Find the pure submodules of a cyclic $mathbb{Z}$-module of order 12
$begingroup$
Let $M = langle alpha rangle$ be a cyclic $mathbb{Z}$-module of order 12.
My goal is to list the pure submodules of $M$. A submodule $N$ is pure in $M$ if for $y in N$ and $a in mathbb{Z}$ and there exists $xin M$ such that $ax=y$, then actually $exists zin N$ such that $az=y$.
I am really rusty on my algebra. My thoughts are this.
$mathbb{Z}$-modules are in correspondence with abelian groups. Since $M$ is cyclic of order 12, then I will stick to what I know and look at $mathbb{Z}_{12}$. The non-trivial subgroups of this are $mathbb{Z}_2, mathbb{Z}_3, mathbb{Z}_4, mathbb{Z}_6$.
I read somewhere that pure submodules are summands of the module $M$. (That is, if $N$ is pure then $exists L$ disjoint from $N$ such that $M = N+L$). And, the converse is true. If $N$ is a summand, then it is pure. The proof is somewhere on this site, but I am having trouble finding it.
From my knowledge of group theory, the only way to write $mathbb{Z}_{12}$ as a direct product $mathbb{Z}_{12} = mathbb{Z}_4 times mathbb{Z}_3$. I suppose this means that the only way to write $M$ as a direct sum of its submodules is in the same way. Thus, the only pure submodules are those corresponding to $mathbb{Z}_4$ and $mathbb{Z}_3$. I guess to be pedantic, $0$ and $M$ are also pure.
Are these thoughts correct?
linear-algebra abstract-algebra
asked Dec 15 '18 at 20:35
PentakiPentaki
1105
$endgroup$
add a comment |
$begingroup$
Let $M = langle alpha rangle$ be a cyclic $mathbb{Z}$-module of order 12.
My goal is to list the pure submodules of $M$. A submodule $N$ is pure in $M$ if for $y in N$ and $a in mathbb{Z}$ and there exists $xin M$ such that $ax=y$, then actually $exists zin N$ such that $az=y$.
I am really rusty on my algebra. My thoughts are this.
$mathbb{Z}$-modules are in correspondence with abelian groups. Since $M$ is cyclic of order 12, then I will stick to what I know and look at $mathbb{Z}_{12}$. The non-trivial subgroups of this are $mathbb{Z}_2, mathbb{Z}_3, mathbb{Z}_4, mathbb{Z}_6$.
I read somewhere that pure submodules are summands of the module $M$. (That is, if $N$ is pure then $exists L$ disjoint from $N$ such that $M = N+L$). And, the converse is true. If $N$ is a summand, then it is pure. The proof is somewhere on this site, but I am having trouble finding it.
From my knowledge of group theory, the only way to write $mathbb{Z}_{12}$ as a direct product $mathbb{Z}_{12} = mathbb{Z}_4 times mathbb{Z}_3$. I suppose this means that the only way to write $M$ as a direct sum of its submodules is in the same way. Thus, the only pure submodules are those corresponding to $mathbb{Z}_4$ and $mathbb{Z}_3$. I guess to be pedantic, $0$ and $M$ are also pure.
Are these thoughts correct?
linear-algebra abstract-algebra
asked Dec 15 '18 at 20:35
PentakiPentaki
1105
$endgroup$
add a comment |
$begingroup$
Let $M = langle alpha rangle$ be a cyclic $mathbb{Z}$-module of order 12.
My goal is to list the pure submodules of $M$. A submodule $N$ is pure in $M$ if for $y in N$ and $a in mathbb{Z}$ and there exists $xin M$ such that $ax=y$, then actually $exists zin N$ such that $az=y$.
I am really rusty on my algebra. My thoughts are this.
$mathbb{Z}$-modules are in correspondence with abelian groups. Since $M$ is cyclic of order 12, then I will stick to what I know and look at $mathbb{Z}_{12}$. The non-trivial subgroups of this are $mathbb{Z}_2, mathbb{Z}_3, mathbb{Z}_4, mathbb{Z}_6$.
I read somewhere that pure submodules are summands of the module $M$. (That is, if $N$ is pure then $exists L$ disjoint from $N$ such that $M = N+L$). And, the converse is true. If $N$ is a summand, then it is pure. The proof is somewhere on this site, but I am having trouble finding it.
From my knowledge of group theory, the only way to write $mathbb{Z}_{12}$ as a direct product $mathbb{Z}_{12} = mathbb{Z}_4 times mathbb{Z}_3$. I suppose this means that the only way to write $M$ as a direct sum of its submodules is in the same way. Thus, the only pure submodules are those corresponding to $mathbb{Z}_4$ and $mathbb{Z}_3$. I guess to be pedantic, $0$ and $M$ are also pure.
Are these thoughts correct?
linear-algebra abstract-algebra
asked Dec 15 '18 at 20:35
PentakiPentaki
1105
$endgroup$
Let $M = langle alpha rangle$ be a cyclic $mathbb{Z}$-module of order 12.
My goal is to list the pure submodules of $M$. A submodule $N$ is pure in $M$ if for $y in N$ and $a in mathbb{Z}$ and there exists $xin M$ such that $ax=y$, then actually $exists zin N$ such that $az=y$.
I am really rusty on my algebra. My thoughts are this.
$mathbb{Z}$-modules are in correspondence with abelian groups. Since $M$ is cyclic of order 12, then I will stick to what I know and look at $mathbb{Z}_{12}$. The non-trivial subgroups of this are $mathbb{Z}_2, mathbb{Z}_3, mathbb{Z}_4, mathbb{Z}_6$.
I read somewhere that pure submodules are summands of the module $M$. (That is, if $N$ is pure then $exists L$ disjoint from $N$ such that $M = N+L$). And, the converse is true. If $N$ is a summand, then it is pure. The proof is somewhere on this site, but I am having trouble finding it.
From my knowledge of group theory, the only way to write $mathbb{Z}_{12}$ as a direct product $mathbb{Z}_{12} = mathbb{Z}_4 times mathbb{Z}_3$. I suppose this means that the only way to write $M$ as a direct sum of its submodules is in the same way. Thus, the only pure submodules are those corresponding to $mathbb{Z}_4$ and $mathbb{Z}_3$. I guess to be pedantic, $0$ and $M$ are also pure.
Are these thoughts correct?
linear-algebra abstract-algebra
linear-algebra abstract-algebra
asked Dec 15 '18 at 20:35
PentakiPentaki
1105
asked Dec 15 '18 at 20:35
PentakiPentaki
1105
asked Dec 15 '18 at 20:35
PentakiPentaki
1105
asked Dec 15 '18 at 20:35
PentakiPentaki
1105
asked Dec 15 '18 at 20:35
PentakiPentaki
1105
1105
add a comment |
add a comment |
1 Answer
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$begingroup$
An easy way to check is to note that $mathbb{Z}_2$ is not pure (with elements of $mathbb{Z}_{12}$ in bold, $2cdotmathbf{3} = mathbf{6}$, but for the two elements of the copy of $mathbb{Z_2}$ we have $2cdotmathbf{0} = mathbf{0} neq mathbf{6}$ and $2cdotmathbf{6} = mathbf{0} neq mathbf{6}$, so there is no element of $mathbb{Z}_2$ that doubles to give $mathbf{6}$), and the same for $mathbb{Z}_6$ ($2 cdot mathbf{3} = mathbf{6}$, but doubling any element of the copy of $mathbb{Z}_{6}$ in $mathbb{Z}_12$ gives either $0$, $4$, or $8$, so there is no element of $mathbb{Z}_6$ that doubles to give $mathbf{6}$.
answered Dec 15 '18 at 20:49
user3482749user3482749
4,296919
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1 Answer
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$begingroup$
An easy way to check is to note that $mathbb{Z}_2$ is not pure (with elements of $mathbb{Z}_{12}$ in bold, $2cdotmathbf{3} = mathbf{6}$, but for the two elements of the copy of $mathbb{Z_2}$ we have $2cdotmathbf{0} = mathbf{0} neq mathbf{6}$ and $2cdotmathbf{6} = mathbf{0} neq mathbf{6}$, so there is no element of $mathbb{Z}_2$ that doubles to give $mathbf{6}$), and the same for $mathbb{Z}_6$ ($2 cdot mathbf{3} = mathbf{6}$, but doubling any element of the copy of $mathbb{Z}_{6}$ in $mathbb{Z}_12$ gives either $0$, $4$, or $8$, so there is no element of $mathbb{Z}_6$ that doubles to give $mathbf{6}$.
answered Dec 15 '18 at 20:49
user3482749user3482749
4,296919
$endgroup$
add a comment |
$begingroup$
An easy way to check is to note that $mathbb{Z}_2$ is not pure (with elements of $mathbb{Z}_{12}$ in bold, $2cdotmathbf{3} = mathbf{6}$, but for the two elements of the copy of $mathbb{Z_2}$ we have $2cdotmathbf{0} = mathbf{0} neq mathbf{6}$ and $2cdotmathbf{6} = mathbf{0} neq mathbf{6}$, so there is no element of $mathbb{Z}_2$ that doubles to give $mathbf{6}$), and the same for $mathbb{Z}_6$ ($2 cdot mathbf{3} = mathbf{6}$, but doubling any element of the copy of $mathbb{Z}_{6}$ in $mathbb{Z}_12$ gives either $0$, $4$, or $8$, so there is no element of $mathbb{Z}_6$ that doubles to give $mathbf{6}$.
answered Dec 15 '18 at 20:49
user3482749user3482749
4,296919
$endgroup$
add a comment |
$begingroup$
An easy way to check is to note that $mathbb{Z}_2$ is not pure (with elements of $mathbb{Z}_{12}$ in bold, $2cdotmathbf{3} = mathbf{6}$, but for the two elements of the copy of $mathbb{Z_2}$ we have $2cdotmathbf{0} = mathbf{0} neq mathbf{6}$ and $2cdotmathbf{6} = mathbf{0} neq mathbf{6}$, so there is no element of $mathbb{Z}_2$ that doubles to give $mathbf{6}$), and the same for $mathbb{Z}_6$ ($2 cdot mathbf{3} = mathbf{6}$, but doubling any element of the copy of $mathbb{Z}_{6}$ in $mathbb{Z}_12$ gives either $0$, $4$, or $8$, so there is no element of $mathbb{Z}_6$ that doubles to give $mathbf{6}$.
answered Dec 15 '18 at 20:49
user3482749user3482749
4,296919
$endgroup$
An easy way to check is to note that $mathbb{Z}_2$ is not pure (with elements of $mathbb{Z}_{12}$ in bold, $2cdotmathbf{3} = mathbf{6}$, but for the two elements of the copy of $mathbb{Z_2}$ we have $2cdotmathbf{0} = mathbf{0} neq mathbf{6}$ and $2cdotmathbf{6} = mathbf{0} neq mathbf{6}$, so there is no element of $mathbb{Z}_2$ that doubles to give $mathbf{6}$), and the same for $mathbb{Z}_6$ ($2 cdot mathbf{3} = mathbf{6}$, but doubling any element of the copy of $mathbb{Z}_{6}$ in $mathbb{Z}_12$ gives either $0$, $4$, or $8$, so there is no element of $mathbb{Z}_6$ that doubles to give $mathbf{6}$.
answered Dec 15 '18 at 20:49
user3482749user3482749
4,296919
answered Dec 15 '18 at 20:49
user3482749user3482749
4,296919
answered Dec 15 '18 at 20:49
user3482749user3482749
4,296919
answered Dec 15 '18 at 20:49
user3482749user3482749
4,296919
4,296919
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add a comment |
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