Gsyfuy

Fourth person (in Slavey language)

The bar has been raised

Why doesn't this Google Translate ad use the word "Translation" instead of "Translate"?

Try Catch Block Affecting a Variable in an Enclosing Scope

How did the power source of Mar-Vell's aircraft end up with her?

Moving plot label

How do I deal with a powergamer in a game full of beginners in a school club?

Look through the portal of every day

Good for you! in Russian

What Happens when Passenger Refuses to Fly Boeing 737 Max?

Is it possible to have an Abelian group under two different binary operations but the binary operations are not distributive?

How much stiffer are 23c tires over 28c?

What to do when during a meeting client people start to fight (even physically) with each others?

How strictly should I take "Candidates must be local"?

Is there an elementary proof that there are infinitely many primes that are *not* completely split in an abelian extension?

Are babies of evil humanoid species inherently evil?

Can someone explain what is being said here in color publishing in the American Mathematical Monthly?

Algorithm to convert a fixed-length string to the smallest possible collision-free representation?

Do I really need to have a scientific explanation for my premise?

Virginia employer terminated employee and wants signing bonus returned

Force user to remove USB token

Accountant/ lawyer will not return my call

Should I tell my boss the work he did was worthless

Is Gradient Descent central to every optimizer?

Norms on fields

Equivalence of Euclidean norm used in Ostrowski's Theoremequivalent norms on vector spaceEquivalence between norms induced by a scalar productStronger, Weaker, and Equivalent Norms on $C[0,1]$.Non-equivalent norms on finite dimensional vector spaces over a non-complete fieldDefinitions of equivalent normsequivalence of norms arising from inner productsSubmultiplicity of Induced NormsCollection of all norms such that a Space is CompleteWhy Do We Only Take Norms Over Real/Complex Numbers?

5

1

$begingroup$

I'm doing an introductory module in number theory, and came across the definition of a norm on a field. It seems to agree with the definition of a norm on a vector space over a field (just view the field as a vector space in the former definition). Then I read Ostrowski's Theorem that says 'every non trivial norm on $mathbb{Q}$ is equivalent either to the standard absoltute value or to the p-adic norm for some prime p. All these norms are inequivalent.'

I don't see how this theorem doesn't contradict the theorem that all norms on a finite dimensional vector space are equivalent.....However I see a difference in the way 'equivalence' is being defined between norms in the case where the norms are on a field as opposed to a general vector space. Everything seems a bit all over the place in my head and I don't know why there is a difference in definition of norms being equivalent on fields and norms on vector spaces when surely a field is a vector space...

If someone could clear things up I'd appreciate it.

norm p-adic-number-theory

share|cite|improve this question

edited 4 hours ago

Displayname

asked 5 hours ago

DisplaynameDisplayname

21511

$endgroup$

add a comment | 

5

1

$begingroup$

I'm doing an introductory module in number theory, and came across the definition of a norm on a field. It seems to agree with the definition of a norm on a vector space over a field (just view the field as a vector space in the former definition). Then I read Ostrowski's Theorem that says 'every non trivial norm on $mathbb{Q}$ is equivalent either to the standard absoltute value or to the p-adic norm for some prime p. All these norms are inequivalent.'

I don't see how this theorem doesn't contradict the theorem that all norms on a finite dimensional vector space are equivalent.....However I see a difference in the way 'equivalence' is being defined between norms in the case where the norms are on a field as opposed to a general vector space. Everything seems a bit all over the place in my head and I don't know why there is a difference in definition of norms being equivalent on fields and norms on vector spaces when surely a field is a vector space...

If someone could clear things up I'd appreciate it.

norm p-adic-number-theory

share|cite|improve this question

edited 4 hours ago

Displayname

asked 5 hours ago

DisplaynameDisplayname

21511

$endgroup$

add a comment | 

$begingroup$

I'm doing an introductory module in number theory, and came across the definition of a norm on a field. It seems to agree with the definition of a norm on a vector space over a field (just view the field as a vector space in the former definition). Then I read Ostrowski's Theorem that says 'every non trivial norm on $mathbb{Q}$ is equivalent either to the standard absoltute value or to the p-adic norm for some prime p. All these norms are inequivalent.'

I don't see how this theorem doesn't contradict the theorem that all norms on a finite dimensional vector space are equivalent.....However I see a difference in the way 'equivalence' is being defined between norms in the case where the norms are on a field as opposed to a general vector space. Everything seems a bit all over the place in my head and I don't know why there is a difference in definition of norms being equivalent on fields and norms on vector spaces when surely a field is a vector space...

If someone could clear things up I'd appreciate it.

norm p-adic-number-theory

share|cite|improve this question

edited 4 hours ago

Displayname

asked 5 hours ago

DisplaynameDisplayname

21511

$endgroup$

share|cite|improve this question

edited 4 hours ago

Displayname

asked 5 hours ago

DisplaynameDisplayname

21511

share|cite|improve this question

edited 4 hours ago

Displayname

asked 5 hours ago

DisplaynameDisplayname

21511

asked 5 hours ago

DisplaynameDisplayname

21511

asked 5 hours ago

DisplaynameDisplayname

21511

2 Answers
2

active

oldest

votes

3

$begingroup$

The difference is how multiplication by scalars is handled.

When considering a norm $VertcdotVert$ on a vector space $V$ over some field $F$, you need to fix the notion of absolute value on the scalar-field $F$, so that the axiom
$$Vert ccdot xVert = vert cvert cdotVert xVert,quad cin F, xin V$$
makes sense.

The statement that 'all norms on a finite dimensional vector space are equivalent' makes use of the assumption that the above axiom is satisfied for the same notion of absolute value. This axiom fails if we try to view the $p$-adic norms as norms on vector spaces over $mathbb{Q}$ with the usual notion of absolute value. E.g. for the $2$-adic norm, we have
$$Vert3cdot 2Vert_2 = Vert 2Vert_2 neq vert 3vertcdotVert 2Vert_2.$$

share|cite|improve this answer

answered 3 hours ago

Eero HakavuoriEero Hakavuori

1,58769

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot for your answer, I've replied to Lubin because he commented as I was reading this, but if you could answer any of my further questions that'd be great
  $endgroup$
  – Displayname
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

I think that everything that @Eero has said is true, but there’s more wrong in the situation than that.

You have not quoted the theorem about norms on a finite-dimensional vector space fully. First, you have to start with a normed field $k$, and ask about the norms on a finite $k$-space that in some sense extend that norm. Second, the theorem does not hold if $k$ is not complete under the given norm.

So there’s no contradiction, because in talking about $Bbb Q$, you have not started with a pre-given norm on a field. Rather, you are asking whether there may be more than one norm on a field that has been given abstractly. Since $Bbb Q$ has many topologies on it, each of them making it a topological field (with continuity of $+$, $times$, and $div$, for instance) and all of them different as topologies, it should not be surprising that the completions are different.

share|cite|improve this answer

answered 3 hours ago

LubinLubin

45.1k44687

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much for your answer, I've encountered norms in another module and after carefully re-reading my notes have realised that the arbitrary field $mathbb{K}$ my notes refer to in that module aren't all that arbitrary, for the notes claim $mathbb{K}$ will either denote $mathbb{C}$ or $mathbb{R}$ and I think I may have mistakenly taken this as any subfield of $mathbb{C}$. I've never thought about what would be the consequences of the normed space being over a field which is not complete w.r.t the absolute value norm...I'm still kind of confused
  $endgroup$
  – Displayname
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My notes (referring to my differentiation module now) have a little side note which says 'Non-archimedean norms are good for number theory but are not 'decent' for our purpose'....I know what a non-archimedean norm is but I don't know why it's not 'decent'....I'm assuming the lecturer has intentionally not said much about this since it may be beyond the scope of the course? (2nd year undergrad) ....back to my original question, what if we consider the vector space $mathbb{Q}$ over the field $mathbb{R}$, is this not valid because the homogeneity condition doesn't hold? (as Eoro pointed out)
  $endgroup$
  – Displayname
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ^ ran out of characters...I mean is the 'normed space' $mathbb{Q}$ over $mathbb{R}$ w.r.t a p-adic norm a normed space (we are now using a field which is complete w.r.t absolute value)
  $endgroup$
  – Displayname
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3145348%2fnorms-on-fields%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

The difference is how multiplication by scalars is handled.

When considering a norm $VertcdotVert$ on a vector space $V$ over some field $F$, you need to fix the notion of absolute value on the scalar-field $F$, so that the axiom
$$Vert ccdot xVert = vert cvert cdotVert xVert,quad cin F, xin V$$
makes sense.

The statement that 'all norms on a finite dimensional vector space are equivalent' makes use of the assumption that the above axiom is satisfied for the same notion of absolute value. This axiom fails if we try to view the $p$-adic norms as norms on vector spaces over $mathbb{Q}$ with the usual notion of absolute value. E.g. for the $2$-adic norm, we have
$$Vert3cdot 2Vert_2 = Vert 2Vert_2 neq vert 3vertcdotVert 2Vert_2.$$

share|cite|improve this answer

answered 3 hours ago

Eero HakavuoriEero Hakavuori

1,58769

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot for your answer, I've replied to Lubin because he commented as I was reading this, but if you could answer any of my further questions that'd be great
  $endgroup$
  – Displayname
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

The difference is how multiplication by scalars is handled.

When considering a norm $VertcdotVert$ on a vector space $V$ over some field $F$, you need to fix the notion of absolute value on the scalar-field $F$, so that the axiom
$$Vert ccdot xVert = vert cvert cdotVert xVert,quad cin F, xin V$$
makes sense.

The statement that 'all norms on a finite dimensional vector space are equivalent' makes use of the assumption that the above axiom is satisfied for the same notion of absolute value. This axiom fails if we try to view the $p$-adic norms as norms on vector spaces over $mathbb{Q}$ with the usual notion of absolute value. E.g. for the $2$-adic norm, we have
$$Vert3cdot 2Vert_2 = Vert 2Vert_2 neq vert 3vertcdotVert 2Vert_2.$$

share|cite|improve this answer

answered 3 hours ago

Eero HakavuoriEero Hakavuori

1,58769

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks a lot for your answer, I've replied to Lubin because he commented as I was reading this, but if you could answer any of my further questions that'd be great
  $endgroup$
  – Displayname
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

The difference is how multiplication by scalars is handled.

When considering a norm $VertcdotVert$ on a vector space $V$ over some field $F$, you need to fix the notion of absolute value on the scalar-field $F$, so that the axiom
$$Vert ccdot xVert = vert cvert cdotVert xVert,quad cin F, xin V$$
makes sense.

The statement that 'all norms on a finite dimensional vector space are equivalent' makes use of the assumption that the above axiom is satisfied for the same notion of absolute value. This axiom fails if we try to view the $p$-adic norms as norms on vector spaces over $mathbb{Q}$ with the usual notion of absolute value. E.g. for the $2$-adic norm, we have
$$Vert3cdot 2Vert_2 = Vert 2Vert_2 neq vert 3vertcdotVert 2Vert_2.$$

share|cite|improve this answer

answered 3 hours ago

Eero HakavuoriEero Hakavuori

1,58769

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Eero HakavuoriEero Hakavuori

1,58769

answered 3 hours ago

Eero HakavuoriEero Hakavuori

1,58769

answered 3 hours ago

Eero HakavuoriEero Hakavuori

1,58769

2

$begingroup$

I think that everything that @Eero has said is true, but there’s more wrong in the situation than that.

You have not quoted the theorem about norms on a finite-dimensional vector space fully. First, you have to start with a normed field $k$, and ask about the norms on a finite $k$-space that in some sense extend that norm. Second, the theorem does not hold if $k$ is not complete under the given norm.

So there’s no contradiction, because in talking about $Bbb Q$, you have not started with a pre-given norm on a field. Rather, you are asking whether there may be more than one norm on a field that has been given abstractly. Since $Bbb Q$ has many topologies on it, each of them making it a topological field (with continuity of $+$, $times$, and $div$, for instance) and all of them different as topologies, it should not be surprising that the completions are different.

share|cite|improve this answer

answered 3 hours ago

LubinLubin

45.1k44687

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much for your answer, I've encountered norms in another module and after carefully re-reading my notes have realised that the arbitrary field $mathbb{K}$ my notes refer to in that module aren't all that arbitrary, for the notes claim $mathbb{K}$ will either denote $mathbb{C}$ or $mathbb{R}$ and I think I may have mistakenly taken this as any subfield of $mathbb{C}$. I've never thought about what would be the consequences of the normed space being over a field which is not complete w.r.t the absolute value norm...I'm still kind of confused
  $endgroup$
  – Displayname
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My notes (referring to my differentiation module now) have a little side note which says 'Non-archimedean norms are good for number theory but are not 'decent' for our purpose'....I know what a non-archimedean norm is but I don't know why it's not 'decent'....I'm assuming the lecturer has intentionally not said much about this since it may be beyond the scope of the course? (2nd year undergrad) ....back to my original question, what if we consider the vector space $mathbb{Q}$ over the field $mathbb{R}$, is this not valid because the homogeneity condition doesn't hold? (as Eoro pointed out)
  $endgroup$
  – Displayname
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ^ ran out of characters...I mean is the 'normed space' $mathbb{Q}$ over $mathbb{R}$ w.r.t a p-adic norm a normed space (we are now using a field which is complete w.r.t absolute value)
  $endgroup$
  – Displayname
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

I think that everything that @Eero has said is true, but there’s more wrong in the situation than that.

You have not quoted the theorem about norms on a finite-dimensional vector space fully. First, you have to start with a normed field $k$, and ask about the norms on a finite $k$-space that in some sense extend that norm. Second, the theorem does not hold if $k$ is not complete under the given norm.

So there’s no contradiction, because in talking about $Bbb Q$, you have not started with a pre-given norm on a field. Rather, you are asking whether there may be more than one norm on a field that has been given abstractly. Since $Bbb Q$ has many topologies on it, each of them making it a topological field (with continuity of $+$, $times$, and $div$, for instance) and all of them different as topologies, it should not be surprising that the completions are different.

share|cite|improve this answer

answered 3 hours ago

LubinLubin

45.1k44687

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much for your answer, I've encountered norms in another module and after carefully re-reading my notes have realised that the arbitrary field $mathbb{K}$ my notes refer to in that module aren't all that arbitrary, for the notes claim $mathbb{K}$ will either denote $mathbb{C}$ or $mathbb{R}$ and I think I may have mistakenly taken this as any subfield of $mathbb{C}$. I've never thought about what would be the consequences of the normed space being over a field which is not complete w.r.t the absolute value norm...I'm still kind of confused
  $endgroup$
  – Displayname
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My notes (referring to my differentiation module now) have a little side note which says 'Non-archimedean norms are good for number theory but are not 'decent' for our purpose'....I know what a non-archimedean norm is but I don't know why it's not 'decent'....I'm assuming the lecturer has intentionally not said much about this since it may be beyond the scope of the course? (2nd year undergrad) ....back to my original question, what if we consider the vector space $mathbb{Q}$ over the field $mathbb{R}$, is this not valid because the homogeneity condition doesn't hold? (as Eoro pointed out)
  $endgroup$
  – Displayname
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ^ ran out of characters...I mean is the 'normed space' $mathbb{Q}$ over $mathbb{R}$ w.r.t a p-adic norm a normed space (we are now using a field which is complete w.r.t absolute value)
  $endgroup$
  – Displayname
  2 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

I think that everything that @Eero has said is true, but there’s more wrong in the situation than that.

You have not quoted the theorem about norms on a finite-dimensional vector space fully. First, you have to start with a normed field $k$, and ask about the norms on a finite $k$-space that in some sense extend that norm. Second, the theorem does not hold if $k$ is not complete under the given norm.

So there’s no contradiction, because in talking about $Bbb Q$, you have not started with a pre-given norm on a field. Rather, you are asking whether there may be more than one norm on a field that has been given abstractly. Since $Bbb Q$ has many topologies on it, each of them making it a topological field (with continuity of $+$, $times$, and $div$, for instance) and all of them different as topologies, it should not be surprising that the completions are different.

share|cite|improve this answer

answered 3 hours ago

LubinLubin

45.1k44687

$endgroup$

share|cite|improve this answer

answered 3 hours ago

LubinLubin

45.1k44687

answered 3 hours ago

LubinLubin

45.1k44687

answered 3 hours ago

LubinLubin

45.1k44687