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 xL and σGal(LK), 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,,αnKsep (Ksep is the separable closure of K)

  • βKsep with

    |βα1|<|βαi|

    for i=2,,n.

Then α1K(β).

Proof. Let L=K(β), L=L(α1,,αn). Then LL is a Galois extension. Let σGal(LL). We have

|βσ(α1)|=|σ(βα1)|=|βα1|

using Lemma 8.2. Hence σ(α1)=α1, so α1K(β).

Proposition 8.4. Assuming that:

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

  • f(X)=i=0naiXiOK[X] a separable irreducible monic polynomial

  • αKsep a root of f

Then there exists 𝜀>0 such that for any g(X)=i=0nbiXiOK[X] monic with |aibi|<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,,αnKsep 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 βK(α1) such that g(β)=0 and |α1|<|g(α1)|. Then

|g(α1)|=|f(α1)|=j=1n|α1αj||α1αi|

for i=2,,n. (Use |α1αi|1 since αi integral). Since |βα1|<|α1αi|=|βαi| using Lemma 1.6, we have that Kummer’s Lemma gives that α1K(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 is the completion of , and is the completion of (i) (with respect to ||).

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

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