Proof that three element vector space cannot be constructed over $mathbb Z_5$












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I'm trying to show that a vector space $V$ with three elements cannot be constructed over the field $mathbb{Z}_5$. The elements of $V$ can be shown to be ${0,1,1^{-1}}$. I don't think the vector space is closed under scalar multiplication, but how do I show this? As far as I know, we could define an operator $cdot$ such that $alpha cdot 1 = beta cdot 1$ for $1 in V$ and $alpha, beta in V$. I need to prove this result using only the definitions of a vector space and of a field.



I appreciate the help.






linear-algebra abstract-algebra







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  • 4




    $begingroup$
    A $k-$dimensional vector space over $mathbb{Z}_5$ has $5^k$ elements.
    $endgroup$
    – saulspatz
    Jan 31 at 16:39










  • $begingroup$
    Yes, but I've yet to show this. I'm looking for a proof based only on the definitions of a vector space and a field.
    $endgroup$
    – Derek Adams
    Jan 31 at 16:41
















0












$begingroup$


I'm trying to show that a vector space $V$ with three elements cannot be constructed over the field $mathbb{Z}_5$. The elements of $V$ can be shown to be ${0,1,1^{-1}}$. I don't think the vector space is closed under scalar multiplication, but how do I show this? As far as I know, we could define an operator $cdot$ such that $alpha cdot 1 = beta cdot 1$ for $1 in V$ and $alpha, beta in V$. I need to prove this result using only the definitions of a vector space and of a field.



I appreciate the help.






linear-algebra abstract-algebra







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    A $k-$dimensional vector space over $mathbb{Z}_5$ has $5^k$ elements.
    $endgroup$
    – saulspatz
    Jan 31 at 16:39










  • $begingroup$
    Yes, but I've yet to show this. I'm looking for a proof based only on the definitions of a vector space and a field.
    $endgroup$
    – Derek Adams
    Jan 31 at 16:41














0












0








0





$begingroup$


I'm trying to show that a vector space $V$ with three elements cannot be constructed over the field $mathbb{Z}_5$. The elements of $V$ can be shown to be ${0,1,1^{-1}}$. I don't think the vector space is closed under scalar multiplication, but how do I show this? As far as I know, we could define an operator $cdot$ such that $alpha cdot 1 = beta cdot 1$ for $1 in V$ and $alpha, beta in V$. I need to prove this result using only the definitions of a vector space and of a field.



I appreciate the help.






linear-algebra abstract-algebra







share|cite|improve this question











$endgroup$




I'm trying to show that a vector space $V$ with three elements cannot be constructed over the field $mathbb{Z}_5$. The elements of $V$ can be shown to be ${0,1,1^{-1}}$. I don't think the vector space is closed under scalar multiplication, but how do I show this? As far as I know, we could define an operator $cdot$ such that $alpha cdot 1 = beta cdot 1$ for $1 in V$ and $alpha, beta in V$. I need to prove this result using only the definitions of a vector space and of a field.



I appreciate the help.






linear-algebra abstract-algebra




linear-algebra abstract-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 31 at 21:32









J. W. Tanner

4,5391320




4,5391320










asked Jan 31 at 16:37









Derek AdamsDerek Adams

567414




567414








  • 4




    $begingroup$
    A $k-$dimensional vector space over $mathbb{Z}_5$ has $5^k$ elements.
    $endgroup$
    – saulspatz
    Jan 31 at 16:39










  • $begingroup$
    Yes, but I've yet to show this. I'm looking for a proof based only on the definitions of a vector space and a field.
    $endgroup$
    – Derek Adams
    Jan 31 at 16:41














  • 4




    $begingroup$
    A $k-$dimensional vector space over $mathbb{Z}_5$ has $5^k$ elements.
    $endgroup$
    – saulspatz
    Jan 31 at 16:39










  • $begingroup$
    Yes, but I've yet to show this. I'm looking for a proof based only on the definitions of a vector space and a field.
    $endgroup$
    – Derek Adams
    Jan 31 at 16:41








4




4




$begingroup$
A $k-$dimensional vector space over $mathbb{Z}_5$ has $5^k$ elements.
$endgroup$
– saulspatz
Jan 31 at 16:39




$begingroup$
A $k-$dimensional vector space over $mathbb{Z}_5$ has $5^k$ elements.
$endgroup$
– saulspatz
Jan 31 at 16:39












$begingroup$
Yes, but I've yet to show this. I'm looking for a proof based only on the definitions of a vector space and a field.
$endgroup$
– Derek Adams
Jan 31 at 16:41




$begingroup$
Yes, but I've yet to show this. I'm looking for a proof based only on the definitions of a vector space and a field.
$endgroup$
– Derek Adams
Jan 31 at 16:41










1 Answer
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$begingroup$

Let $Xne0$ be an element of the space. Let $a,binmathbb{Z}_5$ with $ane b$. If $aX=bX$ then $(a-b)X=0$ and multiplying both sides by $(a-b)^{-1}$ gives $X=0$ contrary to assumption.



Thus, $0, X, 2X, 3X, 4X$ are all distinct.






share|cite|improve this answer











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    1 Answer
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    1 Answer
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    $begingroup$

    Let $Xne0$ be an element of the space. Let $a,binmathbb{Z}_5$ with $ane b$. If $aX=bX$ then $(a-b)X=0$ and multiplying both sides by $(a-b)^{-1}$ gives $X=0$ contrary to assumption.



    Thus, $0, X, 2X, 3X, 4X$ are all distinct.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Let $Xne0$ be an element of the space. Let $a,binmathbb{Z}_5$ with $ane b$. If $aX=bX$ then $(a-b)X=0$ and multiplying both sides by $(a-b)^{-1}$ gives $X=0$ contrary to assumption.



      Thus, $0, X, 2X, 3X, 4X$ are all distinct.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Let $Xne0$ be an element of the space. Let $a,binmathbb{Z}_5$ with $ane b$. If $aX=bX$ then $(a-b)X=0$ and multiplying both sides by $(a-b)^{-1}$ gives $X=0$ contrary to assumption.



        Thus, $0, X, 2X, 3X, 4X$ are all distinct.






        share|cite|improve this answer











        $endgroup$



        Let $Xne0$ be an element of the space. Let $a,binmathbb{Z}_5$ with $ane b$. If $aX=bX$ then $(a-b)X=0$ and multiplying both sides by $(a-b)^{-1}$ gives $X=0$ contrary to assumption.



        Thus, $0, X, 2X, 3X, 4X$ are all distinct.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 31 at 16:59

























        answered Jan 31 at 16:52









        saulspatzsaulspatz

        17.2k31435




        17.2k31435






























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