Is $mathbb{Q}$ a vector space over $mathbb{Z}$ or $mathbb{Q}$?
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Is $mathbb{Q}^n$ a vector space over $mathbb{Z}$ or over $mathbb{Q}$?
$mathbb{Q}^n$ is clearly not a vector space over $mathbb{R}$, because scalar multiplication of some $q in mathbb{Q}$ by $pi$ renders $pi q$, which is irrational.
linear-algebra vector-spaces vectors
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add a comment |
$begingroup$
Is $mathbb{Q}^n$ a vector space over $mathbb{Z}$ or over $mathbb{Q}$?
$mathbb{Q}^n$ is clearly not a vector space over $mathbb{R}$, because scalar multiplication of some $q in mathbb{Q}$ by $pi$ renders $pi q$, which is irrational.
linear-algebra vector-spaces vectors
$endgroup$
9
$begingroup$
It doesn't make sense to be a vector space over $mathbb Z$, because $mathbb Z$ is not a field. However, it is a vector space over $mathbb Q$.
$endgroup$
– Ashwin Trisal
Jan 29 at 7:55
add a comment |
$begingroup$
Is $mathbb{Q}^n$ a vector space over $mathbb{Z}$ or over $mathbb{Q}$?
$mathbb{Q}^n$ is clearly not a vector space over $mathbb{R}$, because scalar multiplication of some $q in mathbb{Q}$ by $pi$ renders $pi q$, which is irrational.
linear-algebra vector-spaces vectors
$endgroup$
Is $mathbb{Q}^n$ a vector space over $mathbb{Z}$ or over $mathbb{Q}$?
$mathbb{Q}^n$ is clearly not a vector space over $mathbb{R}$, because scalar multiplication of some $q in mathbb{Q}$ by $pi$ renders $pi q$, which is irrational.
linear-algebra vector-spaces vectors
linear-algebra vector-spaces vectors
edited Jan 29 at 9:59
stressed out
6,5431939
6,5431939
asked Jan 29 at 7:53
PeterPeter
323
323
9
$begingroup$
It doesn't make sense to be a vector space over $mathbb Z$, because $mathbb Z$ is not a field. However, it is a vector space over $mathbb Q$.
$endgroup$
– Ashwin Trisal
Jan 29 at 7:55
add a comment |
9
$begingroup$
It doesn't make sense to be a vector space over $mathbb Z$, because $mathbb Z$ is not a field. However, it is a vector space over $mathbb Q$.
$endgroup$
– Ashwin Trisal
Jan 29 at 7:55
9
9
$begingroup$
It doesn't make sense to be a vector space over $mathbb Z$, because $mathbb Z$ is not a field. However, it is a vector space over $mathbb Q$.
$endgroup$
– Ashwin Trisal
Jan 29 at 7:55
$begingroup$
It doesn't make sense to be a vector space over $mathbb Z$, because $mathbb Z$ is not a field. However, it is a vector space over $mathbb Q$.
$endgroup$
– Ashwin Trisal
Jan 29 at 7:55
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You can't have a vector space over $Bbb Z$. By definition, a vector space is required to be over a field. If you take away the field requirement, what you're left with is calles a module. And yes, $Bbb Q^n$ is definitely a $Bbb Z$-module.
That being said, there is a list of $10$ requirements (or thereabouts, it varies slightly) for a space like $Bbb Q^n$ to be a vector space over a field like $Bbb Q$. All of them should be easily verifiable. So yes, $Bbb Q^n$ is a vector space over $Bbb Q$.
$endgroup$
$begingroup$
Note that nothing here is said about the operations on the vector spaces. Technically that's bad form. A vector space isn't just a set and a base field, it's a set and a base field together with two operations. However, in the case of $Bbb Z, Bbb Q$ and $Bbb Q^n$, there are standard operations, and unless otherwise specified, you can be pretty certain that that's what is meant.
$endgroup$
– Arthur
Jan 29 at 8:39
add a comment |
$begingroup$
$mathbb{Q}^n$ can't be a vector space over $mathbb{Z}$ since $mathbb{Z}$ isn't a field.
To see that $mathbb{Q}^n$ is a vector space over $mathbb{Q}$ you could first prove that every field $mathbb{F}$ is a vector space over itself and then since the product of vector spaces is a vector space $mathbb{F}^n$ is a vector space over $mathbb{F}$ and so $mathbb{Q}^n$ is a vector space over $mathbb{Q}$ since $mathbb{Q}$ is a field.
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add a comment |
$begingroup$
$Bbb Z$ is a ring but not a field. See modules over rings for a generalization of vector spaces over fields.
$Bbb Q$ would be a $1$-dimensional vector space over $Bbb Q$. $Bbb Q^n$ an $n$-dimensional one.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can't have a vector space over $Bbb Z$. By definition, a vector space is required to be over a field. If you take away the field requirement, what you're left with is calles a module. And yes, $Bbb Q^n$ is definitely a $Bbb Z$-module.
That being said, there is a list of $10$ requirements (or thereabouts, it varies slightly) for a space like $Bbb Q^n$ to be a vector space over a field like $Bbb Q$. All of them should be easily verifiable. So yes, $Bbb Q^n$ is a vector space over $Bbb Q$.
$endgroup$
$begingroup$
Note that nothing here is said about the operations on the vector spaces. Technically that's bad form. A vector space isn't just a set and a base field, it's a set and a base field together with two operations. However, in the case of $Bbb Z, Bbb Q$ and $Bbb Q^n$, there are standard operations, and unless otherwise specified, you can be pretty certain that that's what is meant.
$endgroup$
– Arthur
Jan 29 at 8:39
add a comment |
$begingroup$
You can't have a vector space over $Bbb Z$. By definition, a vector space is required to be over a field. If you take away the field requirement, what you're left with is calles a module. And yes, $Bbb Q^n$ is definitely a $Bbb Z$-module.
That being said, there is a list of $10$ requirements (or thereabouts, it varies slightly) for a space like $Bbb Q^n$ to be a vector space over a field like $Bbb Q$. All of them should be easily verifiable. So yes, $Bbb Q^n$ is a vector space over $Bbb Q$.
$endgroup$
$begingroup$
Note that nothing here is said about the operations on the vector spaces. Technically that's bad form. A vector space isn't just a set and a base field, it's a set and a base field together with two operations. However, in the case of $Bbb Z, Bbb Q$ and $Bbb Q^n$, there are standard operations, and unless otherwise specified, you can be pretty certain that that's what is meant.
$endgroup$
– Arthur
Jan 29 at 8:39
add a comment |
$begingroup$
You can't have a vector space over $Bbb Z$. By definition, a vector space is required to be over a field. If you take away the field requirement, what you're left with is calles a module. And yes, $Bbb Q^n$ is definitely a $Bbb Z$-module.
That being said, there is a list of $10$ requirements (or thereabouts, it varies slightly) for a space like $Bbb Q^n$ to be a vector space over a field like $Bbb Q$. All of them should be easily verifiable. So yes, $Bbb Q^n$ is a vector space over $Bbb Q$.
$endgroup$
You can't have a vector space over $Bbb Z$. By definition, a vector space is required to be over a field. If you take away the field requirement, what you're left with is calles a module. And yes, $Bbb Q^n$ is definitely a $Bbb Z$-module.
That being said, there is a list of $10$ requirements (or thereabouts, it varies slightly) for a space like $Bbb Q^n$ to be a vector space over a field like $Bbb Q$. All of them should be easily verifiable. So yes, $Bbb Q^n$ is a vector space over $Bbb Q$.
edited Jan 29 at 8:36
answered Jan 29 at 7:56
ArthurArthur
117k7116200
117k7116200
$begingroup$
Note that nothing here is said about the operations on the vector spaces. Technically that's bad form. A vector space isn't just a set and a base field, it's a set and a base field together with two operations. However, in the case of $Bbb Z, Bbb Q$ and $Bbb Q^n$, there are standard operations, and unless otherwise specified, you can be pretty certain that that's what is meant.
$endgroup$
– Arthur
Jan 29 at 8:39
add a comment |
$begingroup$
Note that nothing here is said about the operations on the vector spaces. Technically that's bad form. A vector space isn't just a set and a base field, it's a set and a base field together with two operations. However, in the case of $Bbb Z, Bbb Q$ and $Bbb Q^n$, there are standard operations, and unless otherwise specified, you can be pretty certain that that's what is meant.
$endgroup$
– Arthur
Jan 29 at 8:39
$begingroup$
Note that nothing here is said about the operations on the vector spaces. Technically that's bad form. A vector space isn't just a set and a base field, it's a set and a base field together with two operations. However, in the case of $Bbb Z, Bbb Q$ and $Bbb Q^n$, there are standard operations, and unless otherwise specified, you can be pretty certain that that's what is meant.
$endgroup$
– Arthur
Jan 29 at 8:39
$begingroup$
Note that nothing here is said about the operations on the vector spaces. Technically that's bad form. A vector space isn't just a set and a base field, it's a set and a base field together with two operations. However, in the case of $Bbb Z, Bbb Q$ and $Bbb Q^n$, there are standard operations, and unless otherwise specified, you can be pretty certain that that's what is meant.
$endgroup$
– Arthur
Jan 29 at 8:39
add a comment |
$begingroup$
$mathbb{Q}^n$ can't be a vector space over $mathbb{Z}$ since $mathbb{Z}$ isn't a field.
To see that $mathbb{Q}^n$ is a vector space over $mathbb{Q}$ you could first prove that every field $mathbb{F}$ is a vector space over itself and then since the product of vector spaces is a vector space $mathbb{F}^n$ is a vector space over $mathbb{F}$ and so $mathbb{Q}^n$ is a vector space over $mathbb{Q}$ since $mathbb{Q}$ is a field.
$endgroup$
add a comment |
$begingroup$
$mathbb{Q}^n$ can't be a vector space over $mathbb{Z}$ since $mathbb{Z}$ isn't a field.
To see that $mathbb{Q}^n$ is a vector space over $mathbb{Q}$ you could first prove that every field $mathbb{F}$ is a vector space over itself and then since the product of vector spaces is a vector space $mathbb{F}^n$ is a vector space over $mathbb{F}$ and so $mathbb{Q}^n$ is a vector space over $mathbb{Q}$ since $mathbb{Q}$ is a field.
$endgroup$
add a comment |
$begingroup$
$mathbb{Q}^n$ can't be a vector space over $mathbb{Z}$ since $mathbb{Z}$ isn't a field.
To see that $mathbb{Q}^n$ is a vector space over $mathbb{Q}$ you could first prove that every field $mathbb{F}$ is a vector space over itself and then since the product of vector spaces is a vector space $mathbb{F}^n$ is a vector space over $mathbb{F}$ and so $mathbb{Q}^n$ is a vector space over $mathbb{Q}$ since $mathbb{Q}$ is a field.
$endgroup$
$mathbb{Q}^n$ can't be a vector space over $mathbb{Z}$ since $mathbb{Z}$ isn't a field.
To see that $mathbb{Q}^n$ is a vector space over $mathbb{Q}$ you could first prove that every field $mathbb{F}$ is a vector space over itself and then since the product of vector spaces is a vector space $mathbb{F}^n$ is a vector space over $mathbb{F}$ and so $mathbb{Q}^n$ is a vector space over $mathbb{Q}$ since $mathbb{Q}$ is a field.
answered Jan 29 at 8:37
PerturbativePerturbative
4,45621553
4,45621553
add a comment |
add a comment |
$begingroup$
$Bbb Z$ is a ring but not a field. See modules over rings for a generalization of vector spaces over fields.
$Bbb Q$ would be a $1$-dimensional vector space over $Bbb Q$. $Bbb Q^n$ an $n$-dimensional one.
$endgroup$
add a comment |
$begingroup$
$Bbb Z$ is a ring but not a field. See modules over rings for a generalization of vector spaces over fields.
$Bbb Q$ would be a $1$-dimensional vector space over $Bbb Q$. $Bbb Q^n$ an $n$-dimensional one.
$endgroup$
add a comment |
$begingroup$
$Bbb Z$ is a ring but not a field. See modules over rings for a generalization of vector spaces over fields.
$Bbb Q$ would be a $1$-dimensional vector space over $Bbb Q$. $Bbb Q^n$ an $n$-dimensional one.
$endgroup$
$Bbb Z$ is a ring but not a field. See modules over rings for a generalization of vector spaces over fields.
$Bbb Q$ would be a $1$-dimensional vector space over $Bbb Q$. $Bbb Q^n$ an $n$-dimensional one.
edited Jan 29 at 10:22
answered Jan 29 at 9:50
Chris CusterChris Custer
14.2k3827
14.2k3827
add a comment |
add a comment |
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$begingroup$
It doesn't make sense to be a vector space over $mathbb Z$, because $mathbb Z$ is not a field. However, it is a vector space over $mathbb Q$.
$endgroup$
– Ashwin Trisal
Jan 29 at 7:55