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Changing scalar product in a vector space cause different dimensions?

1

$begingroup$

I'm looking for an abelian group $(G,+)$ and a field$(F,+,×)$ and two scalar products $._1$ and $._2$ such that vector spaces caused by first and second scalar products on $G$ over $F$, have different dimensions.

I'm intrested in an example with an infinite field, but at first, any example would be useful (if exists! If there is no such an example, it would be great to give a proof).

linear-algebra vector-spaces

share|cite|improve this question

asked Mar 23 at 4:25

Sim000Sim000

707

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Can an abelian group be turned into two nonisomorphic vector spaces by different actions of the same field?
  $endgroup$
  – Captain Lama
  Mar 23 at 4:58
  

  
  
  
  

add a comment | 

1

$begingroup$

I'm looking for an abelian group $(G,+)$ and a field$(F,+,×)$ and two scalar products $._1$ and $._2$ such that vector spaces caused by first and second scalar products on $G$ over $F$, have different dimensions.

I'm intrested in an example with an infinite field, but at first, any example would be useful (if exists! If there is no such an example, it would be great to give a proof).

linear-algebra vector-spaces

share|cite|improve this question

asked Mar 23 at 4:25

Sim000Sim000

707

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Can an abelian group be turned into two nonisomorphic vector spaces by different actions of the same field?
  $endgroup$
  – Captain Lama
  Mar 23 at 4:58
  

  
  
  
  

add a comment | 

$begingroup$

I'm looking for an abelian group $(G,+)$ and a field$(F,+,×)$ and two scalar products $._1$ and $._2$ such that vector spaces caused by first and second scalar products on $G$ over $F$, have different dimensions.

I'm intrested in an example with an infinite field, but at first, any example would be useful (if exists! If there is no such an example, it would be great to give a proof).

linear-algebra vector-spaces

share|cite|improve this question

asked Mar 23 at 4:25

Sim000Sim000

707

$endgroup$

share|cite|improve this question

asked Mar 23 at 4:25

Sim000Sim000

707

share|cite|improve this question

asked Mar 23 at 4:25

Sim000Sim000

707

asked Mar 23 at 4:25

Sim000Sim000

707

asked Mar 23 at 4:25

Sim000Sim000

707

1 Answer
1

active

oldest

votes

1

$begingroup$

Note any field $k$ is a $1$-dimensional vector space over itself using the operation $ccdot v = cv$. Now let $k$ be a field equipped with an injective, but not surjective, ring homomorphism $fcolon k to k$. Now $k$ is also a vector space using the operation $ccdot v = f(c)v$. As $f$ is not surjective take $x in k$ outside the image, then $1, x$ is linearly independent, for otherwise we would have

$$0 = acdot 1 + bcdot x = f(a) + f(b)x$$

and so either $b = 0$, implying $f(a) = 0$, hence $a = 0$ by injectivity, or $b neq 0$ and $x = f(-fracab)$, contradicting our assumption that $x$ is not in the image. So with this new scalar multiplication $k$ has dimension at least $2$.

For an example of a field $k$ with such a map $f$ how about the field $k = mathbb R(x_1, x_2, ldots)$ of rational polynomials in countably many variables and $f$ defined by $f(x_i) = x_i + 1$ for all $i$.

share|cite|improve this answer

answered Mar 23 at 5:04

JimJim

24.6k23370

$endgroup$

add a comment | 

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1

$begingroup$

Note any field $k$ is a $1$-dimensional vector space over itself using the operation $ccdot v = cv$. Now let $k$ be a field equipped with an injective, but not surjective, ring homomorphism $fcolon k to k$. Now $k$ is also a vector space using the operation $ccdot v = f(c)v$. As $f$ is not surjective take $x in k$ outside the image, then $1, x$ is linearly independent, for otherwise we would have

$$0 = acdot 1 + bcdot x = f(a) + f(b)x$$

and so either $b = 0$, implying $f(a) = 0$, hence $a = 0$ by injectivity, or $b neq 0$ and $x = f(-fracab)$, contradicting our assumption that $x$ is not in the image. So with this new scalar multiplication $k$ has dimension at least $2$.

For an example of a field $k$ with such a map $f$ how about the field $k = mathbb R(x_1, x_2, ldots)$ of rational polynomials in countably many variables and $f$ defined by $f(x_i) = x_i + 1$ for all $i$.

share|cite|improve this answer

answered Mar 23 at 5:04

JimJim

24.6k23370

$endgroup$

add a comment | 

1

$begingroup$

Note any field $k$ is a $1$-dimensional vector space over itself using the operation $ccdot v = cv$. Now let $k$ be a field equipped with an injective, but not surjective, ring homomorphism $fcolon k to k$. Now $k$ is also a vector space using the operation $ccdot v = f(c)v$. As $f$ is not surjective take $x in k$ outside the image, then $1, x$ is linearly independent, for otherwise we would have

$$0 = acdot 1 + bcdot x = f(a) + f(b)x$$

and so either $b = 0$, implying $f(a) = 0$, hence $a = 0$ by injectivity, or $b neq 0$ and $x = f(-fracab)$, contradicting our assumption that $x$ is not in the image. So with this new scalar multiplication $k$ has dimension at least $2$.

For an example of a field $k$ with such a map $f$ how about the field $k = mathbb R(x_1, x_2, ldots)$ of rational polynomials in countably many variables and $f$ defined by $f(x_i) = x_i + 1$ for all $i$.

share|cite|improve this answer

answered Mar 23 at 5:04

JimJim

24.6k23370

$endgroup$

add a comment | 

$begingroup$

Note any field $k$ is a $1$-dimensional vector space over itself using the operation $ccdot v = cv$. Now let $k$ be a field equipped with an injective, but not surjective, ring homomorphism $fcolon k to k$. Now $k$ is also a vector space using the operation $ccdot v = f(c)v$. As $f$ is not surjective take $x in k$ outside the image, then $1, x$ is linearly independent, for otherwise we would have

$$0 = acdot 1 + bcdot x = f(a) + f(b)x$$

and so either $b = 0$, implying $f(a) = 0$, hence $a = 0$ by injectivity, or $b neq 0$ and $x = f(-fracab)$, contradicting our assumption that $x$ is not in the image. So with this new scalar multiplication $k$ has dimension at least $2$.

For an example of a field $k$ with such a map $f$ how about the field $k = mathbb R(x_1, x_2, ldots)$ of rational polynomials in countably many variables and $f$ defined by $f(x_i) = x_i + 1$ for all $i$.

share|cite|improve this answer

answered Mar 23 at 5:04

JimJim

24.6k23370

$endgroup$

share|cite|improve this answer

answered Mar 23 at 5:04

JimJim

24.6k23370

answered Mar 23 at 5:04

JimJim

24.6k23370

answered Mar 23 at 5:04

JimJim

24.6k23370