What is $Ext^1(overline{mathbb{Q}}^times, overline{mathbb{Q}})$ in abelian groups?












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$begingroup$


I want to find a way to describe all the extensions of $overline{mathbb{Q}}^times$ by $overline{mathbb{Q}}$, i.e., all the abelian groups $A$ (and the maps $alpha$ and $beta$) that fit into the exact sequence of abelian groups



$$0 to overline{mathbb{Q}} overset{alpha}{to} A overset{beta}{to} overline{mathbb{Q}}^times to 0.$$



Of course $A:=overline{mathbb{Q}} times overline{mathbb{Q}}^times$ with the obvious maps works, but that's always the case. I was thinking about using the exponential map, but the problem is that the exponential of an algebraic number is in general not algebraic...



I could also compute $Ext^1(overline{mathbb{Q}}^times, overline{mathbb{Q}})$ using resolutions, but I am not sure whether this is easier...






group-theory algebraic-number-theory homological-algebra abelian-groups exact-sequence







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    0












    $begingroup$


    I want to find a way to describe all the extensions of $overline{mathbb{Q}}^times$ by $overline{mathbb{Q}}$, i.e., all the abelian groups $A$ (and the maps $alpha$ and $beta$) that fit into the exact sequence of abelian groups



    $$0 to overline{mathbb{Q}} overset{alpha}{to} A overset{beta}{to} overline{mathbb{Q}}^times to 0.$$



    Of course $A:=overline{mathbb{Q}} times overline{mathbb{Q}}^times$ with the obvious maps works, but that's always the case. I was thinking about using the exponential map, but the problem is that the exponential of an algebraic number is in general not algebraic...



    I could also compute $Ext^1(overline{mathbb{Q}}^times, overline{mathbb{Q}})$ using resolutions, but I am not sure whether this is easier...






    group-theory algebraic-number-theory homological-algebra abelian-groups exact-sequence







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I want to find a way to describe all the extensions of $overline{mathbb{Q}}^times$ by $overline{mathbb{Q}}$, i.e., all the abelian groups $A$ (and the maps $alpha$ and $beta$) that fit into the exact sequence of abelian groups



      $$0 to overline{mathbb{Q}} overset{alpha}{to} A overset{beta}{to} overline{mathbb{Q}}^times to 0.$$



      Of course $A:=overline{mathbb{Q}} times overline{mathbb{Q}}^times$ with the obvious maps works, but that's always the case. I was thinking about using the exponential map, but the problem is that the exponential of an algebraic number is in general not algebraic...



      I could also compute $Ext^1(overline{mathbb{Q}}^times, overline{mathbb{Q}})$ using resolutions, but I am not sure whether this is easier...






      group-theory algebraic-number-theory homological-algebra abelian-groups exact-sequence







      share|cite|improve this question









      $endgroup$




      I want to find a way to describe all the extensions of $overline{mathbb{Q}}^times$ by $overline{mathbb{Q}}$, i.e., all the abelian groups $A$ (and the maps $alpha$ and $beta$) that fit into the exact sequence of abelian groups



      $$0 to overline{mathbb{Q}} overset{alpha}{to} A overset{beta}{to} overline{mathbb{Q}}^times to 0.$$



      Of course $A:=overline{mathbb{Q}} times overline{mathbb{Q}}^times$ with the obvious maps works, but that's always the case. I was thinking about using the exponential map, but the problem is that the exponential of an algebraic number is in general not algebraic...



      I could also compute $Ext^1(overline{mathbb{Q}}^times, overline{mathbb{Q}})$ using resolutions, but I am not sure whether this is easier...






      group-theory algebraic-number-theory homological-algebra abelian-groups exact-sequence




      group-theory algebraic-number-theory homological-algebra abelian-groups exact-sequence






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      asked Jan 21 at 10:18









      57Jimmy57Jimmy

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          $begingroup$

          Both groups are divisible, hence injective. Since $bar{mathbb Q}$ is injective $Ext^1(-,bar{mathbb Q}) = 0$, and there are no other extensions. (There are actually no extensions the other way either.)






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            1 Answer
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            active

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            6












            $begingroup$

            Both groups are divisible, hence injective. Since $bar{mathbb Q}$ is injective $Ext^1(-,bar{mathbb Q}) = 0$, and there are no other extensions. (There are actually no extensions the other way either.)






            share|cite|improve this answer









            $endgroup$


















              6












              $begingroup$

              Both groups are divisible, hence injective. Since $bar{mathbb Q}$ is injective $Ext^1(-,bar{mathbb Q}) = 0$, and there are no other extensions. (There are actually no extensions the other way either.)






              share|cite|improve this answer









              $endgroup$
















                6












                6








                6





                $begingroup$

                Both groups are divisible, hence injective. Since $bar{mathbb Q}$ is injective $Ext^1(-,bar{mathbb Q}) = 0$, and there are no other extensions. (There are actually no extensions the other way either.)






                share|cite|improve this answer









                $endgroup$



                Both groups are divisible, hence injective. Since $bar{mathbb Q}$ is injective $Ext^1(-,bar{mathbb Q}) = 0$, and there are no other extensions. (There are actually no extensions the other way either.)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 21 at 10:40









                BenBen

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