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Turning an Abelian Group into a Vector Space

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5

3

$begingroup$

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

Suppose $mathbbF$ is a field, and that $X$ is the multiplicative group of $mathbbF$.

Let us write this abelian group $X$ additively. To be precise the set is $mathbbFsetminus0$, the zero element is $hat0:=1$, 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 $mathbbF$-scalar multiplication on $V$ so that $V$ becomes an $mathbbF$-vector space?

By looking at $-1$ in $mathbbF$, which satisfies $(-1)hat+(-1)=(-1)^2=1=hat0$ we see that if there is to be any chance of turning $V$ into a vector space then the field $mathbbF$ must have characteristic $2$. More than that, a similar argument on $(2^k -1)$-th roots of unity will show that $mathbbF$ has no finite subfields $mathbbF_2^k$ except $mathbbF_2$. Beyond that I cannot go.

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

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edited 7 hours ago

Omnomnomnom

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asked 9 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 $mathbbF$ is a field, and that $X$ is the multiplicative group of $mathbbF$.

Let us write this abelian group $X$ additively. To be precise the set is $mathbbFsetminus0$, the zero element is $hat0:=1$, 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 $mathbbF$-scalar multiplication on $V$ so that $V$ becomes an $mathbbF$-vector space?

By looking at $-1$ in $mathbbF$, which satisfies $(-1)hat+(-1)=(-1)^2=1=hat0$ we see that if there is to be any chance of turning $V$ into a vector space then the field $mathbbF$ must have characteristic $2$. More than that, a similar argument on $(2^k -1)$-th roots of unity will show that $mathbbF$ has no finite subfields $mathbbF_2^k$ except $mathbbF_2$. Beyond that I cannot go.

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

share|cite|improve this question

edited 7 hours ago

Omnomnomnom

133k77 gold badges9898 silver badges198198 bronze badges

asked 9 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 $mathbbF$ is a field, and that $X$ is the multiplicative group of $mathbbF$.

Let us write this abelian group $X$ additively. To be precise the set is $mathbbFsetminus0$, the zero element is $hat0:=1$, 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 $mathbbF$-scalar multiplication on $V$ so that $V$ becomes an $mathbbF$-vector space?

By looking at $-1$ in $mathbbF$, which satisfies $(-1)hat+(-1)=(-1)^2=1=hat0$ we see that if there is to be any chance of turning $V$ into a vector space then the field $mathbbF$ must have characteristic $2$. More than that, a similar argument on $(2^k -1)$-th roots of unity will show that $mathbbF$ has no finite subfields $mathbbF_2^k$ except $mathbbF_2$. Beyond that I cannot go.

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

share|cite|improve this question

edited 7 hours ago

Omnomnomnom

133k77 gold badges9898 silver badges198198 bronze badges

asked 9 hours ago

ancientmathematicianancientmathematician

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$endgroup$

share|cite|improve this question

edited 7 hours ago

Omnomnomnom

133k77 gold badges9898 silver badges198198 bronze badges

asked 9 hours ago

ancientmathematicianancientmathematician

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share|cite|improve this question

edited 7 hours ago

Omnomnomnom

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asked 9 hours ago

ancientmathematicianancientmathematician

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edited 7 hours ago

Omnomnomnom

133k77 gold badges9898 silver badges198198 bronze badges

edited 7 hours ago

Omnomnomnom

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asked 9 hours ago

ancientmathematicianancientmathematician

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asked 9 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 7 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
  7 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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Omnomnomnom But for $mathbbF_2^k$ and $V=mathbbF_2^ksetminus0$ under multiplication, it is zero: every element of $V$ satisfies $v^2^k=v$, hence the generators are all zero: $votimesalpha = v^2^kotimesalpha = votimes(2^kalpha) = votimes 0 = mathbf0$.
  $endgroup$
  – Arturo Magidin
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @jyrkilahtonen : I think that's the case : the notation $mathbb F$ is used for both
  $endgroup$
  – Max
  7 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.

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edited 7 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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Max I see no justification for my statement, now that I think about it. I'll remove it.
  $endgroup$
  – Omnomnomnom
  7 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 7 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
  7 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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Omnomnomnom But for $mathbbF_2^k$ and $V=mathbbF_2^ksetminus0$ under multiplication, it is zero: every element of $V$ satisfies $v^2^k=v$, hence the generators are all zero: $votimesalpha = v^2^kotimesalpha = votimes(2^kalpha) = votimes 0 = mathbf0$.
  $endgroup$
  – Arturo Magidin
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @jyrkilahtonen : I think that's the case : the notation $mathbb F$ is used for both
  $endgroup$
  – Max
  7 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 7 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
  7 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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Omnomnomnom But for $mathbbF_2^k$ and $V=mathbbF_2^ksetminus0$ under multiplication, it is zero: every element of $V$ satisfies $v^2^k=v$, hence the generators are all zero: $votimesalpha = v^2^kotimesalpha = votimes(2^kalpha) = votimes 0 = mathbf0$.
  $endgroup$
  – Arturo Magidin
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @jyrkilahtonen : I think that's the case : the notation $mathbb F$ is used for both
  $endgroup$
  – Max
  7 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 7 hours ago

MaxMax

23.2k11 gold badge1212 silver badges5454 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

MaxMax

23.2k11 gold badge1212 silver badges5454 bronze badges

answered 7 hours ago

MaxMax

23.2k11 gold badge1212 silver badges5454 bronze badges

answered 7 hours ago

MaxMax

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