Homomorphisms and Tensor Products












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Suppose $A$ and $B$ are finitely generated $mathbb{Z}$-modules. Then if $ operatorname{Hom}_mathbb{Z}(A,B) neq 0$ and $ operatorname{Hom}_mathbb{Z}(B,A)=0$ then $B otimes mathbb{Q} =0 $ and $Aotimes mathbb{Q} neq 0$. Can u guys help me out. I've tried using the fact that any element in the tensor product is of the form $sum botimes q$ and the properties of the module homomorphisms but I'm kinda stuck , should I use the fact they $A,B cong mathbb{Z}/(d_1)opluscdots oplus mathbb{Z}/(d_k)$ ?






abstract-algebra modules







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    $begingroup$


    Suppose $A$ and $B$ are finitely generated $mathbb{Z}$-modules. Then if $ operatorname{Hom}_mathbb{Z}(A,B) neq 0$ and $ operatorname{Hom}_mathbb{Z}(B,A)=0$ then $B otimes mathbb{Q} =0 $ and $Aotimes mathbb{Q} neq 0$. Can u guys help me out. I've tried using the fact that any element in the tensor product is of the form $sum botimes q$ and the properties of the module homomorphisms but I'm kinda stuck , should I use the fact they $A,B cong mathbb{Z}/(d_1)opluscdots oplus mathbb{Z}/(d_k)$ ?






    abstract-algebra modules







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Suppose $A$ and $B$ are finitely generated $mathbb{Z}$-modules. Then if $ operatorname{Hom}_mathbb{Z}(A,B) neq 0$ and $ operatorname{Hom}_mathbb{Z}(B,A)=0$ then $B otimes mathbb{Q} =0 $ and $Aotimes mathbb{Q} neq 0$. Can u guys help me out. I've tried using the fact that any element in the tensor product is of the form $sum botimes q$ and the properties of the module homomorphisms but I'm kinda stuck , should I use the fact they $A,B cong mathbb{Z}/(d_1)opluscdots oplus mathbb{Z}/(d_k)$ ?






      abstract-algebra modules







      share|cite|improve this question











      $endgroup$




      Suppose $A$ and $B$ are finitely generated $mathbb{Z}$-modules. Then if $ operatorname{Hom}_mathbb{Z}(A,B) neq 0$ and $ operatorname{Hom}_mathbb{Z}(B,A)=0$ then $B otimes mathbb{Q} =0 $ and $Aotimes mathbb{Q} neq 0$. Can u guys help me out. I've tried using the fact that any element in the tensor product is of the form $sum botimes q$ and the properties of the module homomorphisms but I'm kinda stuck , should I use the fact they $A,B cong mathbb{Z}/(d_1)opluscdots oplus mathbb{Z}/(d_k)$ ?






      abstract-algebra modules




      abstract-algebra modules






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Jan 27 at 22:41









      Bernard

      123k741117




      123k741117










      asked Jan 27 at 22:27









      Pedro SantosPedro Santos

      1609




      1609






















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          $begingroup$

          Hint: You may write
          $$ A cong bigoplus_{j=1}^m A_j, qquad B cong bigoplus_{k=1}^n B_k$$
          where $A_j, B_j$ are either isomorphic to $mathbb{Z}^{d_j}$ or $mathbb{Z}/p_j^{r_j}mathbb{Z}$. Using
          $$operatorname{Hom}_{mathbb{Z}} left(bigoplus_{j=1}^m A_j, bigoplus_{k=1}^n B_kright) cong bigoplus_{1leq j leq m, 1leq k leq n} operatorname{Hom}_{mathbb{Z}}(A_j, B_k)$$
          it suffices to check when $operatorname{Hom}_{mathbb{Z}}(A_j, B_k) $ is trivial (or not).



          After that we can use
          $$ A otimes mathbb{Q} cong bigoplus_{j=1}^m (A_j otimes mathbb{Q}), qquad B otimes mathbb{Q} cong bigoplus_{k=1}^n (B_k otimes mathbb{Q}) $$
          Furthermore, we have for $dneq 0$ (just use the fact that $[a]otimes c= [da]otimes (c/d)=0$)
          $$ (mathbb{Z}/d mathbb{Z}) otimes mathbb{Q} cong 0$$
          and (using $z otimes q = 1otimes (zg)$)
          $$ mathbb{Z} otimes mathbb{Q} cong mathbb{Q} qquad text{and thus} qquad mathbb{Z}^d otimes mathbb{Q} cong mathbb{Q}^d.$$






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            active

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            2












            $begingroup$

            Hint: You may write
            $$ A cong bigoplus_{j=1}^m A_j, qquad B cong bigoplus_{k=1}^n B_k$$
            where $A_j, B_j$ are either isomorphic to $mathbb{Z}^{d_j}$ or $mathbb{Z}/p_j^{r_j}mathbb{Z}$. Using
            $$operatorname{Hom}_{mathbb{Z}} left(bigoplus_{j=1}^m A_j, bigoplus_{k=1}^n B_kright) cong bigoplus_{1leq j leq m, 1leq k leq n} operatorname{Hom}_{mathbb{Z}}(A_j, B_k)$$
            it suffices to check when $operatorname{Hom}_{mathbb{Z}}(A_j, B_k) $ is trivial (or not).



            After that we can use
            $$ A otimes mathbb{Q} cong bigoplus_{j=1}^m (A_j otimes mathbb{Q}), qquad B otimes mathbb{Q} cong bigoplus_{k=1}^n (B_k otimes mathbb{Q}) $$
            Furthermore, we have for $dneq 0$ (just use the fact that $[a]otimes c= [da]otimes (c/d)=0$)
            $$ (mathbb{Z}/d mathbb{Z}) otimes mathbb{Q} cong 0$$
            and (using $z otimes q = 1otimes (zg)$)
            $$ mathbb{Z} otimes mathbb{Q} cong mathbb{Q} qquad text{and thus} qquad mathbb{Z}^d otimes mathbb{Q} cong mathbb{Q}^d.$$






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Hint: You may write
              $$ A cong bigoplus_{j=1}^m A_j, qquad B cong bigoplus_{k=1}^n B_k$$
              where $A_j, B_j$ are either isomorphic to $mathbb{Z}^{d_j}$ or $mathbb{Z}/p_j^{r_j}mathbb{Z}$. Using
              $$operatorname{Hom}_{mathbb{Z}} left(bigoplus_{j=1}^m A_j, bigoplus_{k=1}^n B_kright) cong bigoplus_{1leq j leq m, 1leq k leq n} operatorname{Hom}_{mathbb{Z}}(A_j, B_k)$$
              it suffices to check when $operatorname{Hom}_{mathbb{Z}}(A_j, B_k) $ is trivial (or not).



              After that we can use
              $$ A otimes mathbb{Q} cong bigoplus_{j=1}^m (A_j otimes mathbb{Q}), qquad B otimes mathbb{Q} cong bigoplus_{k=1}^n (B_k otimes mathbb{Q}) $$
              Furthermore, we have for $dneq 0$ (just use the fact that $[a]otimes c= [da]otimes (c/d)=0$)
              $$ (mathbb{Z}/d mathbb{Z}) otimes mathbb{Q} cong 0$$
              and (using $z otimes q = 1otimes (zg)$)
              $$ mathbb{Z} otimes mathbb{Q} cong mathbb{Q} qquad text{and thus} qquad mathbb{Z}^d otimes mathbb{Q} cong mathbb{Q}^d.$$






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Hint: You may write
                $$ A cong bigoplus_{j=1}^m A_j, qquad B cong bigoplus_{k=1}^n B_k$$
                where $A_j, B_j$ are either isomorphic to $mathbb{Z}^{d_j}$ or $mathbb{Z}/p_j^{r_j}mathbb{Z}$. Using
                $$operatorname{Hom}_{mathbb{Z}} left(bigoplus_{j=1}^m A_j, bigoplus_{k=1}^n B_kright) cong bigoplus_{1leq j leq m, 1leq k leq n} operatorname{Hom}_{mathbb{Z}}(A_j, B_k)$$
                it suffices to check when $operatorname{Hom}_{mathbb{Z}}(A_j, B_k) $ is trivial (or not).



                After that we can use
                $$ A otimes mathbb{Q} cong bigoplus_{j=1}^m (A_j otimes mathbb{Q}), qquad B otimes mathbb{Q} cong bigoplus_{k=1}^n (B_k otimes mathbb{Q}) $$
                Furthermore, we have for $dneq 0$ (just use the fact that $[a]otimes c= [da]otimes (c/d)=0$)
                $$ (mathbb{Z}/d mathbb{Z}) otimes mathbb{Q} cong 0$$
                and (using $z otimes q = 1otimes (zg)$)
                $$ mathbb{Z} otimes mathbb{Q} cong mathbb{Q} qquad text{and thus} qquad mathbb{Z}^d otimes mathbb{Q} cong mathbb{Q}^d.$$






                share|cite|improve this answer











                $endgroup$



                Hint: You may write
                $$ A cong bigoplus_{j=1}^m A_j, qquad B cong bigoplus_{k=1}^n B_k$$
                where $A_j, B_j$ are either isomorphic to $mathbb{Z}^{d_j}$ or $mathbb{Z}/p_j^{r_j}mathbb{Z}$. Using
                $$operatorname{Hom}_{mathbb{Z}} left(bigoplus_{j=1}^m A_j, bigoplus_{k=1}^n B_kright) cong bigoplus_{1leq j leq m, 1leq k leq n} operatorname{Hom}_{mathbb{Z}}(A_j, B_k)$$
                it suffices to check when $operatorname{Hom}_{mathbb{Z}}(A_j, B_k) $ is trivial (or not).



                After that we can use
                $$ A otimes mathbb{Q} cong bigoplus_{j=1}^m (A_j otimes mathbb{Q}), qquad B otimes mathbb{Q} cong bigoplus_{k=1}^n (B_k otimes mathbb{Q}) $$
                Furthermore, we have for $dneq 0$ (just use the fact that $[a]otimes c= [da]otimes (c/d)=0$)
                $$ (mathbb{Z}/d mathbb{Z}) otimes mathbb{Q} cong 0$$
                and (using $z otimes q = 1otimes (zg)$)
                $$ mathbb{Z} otimes mathbb{Q} cong mathbb{Q} qquad text{and thus} qquad mathbb{Z}^d otimes mathbb{Q} cong mathbb{Q}^d.$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 29 at 13:49









                Fabio Lucchini

                9,51111426




                9,51111426










                answered Jan 27 at 23:04









                Severin SchravenSeverin Schraven

                6,5751935




                6,5751935






























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