Is $mathbb{Q}$ a vector space over $mathbb{Z}$ or $mathbb{Q}$?












4












$begingroup$


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







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$endgroup$








  • 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
















4












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














4












4








4


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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














  • 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










3 Answers
3






active

oldest

votes


















11












$begingroup$

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 $Bbb Q$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















1












$begingroup$

$mathbb{Q}^n$ 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.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $Bbb Z$ 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.






    share|cite|improve this answer











    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      11












      $begingroup$

      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 $Bbb Q$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        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
















      11












      $begingroup$

      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 $Bbb Q$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        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














      11












      11








      11





      $begingroup$

      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 $Bbb Q$.






      share|cite|improve this answer











      $endgroup$



      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 $Bbb Q$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      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


















      • $begingroup$
        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











      1












      $begingroup$

      $mathbb{Q}^n$ 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.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        $mathbb{Q}^n$ 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.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          $mathbb{Q}^n$ 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.






          share|cite|improve this answer









          $endgroup$



          $mathbb{Q}^n$ 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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 29 at 8:37









          PerturbativePerturbative

          4,45621553




          4,45621553























              1












              $begingroup$

              $Bbb Z$ 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.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                $Bbb Z$ 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.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $Bbb Z$ 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.






                  share|cite|improve this answer











                  $endgroup$



                  $Bbb Z$ 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.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 29 at 10:22

























                  answered Jan 29 at 9:50









                  Chris CusterChris Custer

                  14.2k3827




                  14.2k3827






























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