Why do we define vector spaces over fields and not over commutative rings with unity?
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I am using commutative ring with unity in the sense that there exists at least one non-zero element in the ring which doesn't have a multiplicative inverse. Can't we define scalar multiplication on a vector space with elements of commutative ring with unity, instead of field?
linear-algebra
asked Nov 16 at 10:28
SALONI SINHA
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up vote
2
down vote
favorite
I am using commutative ring with unity in the sense that there exists at least one non-zero element in the ring which doesn't have a multiplicative inverse. Can't we define scalar multiplication on a vector space with elements of commutative ring with unity, instead of field?
linear-algebra
asked Nov 16 at 10:28
SALONI SINHA
162
New contributor
SALONI SINHA is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2
Yes we can, but it has slightly different properties, and is called a module over the ring.
– Berci
Nov 16 at 10:31
1
Yes we can. But some things break down. For example we cannot prove that $acdot x=0$ implies that either $x=0$ or $a=0$ because we can no longer multiply by $1/a$. This has implications elsewhere. For example, a non-zero vector need not form a linearly independent set. Consequently the concepts of basis and dimension need modifications, and won't play out as nicely as they do over a field. You cannot necessarily find linearly independent sets of generators (even assuming a finite generator set).
– Jyrki Lahtonen
Nov 16 at 12:29
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am using commutative ring with unity in the sense that there exists at least one non-zero element in the ring which doesn't have a multiplicative inverse. Can't we define scalar multiplication on a vector space with elements of commutative ring with unity, instead of field?
linear-algebra
asked Nov 16 at 10:28
SALONI SINHA
162
New contributor
SALONI SINHA is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am using commutative ring with unity in the sense that there exists at least one non-zero element in the ring which doesn't have a multiplicative inverse. Can't we define scalar multiplication on a vector space with elements of commutative ring with unity, instead of field?
linear-algebra
linear-algebra
asked Nov 16 at 10:28
SALONI SINHA
162
New contributor
SALONI SINHA is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 10:28
SALONI SINHA
162
New contributor
SALONI SINHA is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 10:28
SALONI SINHA
162
New contributor
SALONI SINHA is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 10:28
SALONI SINHA
162
asked Nov 16 at 10:28
SALONI SINHA
162
162
New contributor
SALONI SINHA is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
SALONI SINHA is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2
Yes we can, but it has slightly different properties, and is called a module over the ring.
– Berci
Nov 16 at 10:31
1
Yes we can. But some things break down. For example we cannot prove that $acdot x=0$ implies that either $x=0$ or $a=0$ because we can no longer multiply by $1/a$. This has implications elsewhere. For example, a non-zero vector need not form a linearly independent set. Consequently the concepts of basis and dimension need modifications, and won't play out as nicely as they do over a field. You cannot necessarily find linearly independent sets of generators (even assuming a finite generator set).
– Jyrki Lahtonen
Nov 16 at 12:29
add a comment |
2
Yes we can, but it has slightly different properties, and is called a module over the ring.
– Berci
Nov 16 at 10:31
1
Yes we can. But some things break down. For example we cannot prove that $acdot x=0$ implies that either $x=0$ or $a=0$ because we can no longer multiply by $1/a$. This has implications elsewhere. For example, a non-zero vector need not form a linearly independent set. Consequently the concepts of basis and dimension need modifications, and won't play out as nicely as they do over a field. You cannot necessarily find linearly independent sets of generators (even assuming a finite generator set).
– Jyrki Lahtonen
Nov 16 at 12:29
2
2
Yes we can, but it has slightly different properties, and is called a module over the ring.
– Berci
Nov 16 at 10:31
Yes we can, but it has slightly different properties, and is called a module over the ring.
– Berci
Nov 16 at 10:31
1
1
Yes we can. But some things break down. For example we cannot prove that $acdot x=0$ implies that either $x=0$ or $a=0$ because we can no longer multiply by $1/a$. This has implications elsewhere. For example, a non-zero vector need not form a linearly independent set. Consequently the concepts of basis and dimension need modifications, and won't play out as nicely as they do over a field. You cannot necessarily find linearly independent sets of generators (even assuming a finite generator set).
– Jyrki Lahtonen
Nov 16 at 12:29
Yes we can. But some things break down. For example we cannot prove that $acdot x=0$ implies that either $x=0$ or $a=0$ because we can no longer multiply by $1/a$. This has implications elsewhere. For example, a non-zero vector need not form a linearly independent set. Consequently the concepts of basis and dimension need modifications, and won't play out as nicely as they do over a field. You cannot necessarily find linearly independent sets of generators (even assuming a finite generator set).
– Jyrki Lahtonen
Nov 16 at 12:29
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
Yes, we can do that. We just don't call them “vector spaces”. We call them “modules” instead.
answered Nov 16 at 10:30
José Carlos Santos
140k19111204
add a comment |
up vote
0
down vote
You could, but there a lot of things you couldn't do with them, for instance normalization.
answered Nov 16 at 10:32
nbubis
26.9k552106
2
You can't always normalize over fields either. Consider $(1, 1)in Bbb Q^2$. And some vector spaces don't even have norms (although it's often possible to construct one).
– Arthur
Nov 16 at 10:33
add a comment |
up vote
0
down vote
If $R$ is a commutative unital ring, then an $R$-module is an abelian group $M$ with a "scalar multiplication" map $R times M to M$, written $r cdot m = rm$, which satisfies
- $(r + s)m = rm + sm$
- $r(m + n) = rm + rn$
- $r(sm) = (rs)m$
If $R$ is a field, then an $R$-module is precisely a vector space over that field, and we get the usual notions of dimension, bases, and so on. However, if $R$ is not a field, then an $R$-module is usually a slightly more complicated thing.
For example if $R = mathbb{Z}$, then $mathbb{Z}, mathbb{Z}^2, ldots$ are all $mathbb{Z}$-modules in the usual way, and any basis of $mathbb{Z}^n$ will have size $n$. However, $mathbb{Z}/3mathbb{Z}$ is also a $mathbb{Z}-module$, and it has no basis. (No element $m$ of $mathbb{Z}/3mathbb{Z}$ is linearly independent, since $3m = 0$). So the theory of modules is more complicated than the theory of vector spaces.
answered Nov 16 at 12:21
Joppy
5,448420
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Yes, we can do that. We just don't call them “vector spaces”. We call them “modules” instead.
answered Nov 16 at 10:30
José Carlos Santos
140k19111204
add a comment |
up vote
3
down vote
Yes, we can do that. We just don't call them “vector spaces”. We call them “modules” instead.
answered Nov 16 at 10:30
José Carlos Santos
140k19111204
add a comment |
up vote
3
down vote
up vote
3
down vote
Yes, we can do that. We just don't call them “vector spaces”. We call them “modules” instead.
answered Nov 16 at 10:30
José Carlos Santos
140k19111204
Yes, we can do that. We just don't call them “vector spaces”. We call them “modules” instead.
answered Nov 16 at 10:30
José Carlos Santos
140k19111204
answered Nov 16 at 10:30
José Carlos Santos
140k19111204
answered Nov 16 at 10:30
José Carlos Santos
140k19111204
answered Nov 16 at 10:30
José Carlos Santos
140k19111204
140k19111204
add a comment |
add a comment |
up vote
0
down vote
You could, but there a lot of things you couldn't do with them, for instance normalization.
answered Nov 16 at 10:32
nbubis
26.9k552106
2
You can't always normalize over fields either. Consider $(1, 1)in Bbb Q^2$. And some vector spaces don't even have norms (although it's often possible to construct one).
– Arthur
Nov 16 at 10:33
add a comment |
up vote
0
down vote
You could, but there a lot of things you couldn't do with them, for instance normalization.
answered Nov 16 at 10:32
nbubis
26.9k552106
2
You can't always normalize over fields either. Consider $(1, 1)in Bbb Q^2$. And some vector spaces don't even have norms (although it's often possible to construct one).
– Arthur
Nov 16 at 10:33
add a comment |
up vote
0
down vote
up vote
0
down vote
You could, but there a lot of things you couldn't do with them, for instance normalization.
answered Nov 16 at 10:32
nbubis
26.9k552106
You could, but there a lot of things you couldn't do with them, for instance normalization.
answered Nov 16 at 10:32
nbubis
26.9k552106
answered Nov 16 at 10:32
nbubis
26.9k552106
answered Nov 16 at 10:32
nbubis
26.9k552106
answered Nov 16 at 10:32
nbubis
26.9k552106
26.9k552106
2
You can't always normalize over fields either. Consider $(1, 1)in Bbb Q^2$. And some vector spaces don't even have norms (although it's often possible to construct one).
– Arthur
Nov 16 at 10:33
add a comment |
2
You can't always normalize over fields either. Consider $(1, 1)in Bbb Q^2$. And some vector spaces don't even have norms (although it's often possible to construct one).
– Arthur
Nov 16 at 10:33
2
2
You can't always normalize over fields either. Consider $(1, 1)in Bbb Q^2$. And some vector spaces don't even have norms (although it's often possible to construct one).
– Arthur
Nov 16 at 10:33
You can't always normalize over fields either. Consider $(1, 1)in Bbb Q^2$. And some vector spaces don't even have norms (although it's often possible to construct one).
– Arthur
Nov 16 at 10:33
add a comment |
up vote
0
down vote
If $R$ is a commutative unital ring, then an $R$-module is an abelian group $M$ with a "scalar multiplication" map $R times M to M$, written $r cdot m = rm$, which satisfies
- $(r + s)m = rm + sm$
- $r(m + n) = rm + rn$
- $r(sm) = (rs)m$
If $R$ is a field, then an $R$-module is precisely a vector space over that field, and we get the usual notions of dimension, bases, and so on. However, if $R$ is not a field, then an $R$-module is usually a slightly more complicated thing.
For example if $R = mathbb{Z}$, then $mathbb{Z}, mathbb{Z}^2, ldots$ are all $mathbb{Z}$-modules in the usual way, and any basis of $mathbb{Z}^n$ will have size $n$. However, $mathbb{Z}/3mathbb{Z}$ is also a $mathbb{Z}-module$, and it has no basis. (No element $m$ of $mathbb{Z}/3mathbb{Z}$ is linearly independent, since $3m = 0$). So the theory of modules is more complicated than the theory of vector spaces.
answered Nov 16 at 12:21
Joppy
5,448420
add a comment |
up vote
0
down vote
If $R$ is a commutative unital ring, then an $R$-module is an abelian group $M$ with a "scalar multiplication" map $R times M to M$, written $r cdot m = rm$, which satisfies
- $(r + s)m = rm + sm$
- $r(m + n) = rm + rn$
- $r(sm) = (rs)m$
If $R$ is a field, then an $R$-module is precisely a vector space over that field, and we get the usual notions of dimension, bases, and so on. However, if $R$ is not a field, then an $R$-module is usually a slightly more complicated thing.
For example if $R = mathbb{Z}$, then $mathbb{Z}, mathbb{Z}^2, ldots$ are all $mathbb{Z}$-modules in the usual way, and any basis of $mathbb{Z}^n$ will have size $n$. However, $mathbb{Z}/3mathbb{Z}$ is also a $mathbb{Z}-module$, and it has no basis. (No element $m$ of $mathbb{Z}/3mathbb{Z}$ is linearly independent, since $3m = 0$). So the theory of modules is more complicated than the theory of vector spaces.
answered Nov 16 at 12:21
Joppy
5,448420
add a comment |
up vote
0
down vote
up vote
0
down vote
If $R$ is a commutative unital ring, then an $R$-module is an abelian group $M$ with a "scalar multiplication" map $R times M to M$, written $r cdot m = rm$, which satisfies
- $(r + s)m = rm + sm$
- $r(m + n) = rm + rn$
- $r(sm) = (rs)m$
If $R$ is a field, then an $R$-module is precisely a vector space over that field, and we get the usual notions of dimension, bases, and so on. However, if $R$ is not a field, then an $R$-module is usually a slightly more complicated thing.
For example if $R = mathbb{Z}$, then $mathbb{Z}, mathbb{Z}^2, ldots$ are all $mathbb{Z}$-modules in the usual way, and any basis of $mathbb{Z}^n$ will have size $n$. However, $mathbb{Z}/3mathbb{Z}$ is also a $mathbb{Z}-module$, and it has no basis. (No element $m$ of $mathbb{Z}/3mathbb{Z}$ is linearly independent, since $3m = 0$). So the theory of modules is more complicated than the theory of vector spaces.
answered Nov 16 at 12:21
Joppy
5,448420
If $R$ is a commutative unital ring, then an $R$-module is an abelian group $M$ with a "scalar multiplication" map $R times M to M$, written $r cdot m = rm$, which satisfies
- $(r + s)m = rm + sm$
- $r(m + n) = rm + rn$
- $r(sm) = (rs)m$
If $R$ is a field, then an $R$-module is precisely a vector space over that field, and we get the usual notions of dimension, bases, and so on. However, if $R$ is not a field, then an $R$-module is usually a slightly more complicated thing.
For example if $R = mathbb{Z}$, then $mathbb{Z}, mathbb{Z}^2, ldots$ are all $mathbb{Z}$-modules in the usual way, and any basis of $mathbb{Z}^n$ will have size $n$. However, $mathbb{Z}/3mathbb{Z}$ is also a $mathbb{Z}-module$, and it has no basis. (No element $m$ of $mathbb{Z}/3mathbb{Z}$ is linearly independent, since $3m = 0$). So the theory of modules is more complicated than the theory of vector spaces.
answered Nov 16 at 12:21
Joppy
5,448420
answered Nov 16 at 12:21
Joppy
5,448420
answered Nov 16 at 12:21
Joppy
5,448420
answered Nov 16 at 12:21
Joppy
5,448420
5,448420
add a comment |
add a comment |
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SALONI SINHA is a new contributor. Be nice, and check out our Code of Conduct.
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Yes we can, but it has slightly different properties, and is called a module over the ring.
– Berci
Nov 16 at 10:31
1
Yes we can. But some things break down. For example we cannot prove that $acdot x=0$ implies that either $x=0$ or $a=0$ because we can no longer multiply by $1/a$. This has implications elsewhere. For example, a non-zero vector need not form a linearly independent set. Consequently the concepts of basis and dimension need modifications, and won't play out as nicely as they do over a field. You cannot necessarily find linearly independent sets of generators (even assuming a finite generator set).
– Jyrki Lahtonen
Nov 16 at 12:29