What is the direct limit of $mathbb Z$ with $f_{ij}(x)=x cdot j$?












2












$begingroup$


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







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 3:28







      Al Jebr

















      asked Jan 4 at 2:46









      Al JebrAl Jebr

      4,39743478




      4,39743478






















          1 Answer
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          3












          $begingroup$

          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 $n!$).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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












          Your Answer





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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          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 $n!$).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          3












          $begingroup$

          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 $n!$).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          3












          3








          3





          $begingroup$

          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 $n!$).






          share|cite|improve this answer









          $endgroup$



          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 $n!$).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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


















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