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Signed discriminant of a cyclotomic field

Statement

Let $f > 1$ be a reduced index, that is $f$ is odd or $4 ∣ f$ , and let $K = Q (ζ_{f})$ for a primitive $f$ -th root of unity $ζ_{f}$ . Then the signed field discriminant is $d_{K} = (- 1)^{φ (f) / 2} \frac{f^{φ (f)}}{\prod_{p ∣ f} p^{φ (f) / (p - 1)}},$ the product being over the primes $p$ dividing $f$ ; for $f = 1$ one has $d_{Q} = 1$ . The conductor theorem later identifies the reduced index $f$ with the intrinsic conductor of $K$ .

Facts & Assumptions

Given: A reduced index $f$ and a primitive $f$ -th root of unity $ζ = ζ_{f}$ in a fixed algebraic closure of $Q$ , with $K : = Q (ζ)$ and $n : = [K : Q] = φ (f)$ . In the main case $f > 1$ , write $f = p_{1}^{a_{1}} \dots p_{r}^{a_{r}}$ with pairwise distinct primes $p_{i}$ and $a_{i} \geq 1$ , put $e_{i} : = φ (p_{i}^{a_{i}})$ and $K_{i} : = Q (ζ_{p_{i}^{a_{i}}})$ , and for $j \leq r$ put $n_{j} : = p_{1}^{a_{1}} \dots p_{j}^{a_{j}}$ and $M_{j} : = K_{1} \dots K_{j}$ .

[F1]

$K_{i}$ has ring of integers $O_{K_{i}} = Z [ζ_{p_{i}^{a_{i}}}]$ with the power basis as an integral basis, and $∣ d_{K_{i}} ∣ = p_{i}^{L_{i}}, L_{i} : = p_{i}^{a_{i} - 1} (a_{i} (p_{i} - 1) - 1) = a_{i} e_{i} - \frac{e_{i}}{p_{i} - 1}$ (Prime-power cyclotomic ring, discriminant support and p factor).

[F2]

Coprime-discriminant compositum: for number fields $A, B$ with $[A B : Q] = [A : Q] [B : Q]$ and $\gcd (d_{A}, d_{B}) = 1$ , the ring of integers satisfies $O_{A B} = O_{A} O_{B}$ and $d_{A B} = d_{A}^{[B : Q]} d_{B}^{[A : Q]}$ (Integral basis and discriminant of a coprime-discriminant compositum).

[F3]

For every $m \geq 1$ , $O_{Q (ζ_{m})} = Z [ζ_{m}]$ and $1, ζ_{m}, \dots, ζ_{m}^{φ (m) - 1}$ is an integral basis (Ring of integers of every cyclotomic field); the discriminant of an order is independent of the chosen integral basis, so it may be computed from this basis (Discriminant of a basis and order, Power-basis and polynomial discriminants).

[F4]

$M_{j} = Q (ζ_{n_{j}})$ inside the fixed algebraic closure, with $[M_{j} : Q] = φ (n_{j}) = \prod_{i \leq j} e_{i}$ : this is the compositum identity $F (μ_{a}) F (μ_{b}) = F (μ_{lcm (a, b)})$ together with irreducibility of the cyclotomic polynomials and multiplicativity of $φ$ on coprime arguments ( $K (μ_{m}) K (μ_{n}) = K (μ_{lcm (m, n)})$ , $Φ_{n}$ is irreducible in $Q [t]$ for every $n \geq 1$ , Euler's totient is multiplicative: $\gcd (m, n) = 1$ implies $φ (m n) = φ (m) φ (n)$ for positive $m, n$ , The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ).

[F5]

For an ordered $Q$ -basis $α_{1}, \dots, α_{n}$ of a number field $L$ with distinct embeddings $σ_{1}, \dots, σ_{n} ⁣ : L \to C$ one has $disc (α_{1}, \dots, α_{n}) = \det (σ_{i} (α_{j}))^{2}$ and the determinant is nonzero (Embedding determinant formula).

[F6]

A real number that is a root of unity is $\pm 1$ , of multiplicative order $1$ or $2$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity). The signature $(r_{1}, r_{2})$ of a number field satisfies $r_{1} + 2 r_{2} = [L : Q]$ (Archimedean embeddings and signature).

Proof

technique · direct

1.1F4given

In the main case $f > 1$ , each $p_{i}^{a_{i}}$ is at least $3$ : if $p_{i}$ is odd this is clear, and if $p_{i} = 2$ then $4 ∣ f$ because $f$ is reduced, so $a_{i} \geq 2$ . Hence each $K_{i}$ has degree $e_{i} \geq 2$ , and $φ (f) = \prod_{i \leq r} e_{i}$ by multiplicativity of $φ$ on the coprime factors $p_{i}^{a_{i}}$ .

1.2F1given

For every $i$ the power-basis computation of [F1] gives $∣ d_{K_{i}} ∣ = p_{i}^{L_{i}}$ with $L_{i} = a_{i} e_{i} - e_{i} / (p_{i} - 1)$ , and the p_i-adic exponent of the claimed formula is $φ (f) a_{i} - φ (f) / (p_{i} - 1)$ .

1.3F6given

The field $K = Q (ζ)$ has no real embedding: if $σ (K) \subseteq R$ for an embedding $σ$ , then $σ (ζ) \in R$ is a root of unity whose order is exactly $f \geq 3$ , because $σ (ζ)^{m} = 1$ if and only if $ζ^{m} = 1$ if and only if $f ∣ m$ ; but by [F6] a real root of unity has order $1$ or $2$ , a contradiction. Hence $r_{1} = 0$ , complex conjugation acts on the $n$ embeddings as a fixed-point-free involution, and $r_{2} = n / 2 = φ (f) / 2$ by [F6].

1.4F3

In the separate case from the Statement with $f = 1$ , one has $K = Q$ , and [F3] gives the integral basis $(1)$ of $O_{Q} = Z$ . Its trace Gram matrix is the $1 \times 1$ matrix $[{Tr}_{Q / Q} (1 \cdot 1)] = [1]$ , so its determinant is $1$ ; by the discriminant definition in [F3], $d_{Q} = 1$ .

2.1F2F4step 1.2

By induction on $j$ , $∣ d_{M_{j}} ∣ = \prod_{i \leq j} ∣ d_{K_{i}} ∣^{φ (n_{j}) / e_{i}}$ : for $j = 1$ , $M_{1} = K_{1}$ and the exponent is $1$ ; for $j \geq 2$ , [F4] gives $[M_{j - 1} : Q] = φ (n_{j - 1})$ and $[K_{j} : Q] = e_{j}$ with $M_{j} = M_{j - 1} K_{j}$ , the discriminants $d_{M_{j - 1}}$ and $d_{K_{j}}$ are coprime because one is $\pm$ a product of powers of $p_{1}, \dots, p_{j - 1}$ and the other is $\pm p_{j}^{L_{j}}$ , and [F2] then gives $∣ d_{M_{j}} ∣ = ∣ d_{M_{j - 1}} ∣^{e_{j}} ∣ d_{K_{j}} ∣^{φ (n_{j - 1})}$ .

2.2F3F5step 1.3

With respect to the integral basis $1, ζ, \dots, ζ^{n - 1}$ of [F3], the determinant $Δ : = \det (σ_{i} (ζ^{j - 1}))$ of [F5] is nonzero and $d_{K} = Δ^{2}$ . Complex conjugation permutes the index set of the embeddings by a product of $n / 2$ transpositions by step 1.3, so $\overline{Δ} = (- 1)^{n / 2} Δ$ and therefore $∣ Δ ∣^{2} = \overline{Δ} Δ = (- 1)^{n / 2} Δ^{2} = (- 1)^{n / 2} d_{K}$ ; since $∣ Δ ∣^{2} > 0$ , the sign of $d_{K}$ is $(- 1)^{n / 2} = (- 1)^{φ (f) / 2}$ .

3.1F4step 1.2step 2.1

Taking absolute values in the induction of step 2.1 with $j = r$ and using $M_{r} = K$ and $φ (n_{r}) = φ (f)$ gives $∣ d_{K} ∣ = \prod_{i \leq r} p_{i}^{L_{i} φ (f) / e_{i}} = \prod_{i \leq r} p_{i}^{φ (f) (a_{i} - 1 / (p_{i} - 1))} = f^{φ (f)} / \prod_{i \leq r} p_{i}^{φ (f) / (p_{i} - 1)}$ .

4.1step 2.2step 3.1step 1.4∎

Combining the sign of step 2.2 with the absolute value of step 3.1 gives $d_{K} = (- 1)^{φ (f) / 2} f^{φ (f)} / \prod_{p ∣ f} p^{φ (f) / (p - 1)}$ for the given reduced index $f > 1$ ; the separate $f = 1$ case has $d_{Q} = 1$ by step 1.4. These are the cases in the Statement.

Remarks

Sign and absolute value are computed separately. The absolute value comes from the prime-power absolute discriminants and the coprime-discriminant compositum formula; the sign comes from the pairing of complex embeddings. Neither the different ideal nor its positive norm is used.
Reduced index. For an unreduced index $2 m$ with $m$ odd one has $Q (ζ_{2 m}) = Q (ζ_{m})$ , and the formula must be applied to the reduced index $m$ ; the statements of this pair therefore exclude $f \equiv 2 (m o d 4)$ .

Depends on

Used by

Arithmetic of Q(zeta₅) Example
Prime decomposition in Q(zeta₁2) Example
Prime decomposition in Q(zeta₈) Example

Dependency tree · two levels

81 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Algebraic Number Theory, Remark 6.6(c) and Proposition 6.2(d) (standard reference, not scraped)
Conrad-Landesman, Math 154 Algebraic Number Theory, Theorem 11.6 and Remark 11.7 (standard reference, not scraped)