used in Proposition 3.2.1.

Lemma3.3.2
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Proposition 3.3.3
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L∃∀N

Cleared form of B_{1,\chi}. For \chi as in the previous lemma, N \cdot B_{1,\chi} \;=\; \sum_{a \in \mathbb{Z}/N\mathbb{Z}} \chi(a)\,a, an identity valid in any commutative \mathbb{Q}-algebra (denominators have been cleared).

Lemma 3.3.1

Lean code for Lemma3.3.21 theorem
  • theoremdefined in BernoulliRegular/BernoulliGeneralized.lean
    complete
    theorem BernoulliRegular.natCast_mul_BernoulliGen_one_of_ne_one.{u_1} {N : }
      {R : Type u_1} [CommRing R] [Algebra  R] [IsDomain R] [NeZero N]
      {χ : DirichletCharacter R N} ( : χ  1) :
      N * BernoulliRegular.BernoulliGen χ 1 =  a, χ a * a.val
    theorem BernoulliRegular.natCast_mul_BernoulliGen_one_of_ne_one.{u_1}
      {N : } {R : Type u_1} [CommRing R]
      [Algebra  R] [IsDomain R] [NeZero N]
      {χ : DirichletCharacter R N}
      ( : χ  1) :
      N * BernoulliRegular.BernoulliGen χ 1 =
         a, χ a * a.val
    **Diekmann eq. 24/27** (T006), cleared form: `N · B_{1, χ} = ∑_a χ(a) · a`
    for non-trivial `χ`. Cleared of denominators so it holds in any commutative
    `ℚ`-algebra. 
Proof for Lemma 3.3.2
uses 0

Multiply the previous identity through by N and use that the algebra map \mathbb{Q} \to R commutes with multiplication by N \in \mathbb{N}.

Proposition3.3.3
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Definition 2.3.1
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used by 0L∃∀N

Parity vanishing at n = 1. Let R be an integral domain in which 2 \ne 0 and let \chi be a non-trivial even Dirichlet character modulo N > 1 valued in R, with N invertible in R. Then B_{1,\chi} = 0.

Lemma 3.3.2 Definition 2.3.1

Lean code for Proposition3.3.31 theorem
  • theoremdefined in BernoulliRegular/BernoulliGeneralized.lean
    complete
    theorem BernoulliRegular.BernoulliGen_one_eq_zero_of_even_ne_one.{u_1} {N : }
      {R : Type u_1} [CommRing R] [Algebra  R] [IsDomain R] [NeZero 2]
      [NeZero N] [Fact (1 < N)] (hN : N  0) {χ : DirichletCharacter R N}
      (hχ_ne_one : χ  1) (hχ_even : χ.Even) :
      BernoulliRegular.BernoulliGen χ 1 = 0
    theorem BernoulliRegular.BernoulliGen_one_eq_zero_of_even_ne_one.{u_1}
      {N : } {R : Type u_1} [CommRing R]
      [Algebra  R] [IsDomain R] [NeZero 2]
      [NeZero N] [Fact (1 < N)] (hN : N  0)
      {χ : DirichletCharacter R N}
      (hχ_ne_one : χ  1) (hχ_even : χ.Even) :
      BernoulliRegular.BernoulliGen χ 1 = 0
    **Diekmann Lemma 29** (T009): for a non-trivial even Dirichlet character
    `χ` of level `N > 1` over a `ℚ`-algebra domain `R` where `(2 : R) ≠ 0`
    and `(N : R) ≠ 0`, `B_{1, χ} = 0`. 
Proof for Proposition 3.3.3
uses 0

Reindexing the sum in Lemma 3.3.2 by a \mapsto -a and using \chi(-a) = \chi(-1) \chi(a) = \chi(a) (because \chi is even), \sum_a \chi(a)\,a \;=\; \sum_a \chi(a)\,(-a).\mathrm{val}, where (-a).\mathrm{val} denotes the canonical representative in \{0,\dots,N-1\}. Adding the two expressions, 2 \sum_a \chi(a)\,a \;=\; \sum_a \chi(a)\,\bigl(a + (-a).\mathrm{val}\bigr). The pairing identity in \mathbb{Z}/N\mathbb{Z} states that a.\mathrm{val} + (-a).\mathrm{val} equals 0 when a = 0 and N otherwise. Substituting, 2 \sum_a \chi(a)\,a \;=\; N \cdot \sum_{a \ne 0} \chi(a) \;=\; N \cdot \Bigl(\sum_a \chi(a) - \chi(0)\Bigr) \;=\; 0, because \chi(0) = 0 and \sum_a \chi(a) = 0 for non-trivial \chi. Since 2 \ne 0 in R, the sum vanishes, and then N \ne 0 forces B_{1,\chi} = 0 via Lemma 3.3.2.

This is the special case at n = 1 of the general parity phenomenon. For a non-trivial Dirichlet character \chi and n \ge 1 one has B_{n,\chi} = 0 unless \chi(-1) = (-1)^n. Indeed, using the defining sum and reindexing by a\mapsto -a, the Bernoulli polynomials satisfy B_n(1-X)=(-1)^n B_n(X), and the character contributes the scalar \chi(-1); hence B_{n,\chi}=\chi(-1)(-1)^n B_{n,\chi}. If \chi(-1)\ne (-1)^n, the scalar multiplying B_{n,\chi} is not 1, so B_{n,\chi}=0.

In particular: if \chi is odd, B_{n,\chi} can be non-zero only for odd n; if \chi is even, only for even n. Downstream this is used in two places: (i) the cleared form of the analytic formula for \hminus runs only over odd characters, because their B_{1,\chi} are the only non-vanishing terms; and (ii) the odd special-value formula depends on the same parity constraint.