8 Global Fields

Definition 8.1 (Global field). A global field is a field which is either:

  • (i) An algebraic number field
  • (ii) A global function field, i.e. a finite extension of 𝔽p(t).

Lemma 8.2. Assuming that:

Then for x∈L and σ∈Gal⁡(L∕K), we have |σ(x)|L=|x|L.

Proof. Since x↦|σ(x)|L is another absolute value on L extending |∙| on K, the result follows from uniqueness of |∙|L. □

Lemma 8.3 (Kummer’s Lemma). Assuming that:

  • (K,|∙|) a complete discretely valued field

  • f(X)∈K[X] a separable irreducible polynomial with roots α1,…,αn∈Ksep (Ksep is the separable closure of K)

  • β∈Ksep with

    |β−α1|<|β−αi|

    for i=2,…,n.

Then α1∈K(β).

Proof. Let L=K(β), L′=L(α1,…,αn). Then L′∕L is a Galois extension. Let σ∈Gal⁡(L′∕L). We have

|β−σ(α1)|=|σ(β−α1)|=|β−α1|

using Lemma 8.2. Hence σ(α1)=α1, so α1∈K(β). □

Proposition 8.4. Assuming that:

  • (F,|∙|) is a complete discretely valued field

  • f(X)=∑⁡i=0naiXi∈OK[X] a separable irreducible monic polynomial

  • α∈Ksep a root of f

Then there exists 𝜀>0 such that for any g(X)=∑⁡i=0nbiXi∈OK[X] monic with |ai−bi|<𝜀 for all i, there exists a root β of g(X) such that K(α)=K(β).

“Nearby polynomials define the same extensions”.

Proof. Let α1,…,αn∈Ksep be the roots of f which are necessarily distinct. Then f′(α1)≠0. We choose 𝜀 sufficiently small such that |g(α1)|<|f′(α1)|2 and |f′(α1)−g′(α1)|<|f′(α1)|. Then we have |g′(α1)|<|f′(α1)|2=|g′(α1)|2 (the equality is by Lemma 1.6).

By Hensel’s Lemma version 1 applied to the field K(α1) there exists β∈K(α1) such that g(β)=0 and |∙−α1|<|g′(α1)|. Then

|g′(α1)|=|f′(α1)|=∏j=1n|α1−αj|≤|α1−αi|

for i=2,…,n. (Use |α1−αi|≤1 since αi integral). Since |β−α1|<|α1−αi|=|β−αi| using Lemma 1.6, we have that Kummer’s Lemma gives that α1∈K(β) and hence K(α1)=K(β). □

Theorem 8.5. Assuming that:

Then K is the completion of a global field.

Proof. Case 1: |∙| is archimedean. Then ℝ is the completion of ℚ, and ℂ is the completion of ℚ(i) (with respect to |∙|∞).

Case 2: |∙| non-archimedean, equal characteristic. Then K≅𝔽q((t)) is the completion of 𝔽q(t) with respect to the t-adic valuation.

Case 3: |∙| non-archimedean mixed characteristic. Then K=ℚp(α), with α a root of a monic irreducible polynomial f(X)∈ℤp[X]. Since ℤ is dense in ℤp, we choose g(X)∈ℤ[X] as in Proposition 8.4. Then K=ℚ(β) with β a root of g(X). Since ℚ(β) dense in ℚp(β)=K, and K is complete, we must have that K is the completion of ℚ(β). □