Norms on fields
$begingroup$
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
$endgroup$
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
$endgroup$
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
$endgroup$
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
norm p-adic-number-theory
edited 11 hours ago
Displayname
asked 12 hours ago
DisplaynameDisplayname
21511
21511
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
$$Vert3cdot 2Vert_2 = Vert 2Vert_2 neq vert 3vertcdotVert 2Vert_2.$$
$endgroup$
$begingroup$
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
9 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.
$endgroup$
$begingroup$
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
9 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
9 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
9 hours ago
$begingroup$
Over $Bbb R$, all vector spaces are archimedean; even if you throw continuity out the window, no $Bbb Q_p$ can be a ring containing $Bbb R$. For, $Bbb R$ contains square roots of all positive integers, while no $p$-adic field does so (not even finite extensions of $Bbb Q_p$).
$endgroup$
– Lubin
3 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$Vert3cdot 2Vert_2 = Vert 2Vert_2 neq vert 3vertcdotVert 2Vert_2.$$
$endgroup$
$begingroup$
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
9 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
$$Vert3cdot 2Vert_2 = Vert 2Vert_2 neq vert 3vertcdotVert 2Vert_2.$$
$endgroup$
$begingroup$
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
9 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
$$Vert3cdot 2Vert_2 = Vert 2Vert_2 neq vert 3vertcdotVert 2Vert_2.$$
$endgroup$
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
$$Vert3cdot 2Vert_2 = Vert 2Vert_2 neq vert 3vertcdotVert 2Vert_2.$$
answered 11 hours ago
Eero HakavuoriEero Hakavuori
1,58769
1,58769
$begingroup$
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
9 hours ago
add a comment |
$begingroup$
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
9 hours ago
$begingroup$
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
9 hours ago
$begingroup$
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
9 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.
$endgroup$
$begingroup$
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
9 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
9 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
9 hours ago
$begingroup$
Over $Bbb R$, all vector spaces are archimedean; even if you throw continuity out the window, no $Bbb Q_p$ can be a ring containing $Bbb R$. For, $Bbb R$ contains square roots of all positive integers, while no $p$-adic field does so (not even finite extensions of $Bbb Q_p$).
$endgroup$
– Lubin
3 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.
$endgroup$
$begingroup$
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
9 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
9 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
9 hours ago
$begingroup$
Over $Bbb R$, all vector spaces are archimedean; even if you throw continuity out the window, no $Bbb Q_p$ can be a ring containing $Bbb R$. For, $Bbb R$ contains square roots of all positive integers, while no $p$-adic field does so (not even finite extensions of $Bbb Q_p$).
$endgroup$
– Lubin
3 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.
$endgroup$
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.
answered 10 hours ago
LubinLubin
45.1k44687
45.1k44687
$begingroup$
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
9 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
9 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
9 hours ago
$begingroup$
Over $Bbb R$, all vector spaces are archimedean; even if you throw continuity out the window, no $Bbb Q_p$ can be a ring containing $Bbb R$. For, $Bbb R$ contains square roots of all positive integers, while no $p$-adic field does so (not even finite extensions of $Bbb Q_p$).
$endgroup$
– Lubin
3 hours ago
add a comment |
$begingroup$
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
9 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
9 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
9 hours ago
$begingroup$
Over $Bbb R$, all vector spaces are archimedean; even if you throw continuity out the window, no $Bbb Q_p$ can be a ring containing $Bbb R$. For, $Bbb R$ contains square roots of all positive integers, while no $p$-adic field does so (not even finite extensions of $Bbb Q_p$).
$endgroup$
– Lubin
3 hours ago
$begingroup$
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
9 hours ago
$begingroup$
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
9 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
9 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
9 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
9 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
9 hours ago
$begingroup$
Over $Bbb R$, all vector spaces are archimedean; even if you throw continuity out the window, no $Bbb Q_p$ can be a ring containing $Bbb R$. For, $Bbb R$ contains square roots of all positive integers, while no $p$-adic field does so (not even finite extensions of $Bbb Q_p$).
$endgroup$
– Lubin
3 hours ago
$begingroup$
Over $Bbb R$, all vector spaces are archimedean; even if you throw continuity out the window, no $Bbb Q_p$ can be a ring containing $Bbb R$. For, $Bbb R$ contains square roots of all positive integers, while no $p$-adic field does so (not even finite extensions of $Bbb Q_p$).
$endgroup$
– Lubin
3 hours ago
add a comment |
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