The decomposition law at an unramified prime of a multiquadratic field #
Let K = ℚ(√d₁, …, √dₙ) be a number field generated over ℚ by square roots r i of integers
d i, and let p be an odd prime dividing none of the d i. The Frobenius at a prime Q of
𝓞 K above p acts on the generators by the Legendre symbols, σ (r i) = (dᵢ/p) · r i
(NumberField.exists_isArithFrobAt_multiquadratic). This file reads off from that action the
complete decomposition type of p in K:
pis unramified inK: the inertia group ofQis trivial, because an element of it that negated somer iwould force2 r i ∈ Q, hencep ∣ 2 dᵢ;- the residue degree
fof every prime abovepis1if everydᵢis a quadratic residue modpand2otherwise, sincefis the order of the Frobenius, an involution that is trivial exactly when every symbol is1; - the number
gof primes abovepsatisfiesg · f = [K : ℚ]. When everydᵢis a residue this is the splitting lawNumberField.ncard_primesOver_multiquadratic_iff(g = [K : ℚ]); otherwiseg = [K : ℚ] / 2, which is2ⁿ⁻¹under square-class independence of the radicands.
The same holds at p = 2 when every dᵢ is 1 modulo 4, with the congruence class of dᵢ
modulo 8 in place of the Legendre symbol. There the integral half-generator (1 + r i) / 2
replaces r i: an inertia element negating r i sends it to 1 - (1 + r i) / 2, and the
difference -r i would then lie in Q, forcing 2 ∣ dᵢ. The Frobenius above 2 is trivial
exactly when every dᵢ is 1 modulo 8 (isArithFrobAt_eq_one_iff_mod_eight), so 2 splits
completely exactly then, and otherwise has residue degree 2 and [K : ℚ] / 2 primes above it.
No squarefreeness of the radicands is assumed: unramifiedness is proved directly from
p ∤ 2 dᵢ, or from dᵢ ≡ 1 (mod 4) at 2, rather than through the discriminant.
Main results #
TauCeti.Multiquadratic.apply_eq_self_of_mem_inertia: inertia at an odd primepfixes every square root of an integer prime top.TauCeti.Multiquadratic.inertia_eq_bot_of_forall_not_dvdandTauCeti.Multiquadratic.isUnramifiedAt_of_forall_not_dvd: an odd prime dividing no radicand has trivial inertia and is unramified.TauCeti.Multiquadratic.inertiaDeg_eq_one_iff_forall_legendreSym_eq_oneandTauCeti.Multiquadratic.inertiaDeg_eq_two_iff_exists_legendreSym_eq_neg_one: the residue degree of a prime above such apis1if everydᵢis a quadratic residue modp, and2if somedᵢis not.TauCeti.Multiquadratic.ncard_primesOver_mul_inertiaDeg_eq_finrank: the number of primes aboveptimes their residue degree is[K : ℚ].TauCeti.Multiquadratic.ncard_primesOver_mul_two_eq_finrank: when somedᵢis a non-residue, there are[K : ℚ] / 2primes abovep.TauCeti.Multiquadratic.eq_two_pow_sub_of_mul_two_pow_eq_finrank: under square-class independence ofnradicands,g · 2ᵏ = [K : ℚ]forcesg = 2ⁿ⁻ᵏ.TauCeti.Multiquadratic.ncard_primesOver_eq_two_pow_sub_one: under square-class independence ofnradicands, that number is2ⁿ⁻¹.TauCeti.Multiquadratic.two_mul_two_pow_le_ncard_primesOver: under square-class independence ofnradicands, an odd prime dividing none of them, with some radicand a residue, has at least2 * 2 ^ (n - 2)primes above it.TauCeti.Multiquadratic.apply_eq_self_of_mem_inertia_of_mod_four_eq_one: inertia above2fixes every square root of an integer that is1modulo4.TauCeti.Multiquadratic.inertia_eq_bot_of_forall_mod_four_eq_oneandTauCeti.Multiquadratic.isUnramifiedAt_of_forall_mod_four_eq_one:2has trivial inertia and is unramified when every radicand is1modulo4.TauCeti.Multiquadratic.inertiaDeg_eq_one_iff_forall_mod_eight_eq_oneandTauCeti.Multiquadratic.inertiaDeg_eq_two_iff_exists_mod_eight_eq_five: the residue degree of a prime above2is1if everydᵢis1modulo8, and2if somedᵢis5modulo8.TauCeti.Multiquadratic.ncard_primesOver_two_mul_inertiaDeg_eq_finrankandTauCeti.Multiquadratic.ncard_primesOver_two_eq_finrank_iff: the prime-count formula at2, and the splitting law:2splits completely iff everydᵢis1modulo8.TauCeti.Multiquadratic.ncard_primesOver_two_mul_two_eq_finrankandTauCeti.Multiquadratic.ncard_primesOver_two_eq_two_pow_sub_one: when somedᵢis5modulo8, there are[K : ℚ] / 2primes above2, which is2ⁿ⁻¹under square-class independence.TauCeti.Multiquadratic.inertiaDeg_dvd_two: at every rational prime, ramified or not, the residue degree divides2.
References #
- D. A. Cox, Primes of the Form x² + ny², §5.B.
- J. Neukirch, Algebraic Number Theory, Chapter I, §9.
Shared unramified-prime facts #
A number field generated over ℚ by square roots of integers is Galois over ℚ.
Bounding a decomposition exponent by the number of radicands. Under square-class
independence, an identity g * 2 ^ k = [K : ℚ] forces k ≤ Nat.card ι.
Reading a prime count off the degree. Under square-class independence of n radicands,
[K : ℚ] = 2 ^ n, so a number g with g * 2 ^ k = [K : ℚ] is 2 ^ (n - k); the equation itself
forces k ≤ n. This is how the decomposition formulas g · f · e = [K : ℚ] are solved for the
number g of primes above a rational prime.
The decomposition law at an odd prime #
Inertia at an odd prime fixes the square roots of integers prime to it. Let Q be a
prime of 𝓞 K above an odd prime p (primality of p is not needed, only p ≠ 2), and let
x ∈ K square to an integer c with p ∤ c. Then every element of the inertia group of Q in
Gal(K/ℚ) fixes x. Indeed τ x = ± x, and τ x = -x would put 2 x in Q, hence
p ∣ 2 c.
An odd prime dividing no radicand has trivial inertia. Let K be generated over ℚ by
square roots r i of integers d i, and let Q be a prime of 𝓞 K above an odd prime p with
p ∤ d i for all i. Then the inertia group of Q in Gal(K/ℚ) is trivial, so p is
unramified in K. No squarefreeness of the d i is needed.
An odd prime dividing no radicand is unramified in the multiquadratic field.
Residue degree one exactly at the residues. Let K be generated over ℚ by square roots
r i of integers d i, and let Q be a prime of 𝓞 K above an odd prime p dividing none of
the d i. Then Q has residue degree 1 iff every d i is a quadratic residue mod p.
Residue degree two exactly at a non-residue. Let K be generated over ℚ by square roots
r i of integers d i, and let Q be a prime of 𝓞 K above an odd prime p dividing none of
the d i. Then Q has residue degree 2 iff some d i is a quadratic non-residue mod p.
Together with inertiaDeg_eq_one_iff_forall_legendreSym_eq_one, the residue degree is always
1 or 2.
The unramified prime-count formula. If K is generated by square roots of integers
d i, and the odd prime p divides none of them, then the number of primes above p times
their common residue degree is [K : ℚ].
The number of primes above an odd prime with a non-residue radicand. Let K be
generated over ℚ by square roots of integers d i, and let p be an odd prime dividing none of
them such that some d i is a quadratic non-residue mod p. Then there are exactly [K : ℚ] / 2
primes of 𝓞 K above p. (When every d i is a residue, p splits completely instead:
NumberField.ncard_primesOver_multiquadratic_iff.)
The number of primes above an odd prime with a non-residue radicand, explicitly. Let K
be generated over ℚ by square roots of n square-class independent integers d i (no nonempty
subset product is a square), and let p be an odd prime dividing none of them such that some
d i is a quadratic non-residue mod p. Then there are exactly 2 ^ (n - 1) primes of 𝓞 K
above p.
Each rational prime has many primes above it in a multiquadratic field. Let
M = ℚ(√d₁, …, √dₙ) be generated by the square roots of square-class independent integers and let
q be an odd prime dividing none of the radicands, with a radicand d i₀ a quadratic residue
modulo q. Then q has at least 2 * 2 ^ (n - 2) primes of 𝓞 M above it: when every radicand
is a residue mod q there are 2 ^ n, and otherwise the residue degree is two, leaving 2 ^ n / 2, where a non-residue radicand is not d i₀ and hence forces n ≥ 2.
The decomposition law at 2 #
When every radicand is 1 modulo 4, the prime 2 is unramified as well, and the dyadic
Frobenius calculation isArithFrobAt_eq_one_iff_mod_eight plays the role of the Legendre symbols:
the residue degree above 2 is 1 if every dᵢ is 1 modulo 8 and 2 otherwise.
Inertia above 2 fixes the square roots of integers that are 1 modulo 4. Let Q be a
prime of 𝓞 K above 2, and let x ∈ K square to an integer c ≡ 1 (mod 4). Then every element
of the inertia group of Q in Gal(K/ℚ) fixes x.
2 has trivial inertia when every radicand is 1 modulo 4. Let K be generated over
ℚ by square roots r i of integers d i ≡ 1 (mod 4), and let Q be a prime of 𝓞 K above
2. Then the inertia group of Q in Gal(K/ℚ) is trivial. No squarefreeness of the d i is
needed.
2 is unramified in the multiquadratic field when every radicand is 1 modulo 4.
Residue degree one above 2 exactly when every radicand is 1 modulo 8. Let K be
generated over ℚ by square roots r i of integers d i ≡ 1 (mod 4), and let Q be a prime of
𝓞 K above 2. Then Q has residue degree 1 iff every d i is 1 modulo 8.
Residue degree two above 2 exactly when some radicand is 5 modulo 8. Let K be
generated over ℚ by square roots r i of integers d i ≡ 1 (mod 4), and let Q be a prime of
𝓞 K above 2. Then Q has residue degree 2 iff some d i is 5 modulo 8. Together with
inertiaDeg_eq_one_iff_forall_mod_eight_eq_one, the residue degree is always 1 or 2.
The prime-count formula at 2. If K is generated by square roots of integers
d i ≡ 1 (mod 4), then the number of primes above 2 times their common residue degree is
[K : ℚ].
The splitting law at 2. Let K be generated over ℚ by square roots of integers
d i ≡ 1 (mod 4). Then 2 splits completely in K (there are [K : ℚ] primes of 𝓞 K
above 2) iff every d i is 1 modulo 8.
The number of primes above 2 when some radicand is 5 modulo 8. Let K be generated
over ℚ by square roots of integers d i ≡ 1 (mod 4), some d i being 5 modulo 8. Then there
are exactly [K : ℚ] / 2 primes of 𝓞 K above 2.
The number of primes above 2 when some radicand is 5 modulo 8, explicitly. Let K
be generated over ℚ by square roots of n square-class independent integers d i ≡ 1 (mod 4)
(no nonempty subset product is a square), some d i being 5 modulo 8. Then there are exactly
2 ^ (n - 1) primes of 𝓞 K above 2.
Residue degrees divide two #
At any rational prime, ramified or not, the residue degree divides the order of a Frobenius, which is an involution.
In an exponent-two Galois number field, the residue degree at any rational prime divides two, without choosing square-root generators.
Residue degrees in a multiquadratic field divide 2. Let K be generated over ℚ by
square roots of integers. Then every prime of 𝓞 K, ramified or not and above any rational prime
p, including p = 2, has residue degree 1 or 2 over p.