Talk:Local field

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)[reply]

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)[reply]

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

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 Q_p 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)[reply]

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)[reply]

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)[reply]

This is wrong. The square of the absolute value does not satisfy the triangle inequality. (2+1)^2=9 > 4+1=2^2+1. It is not an absolute value in the ordinary sense. — Preceding unsigned comment added by 118.253.4.58 (talk) 04:37, 13 October 2014 (UTC)[reply]

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)[reply]

Needs a definition!

The Introduction includes this as its first sentence:

"In mathematics, a local field is a special type of field that is a locally compact topological field with respect to a non-discrete topology."

This only states that a local field is a special type of field that also has a certain property. This is not a definition, even if it was intended to be.

This informal description is fine for the Introduction, but there should be a rigorous definition that doesn't merely state that a local field is a certain kind of special type of field!

Preferably in the first section after the Introduction.

To make matters worse, at the end of the introduction, this passage occurs:

"There is an equivalent definition of non-archimedean local field: it is a field that is complete with respect to a discrete valuation and whose residue field is finite. However, some authors consider a more general notion, requiring only that the residue field be perfect, not necessarily finite. This article uses the former definition."

The phrase "equivalent definition" implies that someone believes a definition has already been presented. But it has not.

To make matters much worse, the phrase "former definition" is included in this passage, even after two definitions are included beyond the nonexistent first definition, making three definitions or nonexistent definitions that have been referred to! Which one is the "former" definition out of these three? Much confusion is sowed by this mess!Daqu (talk) 20:15, 21 March 2016 (UTC)[reply]

This absolute value does not satisfy the triangle inequality in the case of C

Hey all, I know this issue has been discussed above in 2014, but since the corresponding passge is still in the article, I would like to repeat it.

In the case of K=Complex Numbers, the "absolute value" defined here is the square of the usual absolute value and as the example above ( (2+1)^2=9 > 4+1=2^2+1) shows, this does not satisfy the triangle inequalitiy. For all other local fields, the formula should be fine (for R we get the usual absolute value and for non-archimedian fields we have an "ultra-triangle inequality" which should also work). I came across this issue while working on the German version of this article. I wanted to include this formula from this article only to realize it is false.

Any ideas how to fix this problem?

Best regards, --CoTangent (talk) 11:02, 22 March 2020 (UTC)[reply]

Inconsistency in the lead

The definition given in the first sentence, "a field K is called a local field if it is complete with respect to a topology induced by a discrete valuation v and if its residue field k is finite" is in contradiction with the rest of the paragraph, in that it excludes the Archimedean local fields that are prominently mentioned in what follows. It is true that that definition can be found in the cited reference (Cassels and Fröhlich, p.129), but the author of that article (J.-P. Serre) does not seem to consider the Archimedean case to fall under the "local field" umbrella, as is clear from the classification that follows. (Amusing side note: I tried to locate a definition of local field in Serre's book "Corps locaux"; not only could I not find one (with modest effort), but I also got the impression that the phrase "corps local" actually does not appear at all in that book, apart from in its title.) So it might be true the the real and complex numbers are the least loved among the local fields, or more accurately the fields that care least about the love the get from those studying local fields, but I think this blatant inconsistency should be mended one way or another. Marc van Leeuwen (talk) 14:35, 22 November 2021 (UTC)[reply]