What is the direct limit of $mathbb Z$ with $f_{ij}(x)=x cdot j$?
$begingroup$
I'm just learning the concept of direct limit. I did problem $8$a fromt Dummit Foote section $7.6$.
As an example, someone suggested I look
$$mathbb{Z} xrightarrow{times 2} mathbb{Z} xrightarrow{times 3} mathbb{Z}xrightarrow{times 4} mathbb{Z}xrightarrow{times 5} mathbb{Z}xrightarrow{times 6} cdots$$
and I should get the rationals.
I've tried playing around a bit, seeing which elements are equivalent in the quotient of the disjoint union modulo the relation, but not sure how to get the rationals.
Any help on how to get the rationals out of this direct limit?
abstract-algebra limits-colimits
asked Jan 4 at 2:46
Al JebrAl Jebr
4,39743478
$endgroup$
add a comment |
$begingroup$
I'm just learning the concept of direct limit. I did problem $8$a fromt Dummit Foote section $7.6$.
As an example, someone suggested I look
$$mathbb{Z} xrightarrow{times 2} mathbb{Z} xrightarrow{times 3} mathbb{Z}xrightarrow{times 4} mathbb{Z}xrightarrow{times 5} mathbb{Z}xrightarrow{times 6} cdots$$
and I should get the rationals.
I've tried playing around a bit, seeing which elements are equivalent in the quotient of the disjoint union modulo the relation, but not sure how to get the rationals.
Any help on how to get the rationals out of this direct limit?
abstract-algebra limits-colimits
asked Jan 4 at 2:46
Al JebrAl Jebr
4,39743478
$endgroup$
add a comment |
$begingroup$
I'm just learning the concept of direct limit. I did problem $8$a fromt Dummit Foote section $7.6$.
As an example, someone suggested I look
$$mathbb{Z} xrightarrow{times 2} mathbb{Z} xrightarrow{times 3} mathbb{Z}xrightarrow{times 4} mathbb{Z}xrightarrow{times 5} mathbb{Z}xrightarrow{times 6} cdots$$
and I should get the rationals.
I've tried playing around a bit, seeing which elements are equivalent in the quotient of the disjoint union modulo the relation, but not sure how to get the rationals.
Any help on how to get the rationals out of this direct limit?
abstract-algebra limits-colimits
asked Jan 4 at 2:46
Al JebrAl Jebr
4,39743478
$endgroup$
I'm just learning the concept of direct limit. I did problem $8$a fromt Dummit Foote section $7.6$.
As an example, someone suggested I look
$$mathbb{Z} xrightarrow{times 2} mathbb{Z} xrightarrow{times 3} mathbb{Z}xrightarrow{times 4} mathbb{Z}xrightarrow{times 5} mathbb{Z}xrightarrow{times 6} cdots$$
and I should get the rationals.
I've tried playing around a bit, seeing which elements are equivalent in the quotient of the disjoint union modulo the relation, but not sure how to get the rationals.
Any help on how to get the rationals out of this direct limit?
abstract-algebra limits-colimits
abstract-algebra limits-colimits
asked Jan 4 at 2:46
Al JebrAl Jebr
4,39743478
asked Jan 4 at 2:46
Al JebrAl Jebr
4,39743478
edited Jan 4 at 3:28
Al Jebr
asked Jan 4 at 2:46
Al JebrAl Jebr
4,39743478
asked Jan 4 at 2:46
Al JebrAl Jebr
4,39743478
asked Jan 4 at 2:46
Al JebrAl Jebr
4,39743478
4,39743478
add a comment |
add a comment |
1 Answer
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$begingroup$
Here is a directed system of abelian groups which is isomorphic to yours:
$$
mathbb{Z} to frac{1}{2}mathbb{Z} to frac{1}{6}mathbb{Z} to ldots to frac{1}{n!} mathbb{Z} to ldots
$$
(here we mean the cyclic subgroup of $mathbb{Q}$ generated by the element $1/n!$). Here, the arrows are just inclusion. To prove this is isomorphic to your sequence, write your sequence under it and the necessary vertical maps to make the diagram commute (which will be multiplication by $n!$).
Now this directed system has $mathbb{Q}$ as a limit. (The inclusion into $mathbb{Q}$ commutes with all the maps, and is the initial thing commuting with all the maps since any element of $mathbb{Q}$ is eventually in $frac{1}{n!}mathbb{Z}$ for some $n!$).
answered Jan 4 at 2:52
hunterhunter
15.5k32640
$endgroup$
$begingroup$
Could you please elaborate on the last paragraph? Not sure what you mean with the sentence in the parenthesis.
$endgroup$
– Al Jebr
Jan 4 at 18:47
$begingroup$
Also, if we have a disjoint union of abelian groups, then I know this is the direct sum of them. So, how do we go from direct sum of the the abelian groups you have listed to modding out by the equivalence relation to get $mathbb Q$?
$endgroup$
– Al Jebr
Jan 4 at 18:48
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
Here is a directed system of abelian groups which is isomorphic to yours:
$$
mathbb{Z} to frac{1}{2}mathbb{Z} to frac{1}{6}mathbb{Z} to ldots to frac{1}{n!} mathbb{Z} to ldots
$$
(here we mean the cyclic subgroup of $mathbb{Q}$ generated by the element $1/n!$). Here, the arrows are just inclusion. To prove this is isomorphic to your sequence, write your sequence under it and the necessary vertical maps to make the diagram commute (which will be multiplication by $n!$).
Now this directed system has $mathbb{Q}$ as a limit. (The inclusion into $mathbb{Q}$ commutes with all the maps, and is the initial thing commuting with all the maps since any element of $mathbb{Q}$ is eventually in $frac{1}{n!}mathbb{Z}$ for some $n!$).
answered Jan 4 at 2:52
hunterhunter
15.5k32640
$endgroup$
$begingroup$
Could you please elaborate on the last paragraph? Not sure what you mean with the sentence in the parenthesis.
$endgroup$
– Al Jebr
Jan 4 at 18:47
$begingroup$
Also, if we have a disjoint union of abelian groups, then I know this is the direct sum of them. So, how do we go from direct sum of the the abelian groups you have listed to modding out by the equivalence relation to get $mathbb Q$?
$endgroup$
– Al Jebr
Jan 4 at 18:48
add a comment |
$begingroup$
Here is a directed system of abelian groups which is isomorphic to yours:
$$
mathbb{Z} to frac{1}{2}mathbb{Z} to frac{1}{6}mathbb{Z} to ldots to frac{1}{n!} mathbb{Z} to ldots
$$
(here we mean the cyclic subgroup of $mathbb{Q}$ generated by the element $1/n!$). Here, the arrows are just inclusion. To prove this is isomorphic to your sequence, write your sequence under it and the necessary vertical maps to make the diagram commute (which will be multiplication by $n!$).
Now this directed system has $mathbb{Q}$ as a limit. (The inclusion into $mathbb{Q}$ commutes with all the maps, and is the initial thing commuting with all the maps since any element of $mathbb{Q}$ is eventually in $frac{1}{n!}mathbb{Z}$ for some $n!$).
answered Jan 4 at 2:52
hunterhunter
15.5k32640
$endgroup$
$begingroup$
Could you please elaborate on the last paragraph? Not sure what you mean with the sentence in the parenthesis.
$endgroup$
– Al Jebr
Jan 4 at 18:47
$begingroup$
Also, if we have a disjoint union of abelian groups, then I know this is the direct sum of them. So, how do we go from direct sum of the the abelian groups you have listed to modding out by the equivalence relation to get $mathbb Q$?
$endgroup$
– Al Jebr
Jan 4 at 18:48
add a comment |
$begingroup$
Here is a directed system of abelian groups which is isomorphic to yours:
$$
mathbb{Z} to frac{1}{2}mathbb{Z} to frac{1}{6}mathbb{Z} to ldots to frac{1}{n!} mathbb{Z} to ldots
$$
(here we mean the cyclic subgroup of $mathbb{Q}$ generated by the element $1/n!$). Here, the arrows are just inclusion. To prove this is isomorphic to your sequence, write your sequence under it and the necessary vertical maps to make the diagram commute (which will be multiplication by $n!$).
Now this directed system has $mathbb{Q}$ as a limit. (The inclusion into $mathbb{Q}$ commutes with all the maps, and is the initial thing commuting with all the maps since any element of $mathbb{Q}$ is eventually in $frac{1}{n!}mathbb{Z}$ for some $n!$).
answered Jan 4 at 2:52
hunterhunter
15.5k32640
$endgroup$
Here is a directed system of abelian groups which is isomorphic to yours:
$$
mathbb{Z} to frac{1}{2}mathbb{Z} to frac{1}{6}mathbb{Z} to ldots to frac{1}{n!} mathbb{Z} to ldots
$$
(here we mean the cyclic subgroup of $mathbb{Q}$ generated by the element $1/n!$). Here, the arrows are just inclusion. To prove this is isomorphic to your sequence, write your sequence under it and the necessary vertical maps to make the diagram commute (which will be multiplication by $n!$).
Now this directed system has $mathbb{Q}$ as a limit. (The inclusion into $mathbb{Q}$ commutes with all the maps, and is the initial thing commuting with all the maps since any element of $mathbb{Q}$ is eventually in $frac{1}{n!}mathbb{Z}$ for some $n!$).
answered Jan 4 at 2:52
hunterhunter
15.5k32640
answered Jan 4 at 2:52
hunterhunter
15.5k32640
answered Jan 4 at 2:52
hunterhunter
15.5k32640
answered Jan 4 at 2:52
hunterhunter
15.5k32640
15.5k32640
$begingroup$
Could you please elaborate on the last paragraph? Not sure what you mean with the sentence in the parenthesis.
$endgroup$
– Al Jebr
Jan 4 at 18:47
$begingroup$
Also, if we have a disjoint union of abelian groups, then I know this is the direct sum of them. So, how do we go from direct sum of the the abelian groups you have listed to modding out by the equivalence relation to get $mathbb Q$?
$endgroup$
– Al Jebr
Jan 4 at 18:48
add a comment |
$begingroup$
Could you please elaborate on the last paragraph? Not sure what you mean with the sentence in the parenthesis.
$endgroup$
– Al Jebr
Jan 4 at 18:47
$begingroup$
Also, if we have a disjoint union of abelian groups, then I know this is the direct sum of them. So, how do we go from direct sum of the the abelian groups you have listed to modding out by the equivalence relation to get $mathbb Q$?
$endgroup$
– Al Jebr
Jan 4 at 18:48
$begingroup$
Could you please elaborate on the last paragraph? Not sure what you mean with the sentence in the parenthesis.
$endgroup$
– Al Jebr
Jan 4 at 18:47
$begingroup$
Could you please elaborate on the last paragraph? Not sure what you mean with the sentence in the parenthesis.
$endgroup$
– Al Jebr
Jan 4 at 18:47
$begingroup$
Also, if we have a disjoint union of abelian groups, then I know this is the direct sum of them. So, how do we go from direct sum of the the abelian groups you have listed to modding out by the equivalence relation to get $mathbb Q$?
$endgroup$
– Al Jebr
Jan 4 at 18:48
$begingroup$
Also, if we have a disjoint union of abelian groups, then I know this is the direct sum of them. So, how do we go from direct sum of the the abelian groups you have listed to modding out by the equivalence relation to get $mathbb Q$?
$endgroup$
– Al Jebr
Jan 4 at 18:48
add a comment |
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