Mixing
Torsion in Cokernel is Annihilated by Integer
$begingroup$
My question refers to an argument in a proof from T. tom Dieck's "Algebraic Topology and Transformation Groups" (Lemma 3; page 2). Here the excerpt and the red tagged step which I don't understand:
We have the inclusion map $alpha: mathbb{Z} hookrightarrow M$ and the multiplication by $n$ map $n: mathbb{Z} to mathbb{Z}, z mapsto n cdot z$ with property that the torsion of the cokernel $M / im(alpha)$ is annilated by $n$.
In the proof there is deduced that neglect the torsion of $M$ then $M = N oplus <x>$.
$<x>$ is a free component with $alpha(1) =mx$, so $im(alpha) subset <x>$. First question: What is $N$? We assumed that torsion of $M$ trivial. Why $M$ can be splitted in direct sum $N oplus <x>$? We nowhere assumed that $M$ is finitely generated so I don't know if we here just can apply the classification theorem for abelian groups.
Let continue: Since $im(alpha)= <mx>$ we have $coker(alpha):=M / im(alpha)= M oplus mathbb{Z}/m$ with torsion $mathbb{Z}/m$.
By assumption the torsion is annilated by $n$, so $n = rm$ for appropriate $r in mathbb{Z}$.
By definition the pushout $K$ is the quotient of $N oplus mathbb{Z} oplus mathbb{Z}/ sim$ with the relation $n(0,0,1) = m(0,1,0)$.
Take into account that here we identified $<x>$ with $<(0,1,0)>$.
Therefore the pushout is isomorphic to $K=N oplus mathbb{Z} oplus mathbb{Z}/<(0, m, n) >$. That's clear.
What I don't understand is why holds $N oplus mathbb{Z} oplus mathbb{Z}/<(0, m, n) > cong M oplus mathbb{Z}/m$.
Or in other words why $mathbb{Z} oplus mathbb{Z}/<(m, n) > cong mathbb{Z}/m$ holds under assumption that $m$ divides $n$?
algebraic-topology modules torsion-groups
asked Jan 21 at 20:52
KarlPeterKarlPeter
4031315
$endgroup$
add a comment |
$begingroup$
My question refers to an argument in a proof from T. tom Dieck's "Algebraic Topology and Transformation Groups" (Lemma 3; page 2). Here the excerpt and the red tagged step which I don't understand:
We have the inclusion map $alpha: mathbb{Z} hookrightarrow M$ and the multiplication by $n$ map $n: mathbb{Z} to mathbb{Z}, z mapsto n cdot z$ with property that the torsion of the cokernel $M / im(alpha)$ is annilated by $n$.
In the proof there is deduced that neglect the torsion of $M$ then $M = N oplus <x>$.
$<x>$ is a free component with $alpha(1) =mx$, so $im(alpha) subset <x>$. First question: What is $N$? We assumed that torsion of $M$ trivial. Why $M$ can be splitted in direct sum $N oplus <x>$? We nowhere assumed that $M$ is finitely generated so I don't know if we here just can apply the classification theorem for abelian groups.
Let continue: Since $im(alpha)= <mx>$ we have $coker(alpha):=M / im(alpha)= M oplus mathbb{Z}/m$ with torsion $mathbb{Z}/m$.
By assumption the torsion is annilated by $n$, so $n = rm$ for appropriate $r in mathbb{Z}$.
By definition the pushout $K$ is the quotient of $N oplus mathbb{Z} oplus mathbb{Z}/ sim$ with the relation $n(0,0,1) = m(0,1,0)$.
Take into account that here we identified $<x>$ with $<(0,1,0)>$.
Therefore the pushout is isomorphic to $K=N oplus mathbb{Z} oplus mathbb{Z}/<(0, m, n) >$. That's clear.
What I don't understand is why holds $N oplus mathbb{Z} oplus mathbb{Z}/<(0, m, n) > cong M oplus mathbb{Z}/m$.
Or in other words why $mathbb{Z} oplus mathbb{Z}/<(m, n) > cong mathbb{Z}/m$ holds under assumption that $m$ divides $n$?
algebraic-topology modules torsion-groups
asked Jan 21 at 20:52
KarlPeterKarlPeter
4031315
$endgroup$
add a comment |
$begingroup$
My question refers to an argument in a proof from T. tom Dieck's "Algebraic Topology and Transformation Groups" (Lemma 3; page 2). Here the excerpt and the red tagged step which I don't understand:
We have the inclusion map $alpha: mathbb{Z} hookrightarrow M$ and the multiplication by $n$ map $n: mathbb{Z} to mathbb{Z}, z mapsto n cdot z$ with property that the torsion of the cokernel $M / im(alpha)$ is annilated by $n$.
In the proof there is deduced that neglect the torsion of $M$ then $M = N oplus <x>$.
$<x>$ is a free component with $alpha(1) =mx$, so $im(alpha) subset <x>$. First question: What is $N$? We assumed that torsion of $M$ trivial. Why $M$ can be splitted in direct sum $N oplus <x>$? We nowhere assumed that $M$ is finitely generated so I don't know if we here just can apply the classification theorem for abelian groups.
Let continue: Since $im(alpha)= <mx>$ we have $coker(alpha):=M / im(alpha)= M oplus mathbb{Z}/m$ with torsion $mathbb{Z}/m$.
By assumption the torsion is annilated by $n$, so $n = rm$ for appropriate $r in mathbb{Z}$.
By definition the pushout $K$ is the quotient of $N oplus mathbb{Z} oplus mathbb{Z}/ sim$ with the relation $n(0,0,1) = m(0,1,0)$.
Take into account that here we identified $<x>$ with $<(0,1,0)>$.
Therefore the pushout is isomorphic to $K=N oplus mathbb{Z} oplus mathbb{Z}/<(0, m, n) >$. That's clear.
What I don't understand is why holds $N oplus mathbb{Z} oplus mathbb{Z}/<(0, m, n) > cong M oplus mathbb{Z}/m$.
Or in other words why $mathbb{Z} oplus mathbb{Z}/<(m, n) > cong mathbb{Z}/m$ holds under assumption that $m$ divides $n$?
algebraic-topology modules torsion-groups
asked Jan 21 at 20:52
KarlPeterKarlPeter
4031315
$endgroup$
My question refers to an argument in a proof from T. tom Dieck's "Algebraic Topology and Transformation Groups" (Lemma 3; page 2). Here the excerpt and the red tagged step which I don't understand:
We have the inclusion map $alpha: mathbb{Z} hookrightarrow M$ and the multiplication by $n$ map $n: mathbb{Z} to mathbb{Z}, z mapsto n cdot z$ with property that the torsion of the cokernel $M / im(alpha)$ is annilated by $n$.
In the proof there is deduced that neglect the torsion of $M$ then $M = N oplus <x>$.
$<x>$ is a free component with $alpha(1) =mx$, so $im(alpha) subset <x>$. First question: What is $N$? We assumed that torsion of $M$ trivial. Why $M$ can be splitted in direct sum $N oplus <x>$? We nowhere assumed that $M$ is finitely generated so I don't know if we here just can apply the classification theorem for abelian groups.
Let continue: Since $im(alpha)= <mx>$ we have $coker(alpha):=M / im(alpha)= M oplus mathbb{Z}/m$ with torsion $mathbb{Z}/m$.
By assumption the torsion is annilated by $n$, so $n = rm$ for appropriate $r in mathbb{Z}$.
By definition the pushout $K$ is the quotient of $N oplus mathbb{Z} oplus mathbb{Z}/ sim$ with the relation $n(0,0,1) = m(0,1,0)$.
Take into account that here we identified $<x>$ with $<(0,1,0)>$.
Therefore the pushout is isomorphic to $K=N oplus mathbb{Z} oplus mathbb{Z}/<(0, m, n) >$. That's clear.
What I don't understand is why holds $N oplus mathbb{Z} oplus mathbb{Z}/<(0, m, n) > cong M oplus mathbb{Z}/m$.
Or in other words why $mathbb{Z} oplus mathbb{Z}/<(m, n) > cong mathbb{Z}/m$ holds under assumption that $m$ divides $n$?
algebraic-topology modules torsion-groups
algebraic-topology modules torsion-groups
asked Jan 21 at 20:52
KarlPeterKarlPeter
4031315
asked Jan 21 at 20:52
KarlPeterKarlPeter
4031315
edited Jan 21 at 21:11
KarlPeter
asked Jan 21 at 20:52
KarlPeterKarlPeter
4031315
asked Jan 21 at 20:52
KarlPeterKarlPeter
4031315
asked Jan 21 at 20:52
KarlPeterKarlPeter
4031315
4031315
add a comment |
add a comment |
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$begingroup$
Your final 2 lines are incorrect. Because $M cong N oplus Bbb Z$, what you are trying to prove is actually that $Bbb Z^2/(m,n) = Bbb Z oplus (Bbb Z/text{gcd}(m,n))$.
First it is convenient to reduce this to one coordinate. Write $d = text{gcd}(m,n)$, so that $m = ad$ and $n = bd$ where $text{gcd}(a,b) = 1$. Then choose $x$ and $y$ with $ax + by = 1$; the matrix $$begin{pmatrix}a & -y\b & xend{pmatrix}$$ is invertible over the integers, and hence ${v_1, v_2} = {(a,b), (x,y)}$ forms a basis for $Bbb Z^2$; in this basis your vector $(m,n)$ simply takes the form $dv_1$.
Of course, $Bbb Z^2/(d,0) = Bbb Z/d oplus Bbb Z$. This is what you wanted.
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$begingroup$
Your final 2 lines are incorrect. Because $M cong N oplus Bbb Z$, what you are trying to prove is actually that $Bbb Z^2/(m,n) = Bbb Z oplus (Bbb Z/text{gcd}(m,n))$.
First it is convenient to reduce this to one coordinate. Write $d = text{gcd}(m,n)$, so that $m = ad$ and $n = bd$ where $text{gcd}(a,b) = 1$. Then choose $x$ and $y$ with $ax + by = 1$; the matrix $$begin{pmatrix}a & -y\b & xend{pmatrix}$$ is invertible over the integers, and hence ${v_1, v_2} = {(a,b), (x,y)}$ forms a basis for $Bbb Z^2$; in this basis your vector $(m,n)$ simply takes the form $dv_1$.
Of course, $Bbb Z^2/(d,0) = Bbb Z/d oplus Bbb Z$. This is what you wanted.
$endgroup$
add a comment |
$begingroup$
Your final 2 lines are incorrect. Because $M cong N oplus Bbb Z$, what you are trying to prove is actually that $Bbb Z^2/(m,n) = Bbb Z oplus (Bbb Z/text{gcd}(m,n))$.
First it is convenient to reduce this to one coordinate. Write $d = text{gcd}(m,n)$, so that $m = ad$ and $n = bd$ where $text{gcd}(a,b) = 1$. Then choose $x$ and $y$ with $ax + by = 1$; the matrix $$begin{pmatrix}a & -y\b & xend{pmatrix}$$ is invertible over the integers, and hence ${v_1, v_2} = {(a,b), (x,y)}$ forms a basis for $Bbb Z^2$; in this basis your vector $(m,n)$ simply takes the form $dv_1$.
Of course, $Bbb Z^2/(d,0) = Bbb Z/d oplus Bbb Z$. This is what you wanted.
$endgroup$
add a comment |
$begingroup$
Your final 2 lines are incorrect. Because $M cong N oplus Bbb Z$, what you are trying to prove is actually that $Bbb Z^2/(m,n) = Bbb Z oplus (Bbb Z/text{gcd}(m,n))$.
First it is convenient to reduce this to one coordinate. Write $d = text{gcd}(m,n)$, so that $m = ad$ and $n = bd$ where $text{gcd}(a,b) = 1$. Then choose $x$ and $y$ with $ax + by = 1$; the matrix $$begin{pmatrix}a & -y\b & xend{pmatrix}$$ is invertible over the integers, and hence ${v_1, v_2} = {(a,b), (x,y)}$ forms a basis for $Bbb Z^2$; in this basis your vector $(m,n)$ simply takes the form $dv_1$.
Of course, $Bbb Z^2/(d,0) = Bbb Z/d oplus Bbb Z$. This is what you wanted.
$endgroup$
Your final 2 lines are incorrect. Because $M cong N oplus Bbb Z$, what you are trying to prove is actually that $Bbb Z^2/(m,n) = Bbb Z oplus (Bbb Z/text{gcd}(m,n))$.
First it is convenient to reduce this to one coordinate. Write $d = text{gcd}(m,n)$, so that $m = ad$ and $n = bd$ where $text{gcd}(a,b) = 1$. Then choose $x$ and $y$ with $ax + by = 1$; the matrix $$begin{pmatrix}a & -y\b & xend{pmatrix}$$ is invertible over the integers, and hence ${v_1, v_2} = {(a,b), (x,y)}$ forms a basis for $Bbb Z^2$; in this basis your vector $(m,n)$ simply takes the form $dv_1$.
Of course, $Bbb Z^2/(d,0) = Bbb Z/d oplus Bbb Z$. This is what you wanted.
answered Jan 21 at 23:24
user98602
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