For any set we can construct a vector space on it
Can we construct a vector space on any finite set? It is clear, that on any finite set can be construct the transposition group $S_n$, where $n$ -- cardinal of elements of this set. I thought about non-zero characteristics fields, but I think that this is impossible, for ex. $mathbb{F}_2$ -- $2$-elements field on ${0,1}$ and we will be have $Id_X = (1_{mathbb{F}_2} + 1_{mathbb{F}_2})*S_1 = S_1 cdot S_1 neq Id_X$, where $Id_X$ -- identity transposition and $S_1$ -- any another transposition, $cdot$ -- composition between transpositions in $S_n$ and $+$ -- additive operation in field, $*$ -- multiplication on scalar.
So what after? How can I check zero characteristics fields?
And what about infinite sets?
linear-algebra abstract-algebra
asked Dec 27 '18 at 2:44
Arsenii
15318
add a comment |
Can we construct a vector space on any finite set? It is clear, that on any finite set can be construct the transposition group $S_n$, where $n$ -- cardinal of elements of this set. I thought about non-zero characteristics fields, but I think that this is impossible, for ex. $mathbb{F}_2$ -- $2$-elements field on ${0,1}$ and we will be have $Id_X = (1_{mathbb{F}_2} + 1_{mathbb{F}_2})*S_1 = S_1 cdot S_1 neq Id_X$, where $Id_X$ -- identity transposition and $S_1$ -- any another transposition, $cdot$ -- composition between transpositions in $S_n$ and $+$ -- additive operation in field, $*$ -- multiplication on scalar.
So what after? How can I check zero characteristics fields?
And what about infinite sets?
linear-algebra abstract-algebra
asked Dec 27 '18 at 2:44
Arsenii
15318
add a comment |
Can we construct a vector space on any finite set? It is clear, that on any finite set can be construct the transposition group $S_n$, where $n$ -- cardinal of elements of this set. I thought about non-zero characteristics fields, but I think that this is impossible, for ex. $mathbb{F}_2$ -- $2$-elements field on ${0,1}$ and we will be have $Id_X = (1_{mathbb{F}_2} + 1_{mathbb{F}_2})*S_1 = S_1 cdot S_1 neq Id_X$, where $Id_X$ -- identity transposition and $S_1$ -- any another transposition, $cdot$ -- composition between transpositions in $S_n$ and $+$ -- additive operation in field, $*$ -- multiplication on scalar.
So what after? How can I check zero characteristics fields?
And what about infinite sets?
linear-algebra abstract-algebra
asked Dec 27 '18 at 2:44
Arsenii
15318
Can we construct a vector space on any finite set? It is clear, that on any finite set can be construct the transposition group $S_n$, where $n$ -- cardinal of elements of this set. I thought about non-zero characteristics fields, but I think that this is impossible, for ex. $mathbb{F}_2$ -- $2$-elements field on ${0,1}$ and we will be have $Id_X = (1_{mathbb{F}_2} + 1_{mathbb{F}_2})*S_1 = S_1 cdot S_1 neq Id_X$, where $Id_X$ -- identity transposition and $S_1$ -- any another transposition, $cdot$ -- composition between transpositions in $S_n$ and $+$ -- additive operation in field, $*$ -- multiplication on scalar.
So what after? How can I check zero characteristics fields?
And what about infinite sets?
linear-algebra abstract-algebra
linear-algebra abstract-algebra
asked Dec 27 '18 at 2:44
Arsenii
15318
asked Dec 27 '18 at 2:44
Arsenii
15318
asked Dec 27 '18 at 2:44
Arsenii
15318
asked Dec 27 '18 at 2:44
Arsenii
15318
asked Dec 27 '18 at 2:44
Arsenii
15318
15318
add a comment |
add a comment |
1 Answer
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No. For instance, there can be no vector space structure on a set of order $6$. No matter the ground field (and you only have two options here, $mathbb{F}_2$ or $mathbb{F}_3$) the arithmetic won't work out on the dimension to make the order of the underlying set $6$. (It should be clear that a vector space $V$ of finite dimension $n$ over a finite field of order $q$ will have $|V|=q^n$.)
answered Dec 27 '18 at 2:57
Randall
9,12611129
For now I think, that if we speaking about vec.space on a finite set $X$ with $n$ elements, we should think about the space with $n$ vectors and the $Set$-functor image of this vec-space, that will be have $n$ elements in it. So if we have set of order 6, we should find 6 element-vec.space but the number of this elements in vec,space depends on the number of the field, bcs $|V| = q^m$, so $n = q^m$, where $q$-- order of field-set and $m$ -- dim of space. And after... Why $n = q^m$ is impossible?
– Arsenii
Dec 27 '18 at 3:10
How can you write $6$ as a power of prime? (Finite fields themselves have order a power of a prime, so $V$ will as well.)
– Randall
Dec 27 '18 at 3:11
yes, okey) thx) perfect! but what about infinite sets?
– Arsenii
Dec 27 '18 at 3:12
That's harder because of cardinality issues. For instance, if $V$ has cardinality of $mathbb{R}$, then choose a bijection $V to mathbb{R}$ and pull the vector space structure on $mathbb{R}$ back to $V$.
– Randall
Dec 27 '18 at 3:12
oh, nice, just copy structure... Thank you so much, you answered my questions)
– Arsenii
Dec 27 '18 at 3:15
|
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1 Answer
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No. For instance, there can be no vector space structure on a set of order $6$. No matter the ground field (and you only have two options here, $mathbb{F}_2$ or $mathbb{F}_3$) the arithmetic won't work out on the dimension to make the order of the underlying set $6$. (It should be clear that a vector space $V$ of finite dimension $n$ over a finite field of order $q$ will have $|V|=q^n$.)
answered Dec 27 '18 at 2:57
Randall
9,12611129
For now I think, that if we speaking about vec.space on a finite set $X$ with $n$ elements, we should think about the space with $n$ vectors and the $Set$-functor image of this vec-space, that will be have $n$ elements in it. So if we have set of order 6, we should find 6 element-vec.space but the number of this elements in vec,space depends on the number of the field, bcs $|V| = q^m$, so $n = q^m$, where $q$-- order of field-set and $m$ -- dim of space. And after... Why $n = q^m$ is impossible?
– Arsenii
Dec 27 '18 at 3:10
How can you write $6$ as a power of prime? (Finite fields themselves have order a power of a prime, so $V$ will as well.)
– Randall
Dec 27 '18 at 3:11
yes, okey) thx) perfect! but what about infinite sets?
– Arsenii
Dec 27 '18 at 3:12
That's harder because of cardinality issues. For instance, if $V$ has cardinality of $mathbb{R}$, then choose a bijection $V to mathbb{R}$ and pull the vector space structure on $mathbb{R}$ back to $V$.
– Randall
Dec 27 '18 at 3:12
oh, nice, just copy structure... Thank you so much, you answered my questions)
– Arsenii
Dec 27 '18 at 3:15
|
show 2 more comments
No. For instance, there can be no vector space structure on a set of order $6$. No matter the ground field (and you only have two options here, $mathbb{F}_2$ or $mathbb{F}_3$) the arithmetic won't work out on the dimension to make the order of the underlying set $6$. (It should be clear that a vector space $V$ of finite dimension $n$ over a finite field of order $q$ will have $|V|=q^n$.)
answered Dec 27 '18 at 2:57
Randall
9,12611129
For now I think, that if we speaking about vec.space on a finite set $X$ with $n$ elements, we should think about the space with $n$ vectors and the $Set$-functor image of this vec-space, that will be have $n$ elements in it. So if we have set of order 6, we should find 6 element-vec.space but the number of this elements in vec,space depends on the number of the field, bcs $|V| = q^m$, so $n = q^m$, where $q$-- order of field-set and $m$ -- dim of space. And after... Why $n = q^m$ is impossible?
– Arsenii
Dec 27 '18 at 3:10
How can you write $6$ as a power of prime? (Finite fields themselves have order a power of a prime, so $V$ will as well.)
– Randall
Dec 27 '18 at 3:11
yes, okey) thx) perfect! but what about infinite sets?
– Arsenii
Dec 27 '18 at 3:12
That's harder because of cardinality issues. For instance, if $V$ has cardinality of $mathbb{R}$, then choose a bijection $V to mathbb{R}$ and pull the vector space structure on $mathbb{R}$ back to $V$.
– Randall
Dec 27 '18 at 3:12
oh, nice, just copy structure... Thank you so much, you answered my questions)
– Arsenii
Dec 27 '18 at 3:15
|
show 2 more comments
No. For instance, there can be no vector space structure on a set of order $6$. No matter the ground field (and you only have two options here, $mathbb{F}_2$ or $mathbb{F}_3$) the arithmetic won't work out on the dimension to make the order of the underlying set $6$. (It should be clear that a vector space $V$ of finite dimension $n$ over a finite field of order $q$ will have $|V|=q^n$.)
answered Dec 27 '18 at 2:57
Randall
9,12611129
No. For instance, there can be no vector space structure on a set of order $6$. No matter the ground field (and you only have two options here, $mathbb{F}_2$ or $mathbb{F}_3$) the arithmetic won't work out on the dimension to make the order of the underlying set $6$. (It should be clear that a vector space $V$ of finite dimension $n$ over a finite field of order $q$ will have $|V|=q^n$.)
answered Dec 27 '18 at 2:57
Randall
9,12611129
answered Dec 27 '18 at 2:57
Randall
9,12611129
answered Dec 27 '18 at 2:57
Randall
9,12611129
answered Dec 27 '18 at 2:57
Randall
9,12611129
9,12611129
For now I think, that if we speaking about vec.space on a finite set $X$ with $n$ elements, we should think about the space with $n$ vectors and the $Set$-functor image of this vec-space, that will be have $n$ elements in it. So if we have set of order 6, we should find 6 element-vec.space but the number of this elements in vec,space depends on the number of the field, bcs $|V| = q^m$, so $n = q^m$, where $q$-- order of field-set and $m$ -- dim of space. And after... Why $n = q^m$ is impossible?
– Arsenii
Dec 27 '18 at 3:10
How can you write $6$ as a power of prime? (Finite fields themselves have order a power of a prime, so $V$ will as well.)
– Randall
Dec 27 '18 at 3:11
yes, okey) thx) perfect! but what about infinite sets?
– Arsenii
Dec 27 '18 at 3:12
That's harder because of cardinality issues. For instance, if $V$ has cardinality of $mathbb{R}$, then choose a bijection $V to mathbb{R}$ and pull the vector space structure on $mathbb{R}$ back to $V$.
– Randall
Dec 27 '18 at 3:12
oh, nice, just copy structure... Thank you so much, you answered my questions)
– Arsenii
Dec 27 '18 at 3:15
|
show 2 more comments
For now I think, that if we speaking about vec.space on a finite set $X$ with $n$ elements, we should think about the space with $n$ vectors and the $Set$-functor image of this vec-space, that will be have $n$ elements in it. So if we have set of order 6, we should find 6 element-vec.space but the number of this elements in vec,space depends on the number of the field, bcs $|V| = q^m$, so $n = q^m$, where $q$-- order of field-set and $m$ -- dim of space. And after... Why $n = q^m$ is impossible?
– Arsenii
Dec 27 '18 at 3:10
How can you write $6$ as a power of prime? (Finite fields themselves have order a power of a prime, so $V$ will as well.)
– Randall
Dec 27 '18 at 3:11
yes, okey) thx) perfect! but what about infinite sets?
– Arsenii
Dec 27 '18 at 3:12
That's harder because of cardinality issues. For instance, if $V$ has cardinality of $mathbb{R}$, then choose a bijection $V to mathbb{R}$ and pull the vector space structure on $mathbb{R}$ back to $V$.
– Randall
Dec 27 '18 at 3:12
oh, nice, just copy structure... Thank you so much, you answered my questions)
– Arsenii
Dec 27 '18 at 3:15
For now I think, that if we speaking about vec.space on a finite set $X$ with $n$ elements, we should think about the space with $n$ vectors and the $Set$-functor image of this vec-space, that will be have $n$ elements in it. So if we have set of order 6, we should find 6 element-vec.space but the number of this elements in vec,space depends on the number of the field, bcs $|V| = q^m$, so $n = q^m$, where $q$-- order of field-set and $m$ -- dim of space. And after... Why $n = q^m$ is impossible?
– Arsenii
Dec 27 '18 at 3:10
For now I think, that if we speaking about vec.space on a finite set $X$ with $n$ elements, we should think about the space with $n$ vectors and the $Set$-functor image of this vec-space, that will be have $n$ elements in it. So if we have set of order 6, we should find 6 element-vec.space but the number of this elements in vec,space depends on the number of the field, bcs $|V| = q^m$, so $n = q^m$, where $q$-- order of field-set and $m$ -- dim of space. And after... Why $n = q^m$ is impossible?
– Arsenii
Dec 27 '18 at 3:10
How can you write $6$ as a power of prime? (Finite fields themselves have order a power of a prime, so $V$ will as well.)
– Randall
Dec 27 '18 at 3:11
How can you write $6$ as a power of prime? (Finite fields themselves have order a power of a prime, so $V$ will as well.)
– Randall
Dec 27 '18 at 3:11
yes, okey) thx) perfect! but what about infinite sets?
– Arsenii
Dec 27 '18 at 3:12
yes, okey) thx) perfect! but what about infinite sets?
– Arsenii
Dec 27 '18 at 3:12
That's harder because of cardinality issues. For instance, if $V$ has cardinality of $mathbb{R}$, then choose a bijection $V to mathbb{R}$ and pull the vector space structure on $mathbb{R}$ back to $V$.
– Randall
Dec 27 '18 at 3:12
That's harder because of cardinality issues. For instance, if $V$ has cardinality of $mathbb{R}$, then choose a bijection $V to mathbb{R}$ and pull the vector space structure on $mathbb{R}$ back to $V$.
– Randall
Dec 27 '18 at 3:12
oh, nice, just copy structure... Thank you so much, you answered my questions)
– Arsenii
Dec 27 '18 at 3:15
oh, nice, just copy structure... Thank you so much, you answered my questions)
– Arsenii
Dec 27 '18 at 3:15
|
show 2 more comments
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