Mixing
Exact sequence corresponding to $bar{phi} in operatorname{Ext}_{Bbb Z}^1(Bbb Z_9, Bbb Z)$ without using Baer...
I found that $operatorname{Ext}_{Bbb Z}^1(Bbb Z_9, Bbb Z) cong Bbb Z_9$ by calculating it as the quotient of two $operatorname{Hom}$-sets.
I am tasked with finding the (equivalence classes of) exact sequences corresponding to the elements in the group above.
I know two, the trivial one with middle module $Bbb Z oplus Z_9$, and the one with middle module $Bbb Z$ with morphisms $mu: Bbb Z to Bbb Z$, $mu(n) = 9n$ and $phi:Bbb Z to Bbb Z_9$, $phi(m) = bar{m}$.
Since $9$ is not prime I don't know what I need to construct the remaining seven exact sequences.
abstract-algebra commutative-algebra exact-sequence derived-functors
asked Nov 21 '18 at 12:57
Lukas Kofler
1
add a comment |
I found that $operatorname{Ext}_{Bbb Z}^1(Bbb Z_9, Bbb Z) cong Bbb Z_9$ by calculating it as the quotient of two $operatorname{Hom}$-sets.
I am tasked with finding the (equivalence classes of) exact sequences corresponding to the elements in the group above.
I know two, the trivial one with middle module $Bbb Z oplus Z_9$, and the one with middle module $Bbb Z$ with morphisms $mu: Bbb Z to Bbb Z$, $mu(n) = 9n$ and $phi:Bbb Z to Bbb Z_9$, $phi(m) = bar{m}$.
Since $9$ is not prime I don't know what I need to construct the remaining seven exact sequences.
abstract-algebra commutative-algebra exact-sequence derived-functors
asked Nov 21 '18 at 12:57
Lukas Kofler
1
add a comment |
I found that $operatorname{Ext}_{Bbb Z}^1(Bbb Z_9, Bbb Z) cong Bbb Z_9$ by calculating it as the quotient of two $operatorname{Hom}$-sets.
I am tasked with finding the (equivalence classes of) exact sequences corresponding to the elements in the group above.
I know two, the trivial one with middle module $Bbb Z oplus Z_9$, and the one with middle module $Bbb Z$ with morphisms $mu: Bbb Z to Bbb Z$, $mu(n) = 9n$ and $phi:Bbb Z to Bbb Z_9$, $phi(m) = bar{m}$.
Since $9$ is not prime I don't know what I need to construct the remaining seven exact sequences.
abstract-algebra commutative-algebra exact-sequence derived-functors
asked Nov 21 '18 at 12:57
Lukas Kofler
1
I found that $operatorname{Ext}_{Bbb Z}^1(Bbb Z_9, Bbb Z) cong Bbb Z_9$ by calculating it as the quotient of two $operatorname{Hom}$-sets.
I am tasked with finding the (equivalence classes of) exact sequences corresponding to the elements in the group above.
I know two, the trivial one with middle module $Bbb Z oplus Z_9$, and the one with middle module $Bbb Z$ with morphisms $mu: Bbb Z to Bbb Z$, $mu(n) = 9n$ and $phi:Bbb Z to Bbb Z_9$, $phi(m) = bar{m}$.
Since $9$ is not prime I don't know what I need to construct the remaining seven exact sequences.
abstract-algebra commutative-algebra exact-sequence derived-functors
abstract-algebra commutative-algebra exact-sequence derived-functors
asked Nov 21 '18 at 12:57
Lukas Kofler
1
asked Nov 21 '18 at 12:57
Lukas Kofler
1
asked Nov 21 '18 at 12:57
Lukas Kofler
1
asked Nov 21 '18 at 12:57
Lukas Kofler
1
asked Nov 21 '18 at 12:57
Lukas Kofler
1
1
add a comment |
add a comment |
1 Answer
1
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oldest
votes
Do you know what a pushout is? The construction you need is
$$
begin{array}{llllllll}
0 & to & mathbb{Z} & stackrel{times 9}to & mathbb{Z} & to & mathbb{Z}/9 & to & 0 \
& & downarrow phi & & downarrow & & downarrow textrm{id} & & \
0 & to & mathbb{Z} & to & P & to & mathbb{Z}/9 & to & 0
end{array}
$$
where $P$ is the pushout of $times 9$ and $phi$, the homomorphism whose ext-class you are working with.
For abelian groups, the pushout can be identified with the quotient of the direct sum $mathbb{Z}oplus mathbb{Z}$ by the subgroup generated by things of the form $(phi(z), -9z)$: you can see immediately that $phi=0$ gives you the trivial extension you mentioned.
answered Nov 21 '18 at 14:56
Matthew Towers
7,45822344
Thank you! Assuming I have $phi(1)=2$, what is the resulting quotient module? I feel like we would get $Bbb Z_{18}$ but then the sequence would fail to be exact.
– Lukas Kofler
Nov 21 '18 at 17:09
1
It won't be $mathbb{Z}_{18}$, as you say: the pushout is the quotient of $mathbb{Z}oplus mathbb{Z}$ by the subgroup generated by $(2,-9)$, which is isomorphic to $mathbb{Z}$ again (if you know about Smith Normal Form you can use that, or it's not hard by hand). Remember that it's not just the groups in the short exact sequence that matter - the maps matter as well.
– Matthew Towers
Nov 21 '18 at 18:01
In the end, I only want $9$ equivalence classes of exact sequences. Which $phi$ give equivalent exact sequences and how do I construct the morphism between two middle modules in such sequences?
– Lukas Kofler
Nov 21 '18 at 19:01
1
Two $phi$s give you equivalent classes iff they have the same cohomology class - you know what the classes look like, because you calculated the ext group. The middle map in the ses comes from the pushout construction: it's projection onto the second factor composed with the map $mathbb{Z} to mathbb{Z}/9$ from the top row of my diagram.
– Matthew Towers
Nov 21 '18 at 19:10
add a comment |
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1 Answer
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1 Answer
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active
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active
oldest
votes
Do you know what a pushout is? The construction you need is
$$
begin{array}{llllllll}
0 & to & mathbb{Z} & stackrel{times 9}to & mathbb{Z} & to & mathbb{Z}/9 & to & 0 \
& & downarrow phi & & downarrow & & downarrow textrm{id} & & \
0 & to & mathbb{Z} & to & P & to & mathbb{Z}/9 & to & 0
end{array}
$$
where $P$ is the pushout of $times 9$ and $phi$, the homomorphism whose ext-class you are working with.
For abelian groups, the pushout can be identified with the quotient of the direct sum $mathbb{Z}oplus mathbb{Z}$ by the subgroup generated by things of the form $(phi(z), -9z)$: you can see immediately that $phi=0$ gives you the trivial extension you mentioned.
answered Nov 21 '18 at 14:56
Matthew Towers
7,45822344
Thank you! Assuming I have $phi(1)=2$, what is the resulting quotient module? I feel like we would get $Bbb Z_{18}$ but then the sequence would fail to be exact.
– Lukas Kofler
Nov 21 '18 at 17:09
1
It won't be $mathbb{Z}_{18}$, as you say: the pushout is the quotient of $mathbb{Z}oplus mathbb{Z}$ by the subgroup generated by $(2,-9)$, which is isomorphic to $mathbb{Z}$ again (if you know about Smith Normal Form you can use that, or it's not hard by hand). Remember that it's not just the groups in the short exact sequence that matter - the maps matter as well.
– Matthew Towers
Nov 21 '18 at 18:01
In the end, I only want $9$ equivalence classes of exact sequences. Which $phi$ give equivalent exact sequences and how do I construct the morphism between two middle modules in such sequences?
– Lukas Kofler
Nov 21 '18 at 19:01
1
Two $phi$s give you equivalent classes iff they have the same cohomology class - you know what the classes look like, because you calculated the ext group. The middle map in the ses comes from the pushout construction: it's projection onto the second factor composed with the map $mathbb{Z} to mathbb{Z}/9$ from the top row of my diagram.
– Matthew Towers
Nov 21 '18 at 19:10
add a comment |
Do you know what a pushout is? The construction you need is
$$
begin{array}{llllllll}
0 & to & mathbb{Z} & stackrel{times 9}to & mathbb{Z} & to & mathbb{Z}/9 & to & 0 \
& & downarrow phi & & downarrow & & downarrow textrm{id} & & \
0 & to & mathbb{Z} & to & P & to & mathbb{Z}/9 & to & 0
end{array}
$$
where $P$ is the pushout of $times 9$ and $phi$, the homomorphism whose ext-class you are working with.
For abelian groups, the pushout can be identified with the quotient of the direct sum $mathbb{Z}oplus mathbb{Z}$ by the subgroup generated by things of the form $(phi(z), -9z)$: you can see immediately that $phi=0$ gives you the trivial extension you mentioned.
answered Nov 21 '18 at 14:56
Matthew Towers
7,45822344
Thank you! Assuming I have $phi(1)=2$, what is the resulting quotient module? I feel like we would get $Bbb Z_{18}$ but then the sequence would fail to be exact.
– Lukas Kofler
Nov 21 '18 at 17:09
1
It won't be $mathbb{Z}_{18}$, as you say: the pushout is the quotient of $mathbb{Z}oplus mathbb{Z}$ by the subgroup generated by $(2,-9)$, which is isomorphic to $mathbb{Z}$ again (if you know about Smith Normal Form you can use that, or it's not hard by hand). Remember that it's not just the groups in the short exact sequence that matter - the maps matter as well.
– Matthew Towers
Nov 21 '18 at 18:01
In the end, I only want $9$ equivalence classes of exact sequences. Which $phi$ give equivalent exact sequences and how do I construct the morphism between two middle modules in such sequences?
– Lukas Kofler
Nov 21 '18 at 19:01
1
Two $phi$s give you equivalent classes iff they have the same cohomology class - you know what the classes look like, because you calculated the ext group. The middle map in the ses comes from the pushout construction: it's projection onto the second factor composed with the map $mathbb{Z} to mathbb{Z}/9$ from the top row of my diagram.
– Matthew Towers
Nov 21 '18 at 19:10
add a comment |
Do you know what a pushout is? The construction you need is
$$
begin{array}{llllllll}
0 & to & mathbb{Z} & stackrel{times 9}to & mathbb{Z} & to & mathbb{Z}/9 & to & 0 \
& & downarrow phi & & downarrow & & downarrow textrm{id} & & \
0 & to & mathbb{Z} & to & P & to & mathbb{Z}/9 & to & 0
end{array}
$$
where $P$ is the pushout of $times 9$ and $phi$, the homomorphism whose ext-class you are working with.
For abelian groups, the pushout can be identified with the quotient of the direct sum $mathbb{Z}oplus mathbb{Z}$ by the subgroup generated by things of the form $(phi(z), -9z)$: you can see immediately that $phi=0$ gives you the trivial extension you mentioned.
answered Nov 21 '18 at 14:56
Matthew Towers
7,45822344
Do you know what a pushout is? The construction you need is
$$
begin{array}{llllllll}
0 & to & mathbb{Z} & stackrel{times 9}to & mathbb{Z} & to & mathbb{Z}/9 & to & 0 \
& & downarrow phi & & downarrow & & downarrow textrm{id} & & \
0 & to & mathbb{Z} & to & P & to & mathbb{Z}/9 & to & 0
end{array}
$$
where $P$ is the pushout of $times 9$ and $phi$, the homomorphism whose ext-class you are working with.
For abelian groups, the pushout can be identified with the quotient of the direct sum $mathbb{Z}oplus mathbb{Z}$ by the subgroup generated by things of the form $(phi(z), -9z)$: you can see immediately that $phi=0$ gives you the trivial extension you mentioned.
answered Nov 21 '18 at 14:56
Matthew Towers
7,45822344
answered Nov 21 '18 at 14:56
Matthew Towers
7,45822344
answered Nov 21 '18 at 14:56
Matthew Towers
7,45822344
answered Nov 21 '18 at 14:56
Matthew Towers
7,45822344
7,45822344
Thank you! Assuming I have $phi(1)=2$, what is the resulting quotient module? I feel like we would get $Bbb Z_{18}$ but then the sequence would fail to be exact.
– Lukas Kofler
Nov 21 '18 at 17:09
1
It won't be $mathbb{Z}_{18}$, as you say: the pushout is the quotient of $mathbb{Z}oplus mathbb{Z}$ by the subgroup generated by $(2,-9)$, which is isomorphic to $mathbb{Z}$ again (if you know about Smith Normal Form you can use that, or it's not hard by hand). Remember that it's not just the groups in the short exact sequence that matter - the maps matter as well.
– Matthew Towers
Nov 21 '18 at 18:01
In the end, I only want $9$ equivalence classes of exact sequences. Which $phi$ give equivalent exact sequences and how do I construct the morphism between two middle modules in such sequences?
– Lukas Kofler
Nov 21 '18 at 19:01
1
Two $phi$s give you equivalent classes iff they have the same cohomology class - you know what the classes look like, because you calculated the ext group. The middle map in the ses comes from the pushout construction: it's projection onto the second factor composed with the map $mathbb{Z} to mathbb{Z}/9$ from the top row of my diagram.
– Matthew Towers
Nov 21 '18 at 19:10
add a comment |
Thank you! Assuming I have $phi(1)=2$, what is the resulting quotient module? I feel like we would get $Bbb Z_{18}$ but then the sequence would fail to be exact.
– Lukas Kofler
Nov 21 '18 at 17:09
1
It won't be $mathbb{Z}_{18}$, as you say: the pushout is the quotient of $mathbb{Z}oplus mathbb{Z}$ by the subgroup generated by $(2,-9)$, which is isomorphic to $mathbb{Z}$ again (if you know about Smith Normal Form you can use that, or it's not hard by hand). Remember that it's not just the groups in the short exact sequence that matter - the maps matter as well.
– Matthew Towers
Nov 21 '18 at 18:01
In the end, I only want $9$ equivalence classes of exact sequences. Which $phi$ give equivalent exact sequences and how do I construct the morphism between two middle modules in such sequences?
– Lukas Kofler
Nov 21 '18 at 19:01
1
Two $phi$s give you equivalent classes iff they have the same cohomology class - you know what the classes look like, because you calculated the ext group. The middle map in the ses comes from the pushout construction: it's projection onto the second factor composed with the map $mathbb{Z} to mathbb{Z}/9$ from the top row of my diagram.
– Matthew Towers
Nov 21 '18 at 19:10
Thank you! Assuming I have $phi(1)=2$, what is the resulting quotient module? I feel like we would get $Bbb Z_{18}$ but then the sequence would fail to be exact.
– Lukas Kofler
Nov 21 '18 at 17:09
Thank you! Assuming I have $phi(1)=2$, what is the resulting quotient module? I feel like we would get $Bbb Z_{18}$ but then the sequence would fail to be exact.
– Lukas Kofler
Nov 21 '18 at 17:09
1
1
It won't be $mathbb{Z}_{18}$, as you say: the pushout is the quotient of $mathbb{Z}oplus mathbb{Z}$ by the subgroup generated by $(2,-9)$, which is isomorphic to $mathbb{Z}$ again (if you know about Smith Normal Form you can use that, or it's not hard by hand). Remember that it's not just the groups in the short exact sequence that matter - the maps matter as well.
– Matthew Towers
Nov 21 '18 at 18:01
It won't be $mathbb{Z}_{18}$, as you say: the pushout is the quotient of $mathbb{Z}oplus mathbb{Z}$ by the subgroup generated by $(2,-9)$, which is isomorphic to $mathbb{Z}$ again (if you know about Smith Normal Form you can use that, or it's not hard by hand). Remember that it's not just the groups in the short exact sequence that matter - the maps matter as well.
– Matthew Towers
Nov 21 '18 at 18:01
In the end, I only want $9$ equivalence classes of exact sequences. Which $phi$ give equivalent exact sequences and how do I construct the morphism between two middle modules in such sequences?
– Lukas Kofler
Nov 21 '18 at 19:01
In the end, I only want $9$ equivalence classes of exact sequences. Which $phi$ give equivalent exact sequences and how do I construct the morphism between two middle modules in such sequences?
– Lukas Kofler
Nov 21 '18 at 19:01
1
1
Two $phi$s give you equivalent classes iff they have the same cohomology class - you know what the classes look like, because you calculated the ext group. The middle map in the ses comes from the pushout construction: it's projection onto the second factor composed with the map $mathbb{Z} to mathbb{Z}/9$ from the top row of my diagram.
– Matthew Towers
Nov 21 '18 at 19:10
Two $phi$s give you equivalent classes iff they have the same cohomology class - you know what the classes look like, because you calculated the ext group. The middle map in the ses comes from the pushout construction: it's projection onto the second factor composed with the map $mathbb{Z} to mathbb{Z}/9$ from the top row of my diagram.
– Matthew Towers
Nov 21 '18 at 19:10
add a comment |
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