Mixing
Proof that three element vector space cannot be constructed over $mathbb Z_5$
$begingroup$
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
edited Jan 31 at 21:32
J. W. Tanner
4,5391320
asked Jan 31 at 16:37
Derek AdamsDerek Adams
567414
$endgroup$
add a comment |
$begingroup$
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
edited Jan 31 at 21:32
J. W. Tanner
4,5391320
asked Jan 31 at 16:37
Derek AdamsDerek Adams
567414
$endgroup$
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
add a comment |
$begingroup$
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
edited Jan 31 at 21:32
J. W. Tanner
4,5391320
asked Jan 31 at 16:37
Derek AdamsDerek Adams
567414
$endgroup$
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
edited Jan 31 at 21:32
J. W. Tanner
4,5391320
asked Jan 31 at 16:37
Derek AdamsDerek Adams
567414
edited Jan 31 at 21:32
J. W. Tanner
4,5391320
asked Jan 31 at 16:37
Derek AdamsDerek Adams
567414
edited Jan 31 at 21:32
J. W. Tanner
4,5391320
edited Jan 31 at 21:32
J. W. Tanner
4,5391320
edited Jan 31 at 21:32
J. W. Tanner
4,5391320
4,5391320
asked Jan 31 at 16:37
Derek AdamsDerek Adams
567414
asked Jan 31 at 16:37
Derek AdamsDerek Adams
567414
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 31 at 16:52
saulspatzsaulspatz
17.2k31435
$endgroup$
add a comment |
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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.
answered Jan 31 at 16:52
saulspatzsaulspatz
17.2k31435
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 31 at 16:52
saulspatzsaulspatz
17.2k31435
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 31 at 16:52
saulspatzsaulspatz
17.2k31435
$endgroup$
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.
answered Jan 31 at 16:52
saulspatzsaulspatz
17.2k31435
edited Jan 31 at 16:59
answered Jan 31 at 16:52
saulspatzsaulspatz
17.2k31435
answered Jan 31 at 16:52
saulspatzsaulspatz
17.2k31435
answered Jan 31 at 16:52
saulspatzsaulspatz
17.2k31435
17.2k31435
add a comment |
add a comment |
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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