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Is there a generating set of $mathbbZ^3$ having cardinality $2$?
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up vote
3
down vote
favorite
Is there a generating set of $mathbbZ^3$ having cardinality $2$?
I believe that the answer is no...
I looked over at possible generating sets, such as $<(1,1,0),(0,1,1)>$, and I can see that there are elements which can't be represented using that set, as since $mathbbZ^3$ can have $3$ different elements, the generating set must be of at least $3$ elements...
And yet, I'm not sure, how to prove this formally?
group-theory
asked Jul 21 at 17:06
ChikChak
685214
add a comment |Â
up vote
3
down vote
favorite
Is there a generating set of $mathbbZ^3$ having cardinality $2$?
I believe that the answer is no...
I looked over at possible generating sets, such as $<(1,1,0),(0,1,1)>$, and I can see that there are elements which can't be represented using that set, as since $mathbbZ^3$ can have $3$ different elements, the generating set must be of at least $3$ elements...
And yet, I'm not sure, how to prove this formally?
group-theory
asked Jul 21 at 17:06
ChikChak
685214
2
No. Such a set would be a basis of $mathbbZ^3 otimes mathbbQ = mathbbQ^3$.
– anomaly
Jul 21 at 17:10
2
It is well known that $Bbb R^3$ is a three-dimensional vector space. Any subset of $Bbb Z^3$ which is generated by two elements will necessarily be a subset of the subspace of $Bbb R^3$ generated by those two elements, and is therefore a subset of only a two-dimensional subspace of $Bbb R^3$, a third dimension being left entirely untouched.
– JMoravitz
Jul 21 at 17:10
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Is there a generating set of $mathbbZ^3$ having cardinality $2$?
I believe that the answer is no...
I looked over at possible generating sets, such as $<(1,1,0),(0,1,1)>$, and I can see that there are elements which can't be represented using that set, as since $mathbbZ^3$ can have $3$ different elements, the generating set must be of at least $3$ elements...
And yet, I'm not sure, how to prove this formally?
group-theory
asked Jul 21 at 17:06
ChikChak
685214
Is there a generating set of $mathbbZ^3$ having cardinality $2$?
I believe that the answer is no...
I looked over at possible generating sets, such as $<(1,1,0),(0,1,1)>$, and I can see that there are elements which can't be represented using that set, as since $mathbbZ^3$ can have $3$ different elements, the generating set must be of at least $3$ elements...
And yet, I'm not sure, how to prove this formally?
group-theory
asked Jul 21 at 17:06
ChikChak
685214
edited Jul 21 at 17:10
asked Jul 21 at 17:06
ChikChak
685214
asked Jul 21 at 17:06
ChikChak
685214
asked Jul 21 at 17:06
ChikChak
685214
685214
2
No. Such a set would be a basis of $mathbbZ^3 otimes mathbbQ = mathbbQ^3$.
– anomaly
Jul 21 at 17:10
2
It is well known that $Bbb R^3$ is a three-dimensional vector space. Any subset of $Bbb Z^3$ which is generated by two elements will necessarily be a subset of the subspace of $Bbb R^3$ generated by those two elements, and is therefore a subset of only a two-dimensional subspace of $Bbb R^3$, a third dimension being left entirely untouched.
– JMoravitz
Jul 21 at 17:10
add a comment |Â
2
No. Such a set would be a basis of $mathbbZ^3 otimes mathbbQ = mathbbQ^3$.
– anomaly
Jul 21 at 17:10
2
It is well known that $Bbb R^3$ is a three-dimensional vector space. Any subset of $Bbb Z^3$ which is generated by two elements will necessarily be a subset of the subspace of $Bbb R^3$ generated by those two elements, and is therefore a subset of only a two-dimensional subspace of $Bbb R^3$, a third dimension being left entirely untouched.
– JMoravitz
Jul 21 at 17:10
2
2
No. Such a set would be a basis of $mathbbZ^3 otimes mathbbQ = mathbbQ^3$.
– anomaly
Jul 21 at 17:10
No. Such a set would be a basis of $mathbbZ^3 otimes mathbbQ = mathbbQ^3$.
– anomaly
Jul 21 at 17:10
2
2
It is well known that $Bbb R^3$ is a three-dimensional vector space. Any subset of $Bbb Z^3$ which is generated by two elements will necessarily be a subset of the subspace of $Bbb R^3$ generated by those two elements, and is therefore a subset of only a two-dimensional subspace of $Bbb R^3$, a third dimension being left entirely untouched.
– JMoravitz
Jul 21 at 17:10
It is well known that $Bbb R^3$ is a three-dimensional vector space. Any subset of $Bbb Z^3$ which is generated by two elements will necessarily be a subset of the subspace of $Bbb R^3$ generated by those two elements, and is therefore a subset of only a two-dimensional subspace of $Bbb R^3$, a third dimension being left entirely untouched.
– JMoravitz
Jul 21 at 17:10
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Suppse that such a set $v,w$ exists. Then there are numbers $a,a',a'',b,b',b''inmathbb Z$ such that$$(1,0,0)=av+bw, (0,1,0)=a'v+b'w,text and (0,0,1)=a''v+b''w.$$But since $(1,0,0),(0,1,0),(0,0,0)$ are $3$ linearly independent elements of $mathbbQ^3$, this would mean that $mathbbQ^3$, which is a $3$-dimensional vector space over $mathbb Q$, would have a generating set with two elements. That is impossible.
answered Jul 21 at 17:11
José Carlos Santos
114k1698177
add a comment |Â
up vote
9
down vote
One trick is to consider the images of your would-be generators
in $G=Bbb Z^3/2Bbb Z^3$. Then $G$ is a vector space of dimension $3$
over $Bbb Z/2Bbb Z$ and cannot be generated by two generators.
answered Jul 21 at 17:10
Lord Shark the Unknown
85.2k950111
clever! $+1 dots$
– Andres Mejia
Jul 21 at 17:12
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Suppse that such a set $v,w$ exists. Then there are numbers $a,a',a'',b,b',b''inmathbb Z$ such that$$(1,0,0)=av+bw, (0,1,0)=a'v+b'w,text and (0,0,1)=a''v+b''w.$$But since $(1,0,0),(0,1,0),(0,0,0)$ are $3$ linearly independent elements of $mathbbQ^3$, this would mean that $mathbbQ^3$, which is a $3$-dimensional vector space over $mathbb Q$, would have a generating set with two elements. That is impossible.
answered Jul 21 at 17:11
José Carlos Santos
114k1698177
add a comment |Â
up vote
4
down vote
accepted
Suppse that such a set $v,w$ exists. Then there are numbers $a,a',a'',b,b',b''inmathbb Z$ such that$$(1,0,0)=av+bw, (0,1,0)=a'v+b'w,text and (0,0,1)=a''v+b''w.$$But since $(1,0,0),(0,1,0),(0,0,0)$ are $3$ linearly independent elements of $mathbbQ^3$, this would mean that $mathbbQ^3$, which is a $3$-dimensional vector space over $mathbb Q$, would have a generating set with two elements. That is impossible.
answered Jul 21 at 17:11
José Carlos Santos
114k1698177
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Suppse that such a set $v,w$ exists. Then there are numbers $a,a',a'',b,b',b''inmathbb Z$ such that$$(1,0,0)=av+bw, (0,1,0)=a'v+b'w,text and (0,0,1)=a''v+b''w.$$But since $(1,0,0),(0,1,0),(0,0,0)$ are $3$ linearly independent elements of $mathbbQ^3$, this would mean that $mathbbQ^3$, which is a $3$-dimensional vector space over $mathbb Q$, would have a generating set with two elements. That is impossible.
answered Jul 21 at 17:11
José Carlos Santos
114k1698177
Suppse that such a set $v,w$ exists. Then there are numbers $a,a',a'',b,b',b''inmathbb Z$ such that$$(1,0,0)=av+bw, (0,1,0)=a'v+b'w,text and (0,0,1)=a''v+b''w.$$But since $(1,0,0),(0,1,0),(0,0,0)$ are $3$ linearly independent elements of $mathbbQ^3$, this would mean that $mathbbQ^3$, which is a $3$-dimensional vector space over $mathbb Q$, would have a generating set with two elements. That is impossible.
answered Jul 21 at 17:11
José Carlos Santos
114k1698177
answered Jul 21 at 17:11
José Carlos Santos
114k1698177
answered Jul 21 at 17:11
José Carlos Santos
114k1698177
answered Jul 21 at 17:11
José Carlos Santos
114k1698177
114k1698177
add a comment |Â
add a comment |Â
up vote
9
down vote
One trick is to consider the images of your would-be generators
in $G=Bbb Z^3/2Bbb Z^3$. Then $G$ is a vector space of dimension $3$
over $Bbb Z/2Bbb Z$ and cannot be generated by two generators.
answered Jul 21 at 17:10
Lord Shark the Unknown
85.2k950111
clever! $+1 dots$
– Andres Mejia
Jul 21 at 17:12
add a comment |Â
up vote
9
down vote
One trick is to consider the images of your would-be generators
in $G=Bbb Z^3/2Bbb Z^3$. Then $G$ is a vector space of dimension $3$
over $Bbb Z/2Bbb Z$ and cannot be generated by two generators.
answered Jul 21 at 17:10
Lord Shark the Unknown
85.2k950111
clever! $+1 dots$
– Andres Mejia
Jul 21 at 17:12
add a comment |Â
up vote
9
down vote
up vote
9
down vote
One trick is to consider the images of your would-be generators
in $G=Bbb Z^3/2Bbb Z^3$. Then $G$ is a vector space of dimension $3$
over $Bbb Z/2Bbb Z$ and cannot be generated by two generators.
answered Jul 21 at 17:10
Lord Shark the Unknown
85.2k950111
One trick is to consider the images of your would-be generators
in $G=Bbb Z^3/2Bbb Z^3$. Then $G$ is a vector space of dimension $3$
over $Bbb Z/2Bbb Z$ and cannot be generated by two generators.
answered Jul 21 at 17:10
Lord Shark the Unknown
85.2k950111
answered Jul 21 at 17:10
Lord Shark the Unknown
85.2k950111
answered Jul 21 at 17:10
Lord Shark the Unknown
85.2k950111
answered Jul 21 at 17:10
Lord Shark the Unknown
85.2k950111
85.2k950111
clever! $+1 dots$
– Andres Mejia
Jul 21 at 17:12
add a comment |Â
clever! $+1 dots$
– Andres Mejia
Jul 21 at 17:12
clever! $+1 dots$
– Andres Mejia
Jul 21 at 17:12
clever! $+1 dots$
– Andres Mejia
Jul 21 at 17:12
add a comment |Â
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2
No. Such a set would be a basis of $mathbbZ^3 otimes mathbbQ = mathbbQ^3$.
– anomaly
Jul 21 at 17:10
2
It is well known that $Bbb R^3$ is a three-dimensional vector space. Any subset of $Bbb Z^3$ which is generated by two elements will necessarily be a subset of the subspace of $Bbb R^3$ generated by those two elements, and is therefore a subset of only a two-dimensional subspace of $Bbb R^3$, a third dimension being left entirely untouched.
– JMoravitz
Jul 21 at 17:10