# Talk:Local field

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Field: Number theory

I changed the definition of discrete valuation so that it directly maps into the integers. Otherwise, we would have to say it maps to an ordered abelian group isomorphic as ordered abelian groups to Z in order to be able speak about the elements with negative valutation, and that seems to be overkill.

Regarding the definition: in http://planetmath.org/encyclopedia/LocalField.html they include R and C as local fields. Is our definition the standard one? http://mathworld.wolfram.com/LocalField.html has yet another definition, which seems redundant however. They also give a different list of examples. AxelBoldt 19:18 Nov 9, 2002 (UTC)

The definition in this article as it is right now is incorrect (it is the definition of a complete nonarchimedean field). One must add, as in the mathworld.wolfram link above that the residue field is finite. Otherwise, $\mathbb{C}((t))$ will be a local field, which it isn't (its maximal ideal is (t), hence its residue field is $\mathbb{C}$ which isn't finite). The better definition (at least in my opinion) is the planetmath definition, though one would normally require that the topology is non-discrete. I'll rewrite this article in the next few days. RobHar 07:46, 1 December 2006 (UTC)

And also the planetmath definition allows R and C to be called local fields, which they should be. RobHar 08:17, 1 December 2006 (UTC)

Done. RobHar 11:40, 1 December 2006 (UTC)

I think that's probably clearer anyway. Good catch on the Laurent/power series mistake also, that was rather a gaffe. R and C are also referred to as local fields in the real numbers article. It depends on the definition: when I saw that and your question I recalled having read a definition of local fields as a locally compact Hausdorff topological field and that was that. I'll do some poking around today and probably add some remarks about the ambiguity of the definition. Also added finite extensions of Qp and finite extensions of function fields over finite fields to the zoo. alodyne

If we go for the "locally compact" definition, we have to be careful however, since the maximal-compact-subring statement won't remain true.

Can every finite extension of the p-adics be turned into a local field? And if so, can it be done in a natural or unique way? AxelBoldt 00:02 Nov 11, 2002 (UTC)

yes, and in a natural way. If L is a finite extension of the p-adics of degree n and v is the valuation on the p-adics, then this valuation can be uniquely extended to a valuation w on L, and L under this valuation will be a local field (note however that if v was the normalized valuation on the p-adics then w will not be the normalized valuation on L, if n>1). In terms of the absolute value, the extension is given as follows
$|a|_w=\sqrt[n]{|N(a)|_v}$
where a is an element of L and N is the norm map from L to the p-adics. This statement is still true (with the same formula holding) if L is an algebraic extension of any field complete with respect to an absolute value. RobHar 05:39, 18 December 2006 (UTC)

Under most usual definitions, the valuation domain R of a valued field K is either R = { x in K : |x| <= 1 } given the Artin-style valuations, or R = { x in K : v(x) >= 0 } given the Krull-style valuations. In this article one has the inequality backwards. Since valuation hasn't been defined yet, it's not wrong (yet), but is probably confusing.

When you say that the valuation takes its values in the integers, that is the Krull-style, v(x), and leaves out the value infinity for v(0). The Artin-style would take its values in the set { p^n : n in Z } union {0}, where p is probably a positive prime integer, but occasionally set to other numbers. These are all nitpicks until a definition has been given. For instance sometimes v is only defined on the nonzero elements of K.

It seems to me that the modulus we define by |a| = mu(a S)/mu(S) where mu is the Haar measure, need not be an absolute value in the usual sense. For example, if we take the field to be the complex numbers then we get that |a| is the square of the usual modulus of a complex number; this does not satisfy the triangle inequality.

What is true however is that the modulus we define by Haar measure satisfies

 |a+b| <= A sup(|a|,|b|)


for some constant A>0.

It would be good if someone who is more knowledgeable than me could clean this point up. As it is written now it is misleading to act as if the modulus we define using Haar measure is an absolute value. Maybe it can be made an absolute value by scaling it by a power? 128.97.19.89 (talk) 17:51, 27 April 2010 (UTC)

That the usual absolute value on C satisfies the triangle inequality follows from the law of cosines. This law clearly states that the square of the usual absolute value on C also satisfies the triangle in equality, so the the modulus defined using a Haar measure is an absolute value. RobHar (talk) 19:31, 27 April 2010 (UTC)

## Where is the definition?

I think this article could be improved a great deal. (I'm a professional pure mathematician, but don't know about local fields.) The main problem is that the page doesn't seem to contain an actual definition of local field.

Maybe the first sentence of the article is intended as a definition? If so, it's certainly not clear that that's the intention: it could just mean that every local field is a locally compact topological field with respect to a non-discrete topology, without that being a sufficient condition. By way of comparison, an article on topological groups might start "A topological group is a topological space that also carries a group structure", although that's not a sufficient condition (as the group operations have to be continuous). It's also not usual for Wikipedia mathematics articles to give the definition in the first sentence.

Somewhere in the article should be a sentence beginning: "Definition: a local field is ..."

Another point that needs clarification is whether being local is a property of a field, or extra structure on a field. The fact that a local field carries a topology suggests that it might be extra structure, so that a field could be a local field in more than one way (just as many groups can be given the structure of a topological group in more than one way). But perhaps it's a theorem that any given field can be a local field in at most one way. I don't know: I couldn't tell from the article.

It would be great if someone with relevant expertise could edit the page to clarify these points. — Preceding unsigned comment added by 81.109.231.166 (talk) 20:18, 18 August 2012 (UTC)