Additive group of local rings












3












$begingroup$


Is there a theory or characterization for those finite $p$-groups that can be considered as the additive group of a finite local commutative ring with identity?






ac.commutative-algebra ra.rings-and-algebras abelian-groups local-rings







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$endgroup$

















    3












    $begingroup$


    Is there a theory or characterization for those finite $p$-groups that can be considered as the additive group of a finite local commutative ring with identity?






    ac.commutative-algebra ra.rings-and-algebras abelian-groups local-rings







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Is there a theory or characterization for those finite $p$-groups that can be considered as the additive group of a finite local commutative ring with identity?






      ac.commutative-algebra ra.rings-and-algebras abelian-groups local-rings







      share|cite|improve this question









      $endgroup$




      Is there a theory or characterization for those finite $p$-groups that can be considered as the additive group of a finite local commutative ring with identity?






      ac.commutative-algebra ra.rings-and-algebras abelian-groups local-rings




      ac.commutative-algebra ra.rings-and-algebras abelian-groups local-rings






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      asked 3 hours ago









      Lisa_KLisa_K

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          $begingroup$

          Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring.



          Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ mathbb{Z}/p^k mathbb{Z} times M $ where $ p^k = exp (G) $. Then $ M $ naturally has the structure of a $ mathbb{Z}/p^k mathbb{Z} $ - module.



          Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.



          Now let's prove it's local with maximal ideal $ { (x,m) mid p $ divides $x } $.



          Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.



          This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Precisely every nonzero such group.
            $endgroup$
            – YCor
            2 hours ago










          • $begingroup$
            @YCor : indeed, let me correct that
            $endgroup$
            – Max
            2 hours ago










          • $begingroup$
            What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
            $endgroup$
            – LSpice
            2 hours ago










          • $begingroup$
            @LSpice : you can see it as $G/(mathbb{Z}/p^kmathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups
            $endgroup$
            – Max
            2 hours ago












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          $begingroup$

          Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring.



          Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ mathbb{Z}/p^k mathbb{Z} times M $ where $ p^k = exp (G) $. Then $ M $ naturally has the structure of a $ mathbb{Z}/p^k mathbb{Z} $ - module.



          Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.



          Now let's prove it's local with maximal ideal $ { (x,m) mid p $ divides $x } $.



          Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.



          This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Precisely every nonzero such group.
            $endgroup$
            – YCor
            2 hours ago










          • $begingroup$
            @YCor : indeed, let me correct that
            $endgroup$
            – Max
            2 hours ago










          • $begingroup$
            What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
            $endgroup$
            – LSpice
            2 hours ago










          • $begingroup$
            @LSpice : you can see it as $G/(mathbb{Z}/p^kmathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups
            $endgroup$
            – Max
            2 hours ago
















          4












          $begingroup$

          Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring.



          Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ mathbb{Z}/p^k mathbb{Z} times M $ where $ p^k = exp (G) $. Then $ M $ naturally has the structure of a $ mathbb{Z}/p^k mathbb{Z} $ - module.



          Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.



          Now let's prove it's local with maximal ideal $ { (x,m) mid p $ divides $x } $.



          Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.



          This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Precisely every nonzero such group.
            $endgroup$
            – YCor
            2 hours ago










          • $begingroup$
            @YCor : indeed, let me correct that
            $endgroup$
            – Max
            2 hours ago










          • $begingroup$
            What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
            $endgroup$
            – LSpice
            2 hours ago










          • $begingroup$
            @LSpice : you can see it as $G/(mathbb{Z}/p^kmathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups
            $endgroup$
            – Max
            2 hours ago














          4












          4








          4





          $begingroup$

          Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring.



          Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ mathbb{Z}/p^k mathbb{Z} times M $ where $ p^k = exp (G) $. Then $ M $ naturally has the structure of a $ mathbb{Z}/p^k mathbb{Z} $ - module.



          Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.



          Now let's prove it's local with maximal ideal $ { (x,m) mid p $ divides $x } $.



          Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.



          This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.






          share|cite|improve this answer











          $endgroup$



          Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring.



          Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ mathbb{Z}/p^k mathbb{Z} times M $ where $ p^k = exp (G) $. Then $ M $ naturally has the structure of a $ mathbb{Z}/p^k mathbb{Z} $ - module.



          Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.



          Now let's prove it's local with maximal ideal $ { (x,m) mid p $ divides $x } $.



          Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.



          This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 hours ago

























          answered 2 hours ago









          MaxMax

          6091619




          6091619








          • 1




            $begingroup$
            Precisely every nonzero such group.
            $endgroup$
            – YCor
            2 hours ago










          • $begingroup$
            @YCor : indeed, let me correct that
            $endgroup$
            – Max
            2 hours ago










          • $begingroup$
            What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
            $endgroup$
            – LSpice
            2 hours ago










          • $begingroup$
            @LSpice : you can see it as $G/(mathbb{Z}/p^kmathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups
            $endgroup$
            – Max
            2 hours ago














          • 1




            $begingroup$
            Precisely every nonzero such group.
            $endgroup$
            – YCor
            2 hours ago










          • $begingroup$
            @YCor : indeed, let me correct that
            $endgroup$
            – Max
            2 hours ago










          • $begingroup$
            What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
            $endgroup$
            – LSpice
            2 hours ago










          • $begingroup$
            @LSpice : you can see it as $G/(mathbb{Z}/p^kmathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups
            $endgroup$
            – Max
            2 hours ago








          1




          1




          $begingroup$
          Precisely every nonzero such group.
          $endgroup$
          – YCor
          2 hours ago




          $begingroup$
          Precisely every nonzero such group.
          $endgroup$
          – YCor
          2 hours ago












          $begingroup$
          @YCor : indeed, let me correct that
          $endgroup$
          – Max
          2 hours ago




          $begingroup$
          @YCor : indeed, let me correct that
          $endgroup$
          – Max
          2 hours ago












          $begingroup$
          What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
          $endgroup$
          – LSpice
          2 hours ago




          $begingroup$
          What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
          $endgroup$
          – LSpice
          2 hours ago












          $begingroup$
          @LSpice : you can see it as $G/(mathbb{Z}/p^kmathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups
          $endgroup$
          – Max
          2 hours ago




          $begingroup$
          @LSpice : you can see it as $G/(mathbb{Z}/p^kmathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups
          $endgroup$
          – Max
          2 hours ago


















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