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Turning an Abelian Group into a Vector Space

Are these vector spaces?Proving a subspace of $mathbb{R}^{2}$ is still a vector space with redefined addition / scalar multiplication operations?Define two differents vector space structures over a field on an abelian groupnumber of differents vector space structures over the same field $mathbb{F}$ on an abelian groupgroups products vs vector space productsCan we define a binary operation on $mathbb Z$ to make it a vector space over $mathbb Q$?Vector spaces: Is (the) scalar multiplication unique?Is a vector space a subset of an abelian group?An abelian group as an $mathbb F_2$-vector spaceCan an abelian group be a real vector space in more than one way?

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5

3

$begingroup$

This question is inspired by the question Are these vector spaces? but is free-standing.

Suppose $mathbb{F}$ is a field, and that $X$ is the multiplicative group of $mathbb{F}$.

Let us write this abelian group $X$ additively. To be precise the set is $mathbb{F}setminus{0}$, the zero element is $hat{0}:=1$, the addition operation is $x hat{+}y:=xy$, and the negative operation is $widehat{-}x:=x^{-1}$.

Let $V$ be the abelian group of $n$-tuples $X^n$ with the usual co-ordinatewise operations.

Question : Can we define an $mathbb{F}$-scalar multiplication on $V$ so that $V$ becomes an $mathbb{F}$-vector space?

By looking at $-1$ in $mathbb{F}$, which satisfies $(-1)hat{+}(-1)=(-1)^2=1=hat{0}$ we see that if there is to be any chance of turning $V$ into a vector space then the field $mathbb{F}$ must have characteristic $2$. More than that, a similar argument on $(2^k -1)$-th roots of unity will show that $mathbb{F}$ has no finite subfields $mathbb{F}_{2^k}$ except $mathbb{F}_2$. Beyond that I cannot go.

linear-algebra abstract-algebra vector-spaces commutative-algebra modules

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edited 11 hours ago

Omnomnomnom

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asked 12 hours ago

ancientmathematicianancientmathematician

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$endgroup$

add a comment | 

5

3

$begingroup$

This question is inspired by the question Are these vector spaces? but is free-standing.

Suppose $mathbb{F}$ is a field, and that $X$ is the multiplicative group of $mathbb{F}$.

Let us write this abelian group $X$ additively. To be precise the set is $mathbb{F}setminus{0}$, the zero element is $hat{0}:=1$, the addition operation is $x hat{+}y:=xy$, and the negative operation is $widehat{-}x:=x^{-1}$.

Let $V$ be the abelian group of $n$-tuples $X^n$ with the usual co-ordinatewise operations.

Question : Can we define an $mathbb{F}$-scalar multiplication on $V$ so that $V$ becomes an $mathbb{F}$-vector space?

By looking at $-1$ in $mathbb{F}$, which satisfies $(-1)hat{+}(-1)=(-1)^2=1=hat{0}$ we see that if there is to be any chance of turning $V$ into a vector space then the field $mathbb{F}$ must have characteristic $2$. More than that, a similar argument on $(2^k -1)$-th roots of unity will show that $mathbb{F}$ has no finite subfields $mathbb{F}_{2^k}$ except $mathbb{F}_2$. Beyond that I cannot go.

linear-algebra abstract-algebra vector-spaces commutative-algebra modules

share|cite|improve this question

edited 11 hours ago

Omnomnomnom

133k77 gold badges9898 silver badges198198 bronze badges

asked 12 hours ago

ancientmathematicianancientmathematician

5,39711 gold badge66 silver badges1717 bronze badges

$endgroup$

add a comment | 

$begingroup$

This question is inspired by the question Are these vector spaces? but is free-standing.

Suppose $mathbb{F}$ is a field, and that $X$ is the multiplicative group of $mathbb{F}$.

Let us write this abelian group $X$ additively. To be precise the set is $mathbb{F}setminus{0}$, the zero element is $hat{0}:=1$, the addition operation is $x hat{+}y:=xy$, and the negative operation is $widehat{-}x:=x^{-1}$.

Let $V$ be the abelian group of $n$-tuples $X^n$ with the usual co-ordinatewise operations.

Question : Can we define an $mathbb{F}$-scalar multiplication on $V$ so that $V$ becomes an $mathbb{F}$-vector space?

By looking at $-1$ in $mathbb{F}$, which satisfies $(-1)hat{+}(-1)=(-1)^2=1=hat{0}$ we see that if there is to be any chance of turning $V$ into a vector space then the field $mathbb{F}$ must have characteristic $2$. More than that, a similar argument on $(2^k -1)$-th roots of unity will show that $mathbb{F}$ has no finite subfields $mathbb{F}_{2^k}$ except $mathbb{F}_2$. Beyond that I cannot go.

linear-algebra abstract-algebra vector-spaces commutative-algebra modules

share|cite|improve this question

edited 11 hours ago

Omnomnomnom

133k77 gold badges9898 silver badges198198 bronze badges

asked 12 hours ago

ancientmathematicianancientmathematician

5,39711 gold badge66 silver badges1717 bronze badges

$endgroup$

share|cite|improve this question

edited 11 hours ago

Omnomnomnom

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asked 12 hours ago

ancientmathematicianancientmathematician

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share|cite|improve this question

edited 11 hours ago

Omnomnomnom

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asked 12 hours ago

ancientmathematicianancientmathematician

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edited 11 hours ago

Omnomnomnom

133k77 gold badges9898 silver badges198198 bronze badges

edited 11 hours ago

Omnomnomnom

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asked 12 hours ago

ancientmathematicianancientmathematician

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asked 12 hours ago

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2 Answers
2

active

oldest

votes

7

$begingroup$

The answer is no except for the trivial case $mathbb F =mathbb F_2$ (in which $V=0$)

Indeed, assume $mathbb F$ is such a field and take $xin mathbb F^*$. In the structure of $V$, we have $2cdot x = 0$, which translates to $x^2= 1$ in $mathbb F$ (for each coordinate).

This implies that any element of $mathbb F$ is a root of $X^2-1$ which has at most to roots : $|mathbb F| leq 2$

share|cite|improve this answer

answered 10 hours ago

MaxMax

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Fantastic! Any clue what $V otimes Bbb F_{2^k}$ looks like? Is it ever non-zero?
  $endgroup$
  – Omnomnomnom
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @omnomnomnom : yes, in fact if $mathbb F$ is e.g. $mathbb F_2 (X) $ then its multiplicative group is free, so $Votimes mathbb F$ is a direct sum of copies of $mathbb F$
  $endgroup$
  – Max
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Omnomnomnom But for $mathbb{F}_{2^k}$ and $V=mathbb{F}_{2^k}setminus{0}$ under multiplication, it is zero: every element of $V$ satisfies $v^{2^k}=v$, hence the generators are all zero: $votimesalpha = v^{2^k}otimesalpha = votimes(2^kalpha) = votimes 0 = mathbf{0}$.
  $endgroup$
  – Arturo Magidin
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great, thanks to both of you
  $endgroup$
  – Omnomnomnom
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @jyrkilahtonen : I think that's the case : the notation $mathbb F$ is used for both
  $endgroup$
  – Max
  10 hours ago
  

  
  
  
  

 | 
show 6 more comments

0

$begingroup$

Not an answer, but hopefully a useful observation. Your question (if I have understood it correctly) can be reframed as follows.

We are given an abelian group $X = Bbb Fsetminus {0}$, which we present as a $Bbb Z$-module where $x+y := xy$ and $nx := x^{n}$. Is it possible to take $V = X^n$ and extend multiplication by $Bbb Z$ into multiplication by $Bbb F$ in such a way that $V$ becomes a $Bbb F$-module (i.e. a vector space over $Bbb F$)?

As Max's comment below observes: any $Bbb F$-space structure $V$ exists corresponds to an $Bbb F$-linear map $V otimes_{Bbb Z} Bbb F to V$. Notably, $V otimes_{Bbb Z} Bbb F$ is the usual extension of scalars.

share|cite|improve this answer

edited 11 hours ago

community wiki

3 revs
Omnomnomnom

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why does $Votimes mathbb F = 0$ ? if it is the case then there is no $mathbb F$-vector space structure on $V$ as such a thing would be a map $Votimes mathbb Fto V$
  $endgroup$
  – Max
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Max I see no justification for my statement, now that I think about it. I'll remove it.
  $endgroup$
  – Omnomnomnom
  11 hours ago
  

  
  
  
  

add a comment | 

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7

$begingroup$

The answer is no except for the trivial case $mathbb F =mathbb F_2$ (in which $V=0$)

Indeed, assume $mathbb F$ is such a field and take $xin mathbb F^*$. In the structure of $V$, we have $2cdot x = 0$, which translates to $x^2= 1$ in $mathbb F$ (for each coordinate).

This implies that any element of $mathbb F$ is a root of $X^2-1$ which has at most to roots : $|mathbb F| leq 2$

share|cite|improve this answer

answered 10 hours ago

MaxMax

23.2k11 gold badge1212 silver badges5454 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Fantastic! Any clue what $V otimes Bbb F_{2^k}$ looks like? Is it ever non-zero?
  $endgroup$
  – Omnomnomnom
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @omnomnomnom : yes, in fact if $mathbb F$ is e.g. $mathbb F_2 (X) $ then its multiplicative group is free, so $Votimes mathbb F$ is a direct sum of copies of $mathbb F$
  $endgroup$
  – Max
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Omnomnomnom But for $mathbb{F}_{2^k}$ and $V=mathbb{F}_{2^k}setminus{0}$ under multiplication, it is zero: every element of $V$ satisfies $v^{2^k}=v$, hence the generators are all zero: $votimesalpha = v^{2^k}otimesalpha = votimes(2^kalpha) = votimes 0 = mathbf{0}$.
  $endgroup$
  – Arturo Magidin
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great, thanks to both of you
  $endgroup$
  – Omnomnomnom
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @jyrkilahtonen : I think that's the case : the notation $mathbb F$ is used for both
  $endgroup$
  – Max
  10 hours ago
  

  
  
  
  

 | 
show 6 more comments

7

$begingroup$

The answer is no except for the trivial case $mathbb F =mathbb F_2$ (in which $V=0$)

Indeed, assume $mathbb F$ is such a field and take $xin mathbb F^*$. In the structure of $V$, we have $2cdot x = 0$, which translates to $x^2= 1$ in $mathbb F$ (for each coordinate).

This implies that any element of $mathbb F$ is a root of $X^2-1$ which has at most to roots : $|mathbb F| leq 2$

share|cite|improve this answer

answered 10 hours ago

MaxMax

23.2k11 gold badge1212 silver badges5454 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Fantastic! Any clue what $V otimes Bbb F_{2^k}$ looks like? Is it ever non-zero?
  $endgroup$
  – Omnomnomnom
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @omnomnomnom : yes, in fact if $mathbb F$ is e.g. $mathbb F_2 (X) $ then its multiplicative group is free, so $Votimes mathbb F$ is a direct sum of copies of $mathbb F$
  $endgroup$
  – Max
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Omnomnomnom But for $mathbb{F}_{2^k}$ and $V=mathbb{F}_{2^k}setminus{0}$ under multiplication, it is zero: every element of $V$ satisfies $v^{2^k}=v$, hence the generators are all zero: $votimesalpha = v^{2^k}otimesalpha = votimes(2^kalpha) = votimes 0 = mathbf{0}$.
  $endgroup$
  – Arturo Magidin
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great, thanks to both of you
  $endgroup$
  – Omnomnomnom
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @jyrkilahtonen : I think that's the case : the notation $mathbb F$ is used for both
  $endgroup$
  – Max
  10 hours ago
  

  
  
  
  

 | 
show 6 more comments

$begingroup$

The answer is no except for the trivial case $mathbb F =mathbb F_2$ (in which $V=0$)

Indeed, assume $mathbb F$ is such a field and take $xin mathbb F^*$. In the structure of $V$, we have $2cdot x = 0$, which translates to $x^2= 1$ in $mathbb F$ (for each coordinate).

This implies that any element of $mathbb F$ is a root of $X^2-1$ which has at most to roots : $|mathbb F| leq 2$

share|cite|improve this answer

answered 10 hours ago

MaxMax

23.2k11 gold badge1212 silver badges5454 bronze badges

$endgroup$

share|cite|improve this answer

answered 10 hours ago

MaxMax

23.2k11 gold badge1212 silver badges5454 bronze badges

answered 10 hours ago

MaxMax

23.2k11 gold badge1212 silver badges5454 bronze badges

answered 10 hours ago

MaxMax

23.2k11 gold badge1212 silver badges5454 bronze badges