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Are two submodules (where one is contained in the other) isomorphic if their quotientmodules are isomorphic?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Error in proof that submodules of f.g. modules are f.g.Pure Submodules and Finitely Presented versus Finitely Generated SubmodulesProving that two quotient modules are isomorphicLet M be an Artinian module over a commutative Artinian unital ring R. Is M necessarily finitely generated?Slick proof of isomorphism of quotients of direct sums of modulesNumber of submodules of a semisimple moduleProduct/coproduct properties: If $N_1simeq N_2$ in some category, then $N_1times N_3simeq N_2times N_3$?A module of a Lie algebra can be decomposed into irreducibles if and only if every submodule has a complement.Confusion over finite generation and noetherian modulesOn showing that $varphi(N)=(N+K)/K$, for modules $N,K<M$, $varphi$ natural map.










4












$begingroup$


Let $M$ be an $R$ module and $N_1 subset N_2$ be submodules of $M$ such that $M / N_1 cong M / N_2$. Can I know conclude $N_1 cong N_2$ or even $N_1 = N_2$? I know that a proper submodule can be isomorphic to the original module (I know some examples when considering vectorspaces) which is why I am almost certain that $N_1 neq N_2$ in the general case, but $N_1 cong N_2$ seems intuitive enough.

I would appreciate any help, since I am relatively new to modules.






vector-spaces modules vector-space-isomorphism module-isomorphism







share|cite|improve this question









$endgroup$
















    4












    $begingroup$


    Let $M$ be an $R$ module and $N_1 subset N_2$ be submodules of $M$ such that $M / N_1 cong M / N_2$. Can I know conclude $N_1 cong N_2$ or even $N_1 = N_2$? I know that a proper submodule can be isomorphic to the original module (I know some examples when considering vectorspaces) which is why I am almost certain that $N_1 neq N_2$ in the general case, but $N_1 cong N_2$ seems intuitive enough.

    I would appreciate any help, since I am relatively new to modules.






    vector-spaces modules vector-space-isomorphism module-isomorphism







    share|cite|improve this question









    $endgroup$














      4












      4








      4





      $begingroup$


      Let $M$ be an $R$ module and $N_1 subset N_2$ be submodules of $M$ such that $M / N_1 cong M / N_2$. Can I know conclude $N_1 cong N_2$ or even $N_1 = N_2$? I know that a proper submodule can be isomorphic to the original module (I know some examples when considering vectorspaces) which is why I am almost certain that $N_1 neq N_2$ in the general case, but $N_1 cong N_2$ seems intuitive enough.

      I would appreciate any help, since I am relatively new to modules.






      vector-spaces modules vector-space-isomorphism module-isomorphism







      share|cite|improve this question









      $endgroup$




      Let $M$ be an $R$ module and $N_1 subset N_2$ be submodules of $M$ such that $M / N_1 cong M / N_2$. Can I know conclude $N_1 cong N_2$ or even $N_1 = N_2$? I know that a proper submodule can be isomorphic to the original module (I know some examples when considering vectorspaces) which is why I am almost certain that $N_1 neq N_2$ in the general case, but $N_1 cong N_2$ seems intuitive enough.

      I would appreciate any help, since I am relatively new to modules.






      vector-spaces modules vector-space-isomorphism module-isomorphism




      vector-spaces modules vector-space-isomorphism module-isomorphism






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 13 hours ago









      user9620780user9620780

      1096




      1096




















          2 Answers
          2






          active

          oldest

          votes


















          8












          $begingroup$

          Nice question. No, this is not true in general, even if $R$ is a field.



          Say the ring is a field, and let $V$ be a vector space over it with uncountable dimension. We can pick two subspaces $W_f$ and $W_infty$ of finite, respectively countable, dimension. Then the quotients will both have the same dimension as $V$ and thus be isomorphic to each other (we are using the axiom of choice here), but the subspaces are not isomorphic to each other.






          share|cite|improve this answer









          $endgroup$








          • 2




            $begingroup$
            You don't need the axiom of choice to make a concrete example out of this idea. Just declare that $V$ is, for example, the space of functions $mathbb Rto F$ with finite support, and let $W_f$ be the trivial subspace and $W_infty$ be the subspace of functions that vanish on $(-infty,0]$. Then an isomorphism between the quotients can easily be written down expicitly.
            $endgroup$
            – Henning Makholm
            8 hours ago



















          4












          $begingroup$

          Let $R$ be a nontrivial (necessarily
          noncommutative) ring with the property that $Rcong Roplus R$ as a left $R$-module. (This is the case if $R$ is the endomorphism
          ring of an infinite-dimensional vector space.)
          Then $R$ has a proper submodule $N$ with $Ncong R$ and $R/N
          cong R$           as $R$-modules. Take $M=R$, $N_1=0$ and $N_2=N$.
          Clearly $N_1notcong N_2$.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            active

            oldest

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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            8












            $begingroup$

            Nice question. No, this is not true in general, even if $R$ is a field.



            Say the ring is a field, and let $V$ be a vector space over it with uncountable dimension. We can pick two subspaces $W_f$ and $W_infty$ of finite, respectively countable, dimension. Then the quotients will both have the same dimension as $V$ and thus be isomorphic to each other (we are using the axiom of choice here), but the subspaces are not isomorphic to each other.






            share|cite|improve this answer









            $endgroup$








            • 2




              $begingroup$
              You don't need the axiom of choice to make a concrete example out of this idea. Just declare that $V$ is, for example, the space of functions $mathbb Rto F$ with finite support, and let $W_f$ be the trivial subspace and $W_infty$ be the subspace of functions that vanish on $(-infty,0]$. Then an isomorphism between the quotients can easily be written down expicitly.
              $endgroup$
              – Henning Makholm
              8 hours ago
















            8












            $begingroup$

            Nice question. No, this is not true in general, even if $R$ is a field.



            Say the ring is a field, and let $V$ be a vector space over it with uncountable dimension. We can pick two subspaces $W_f$ and $W_infty$ of finite, respectively countable, dimension. Then the quotients will both have the same dimension as $V$ and thus be isomorphic to each other (we are using the axiom of choice here), but the subspaces are not isomorphic to each other.






            share|cite|improve this answer









            $endgroup$








            • 2




              $begingroup$
              You don't need the axiom of choice to make a concrete example out of this idea. Just declare that $V$ is, for example, the space of functions $mathbb Rto F$ with finite support, and let $W_f$ be the trivial subspace and $W_infty$ be the subspace of functions that vanish on $(-infty,0]$. Then an isomorphism between the quotients can easily be written down expicitly.
              $endgroup$
              – Henning Makholm
              8 hours ago














            8












            8








            8





            $begingroup$

            Nice question. No, this is not true in general, even if $R$ is a field.



            Say the ring is a field, and let $V$ be a vector space over it with uncountable dimension. We can pick two subspaces $W_f$ and $W_infty$ of finite, respectively countable, dimension. Then the quotients will both have the same dimension as $V$ and thus be isomorphic to each other (we are using the axiom of choice here), but the subspaces are not isomorphic to each other.






            share|cite|improve this answer









            $endgroup$



            Nice question. No, this is not true in general, even if $R$ is a field.



            Say the ring is a field, and let $V$ be a vector space over it with uncountable dimension. We can pick two subspaces $W_f$ and $W_infty$ of finite, respectively countable, dimension. Then the quotients will both have the same dimension as $V$ and thus be isomorphic to each other (we are using the axiom of choice here), but the subspaces are not isomorphic to each other.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 13 hours ago









            hunterhunter

            15.9k32643




            15.9k32643







            • 2




              $begingroup$
              You don't need the axiom of choice to make a concrete example out of this idea. Just declare that $V$ is, for example, the space of functions $mathbb Rto F$ with finite support, and let $W_f$ be the trivial subspace and $W_infty$ be the subspace of functions that vanish on $(-infty,0]$. Then an isomorphism between the quotients can easily be written down expicitly.
              $endgroup$
              – Henning Makholm
              8 hours ago













            • 2




              $begingroup$
              You don't need the axiom of choice to make a concrete example out of this idea. Just declare that $V$ is, for example, the space of functions $mathbb Rto F$ with finite support, and let $W_f$ be the trivial subspace and $W_infty$ be the subspace of functions that vanish on $(-infty,0]$. Then an isomorphism between the quotients can easily be written down expicitly.
              $endgroup$
              – Henning Makholm
              8 hours ago








            2




            2




            $begingroup$
            You don't need the axiom of choice to make a concrete example out of this idea. Just declare that $V$ is, for example, the space of functions $mathbb Rto F$ with finite support, and let $W_f$ be the trivial subspace and $W_infty$ be the subspace of functions that vanish on $(-infty,0]$. Then an isomorphism between the quotients can easily be written down expicitly.
            $endgroup$
            – Henning Makholm
            8 hours ago





            $begingroup$
            You don't need the axiom of choice to make a concrete example out of this idea. Just declare that $V$ is, for example, the space of functions $mathbb Rto F$ with finite support, and let $W_f$ be the trivial subspace and $W_infty$ be the subspace of functions that vanish on $(-infty,0]$. Then an isomorphism between the quotients can easily be written down expicitly.
            $endgroup$
            – Henning Makholm
            8 hours ago












            4












            $begingroup$

            Let $R$ be a nontrivial (necessarily
            noncommutative) ring with the property that $Rcong Roplus R$ as a left $R$-module. (This is the case if $R$ is the endomorphism
            ring of an infinite-dimensional vector space.)
            Then $R$ has a proper submodule $N$ with $Ncong R$ and $R/N
            cong R$             as $R$-modules. Take $M=R$, $N_1=0$ and $N_2=N$.
            Clearly $N_1notcong N_2$.






            share|cite|improve this answer









            $endgroup$

















              4












              $begingroup$

              Let $R$ be a nontrivial (necessarily
              noncommutative) ring with the property that $Rcong Roplus R$ as a left $R$-module. (This is the case if $R$ is the endomorphism
              ring of an infinite-dimensional vector space.)
              Then $R$ has a proper submodule $N$ with $Ncong R$ and $R/N
              cong R$               as $R$-modules. Take $M=R$, $N_1=0$ and $N_2=N$.
              Clearly $N_1notcong N_2$.






              share|cite|improve this answer









              $endgroup$















                4












                4








                4





                $begingroup$

                Let $R$ be a nontrivial (necessarily
                noncommutative) ring with the property that $Rcong Roplus R$ as a left $R$-module. (This is the case if $R$ is the endomorphism
                ring of an infinite-dimensional vector space.)
                Then $R$ has a proper submodule $N$ with $Ncong R$ and $R/N
                cong R$                 as $R$-modules. Take $M=R$, $N_1=0$ and $N_2=N$.
                Clearly $N_1notcong N_2$.






                share|cite|improve this answer









                $endgroup$



                Let $R$ be a nontrivial (necessarily
                noncommutative) ring with the property that $Rcong Roplus R$ as a left $R$-module. (This is the case if $R$ is the endomorphism
                ring of an infinite-dimensional vector space.)
                Then $R$ has a proper submodule $N$ with $Ncong R$ and $R/N
                cong R$                 as $R$-modules. Take $M=R$, $N_1=0$ and $N_2=N$.
                Clearly $N_1notcong N_2$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 13 hours ago









                Lord Shark the UnknownLord Shark the Unknown

                109k1163136




                109k1163136



























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