On showing that $varphi(N)=(N+K)/K$, for modules $N,K<M$, $varphi$ natural map.











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In this question modules are A-modules, where A is a unitary commutative ring



This seems trivial, but beginning with module theory, I was trying to see why $varphi(N)=(N+K)/K$, for the natural A-module homomorphism $varphi:Mto M/K$, and submodules $N,Ksubset M$.



In doing so, I got a bit confused over why we don't just have $(N+K)/K=N/K$. My thoughts are these:



$N+K={n+k:nin N, kin K}$, and $(N+K)/K={(n+k)+K:(n+k)in (N+K)}$, $N/K={n+K:nin N}$, as sets.



And isn't $(n+k)+K=n+K$, for some $nin N, kin K$?



And then, for the quotients we have equivalence relations, so that



$N/K nibar{n_1}=bar{n_2}Leftrightarrow n_1 + K=n_2+KLeftrightarrow (n_1-n_2)in K$.



Well, we also have



$(N+K)/K nioverline{n_1+k_1}=overline{n_2+k_2}Leftrightarrow n_1 + k_1 + K=n_2+k_2 +KLeftrightarrow (n_1-n_2)+(k_1-k_2)in K Leftrightarrow (n_1-n_2)in K$.



Thus, intuitively at least, it seems we should have $(N+K)/K=N/K$.



Now,



Question1: Where have I gone wrong, misunderstood or forgotten something?



Question2: Will someone provide a proof sketch of the statement in the title?










share|cite|improve this question




























    up vote
    1
    down vote

    favorite












    In this question modules are A-modules, where A is a unitary commutative ring



    This seems trivial, but beginning with module theory, I was trying to see why $varphi(N)=(N+K)/K$, for the natural A-module homomorphism $varphi:Mto M/K$, and submodules $N,Ksubset M$.



    In doing so, I got a bit confused over why we don't just have $(N+K)/K=N/K$. My thoughts are these:



    $N+K={n+k:nin N, kin K}$, and $(N+K)/K={(n+k)+K:(n+k)in (N+K)}$, $N/K={n+K:nin N}$, as sets.



    And isn't $(n+k)+K=n+K$, for some $nin N, kin K$?



    And then, for the quotients we have equivalence relations, so that



    $N/K nibar{n_1}=bar{n_2}Leftrightarrow n_1 + K=n_2+KLeftrightarrow (n_1-n_2)in K$.



    Well, we also have



    $(N+K)/K nioverline{n_1+k_1}=overline{n_2+k_2}Leftrightarrow n_1 + k_1 + K=n_2+k_2 +KLeftrightarrow (n_1-n_2)+(k_1-k_2)in K Leftrightarrow (n_1-n_2)in K$.



    Thus, intuitively at least, it seems we should have $(N+K)/K=N/K$.



    Now,



    Question1: Where have I gone wrong, misunderstood or forgotten something?



    Question2: Will someone provide a proof sketch of the statement in the title?










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      In this question modules are A-modules, where A is a unitary commutative ring



      This seems trivial, but beginning with module theory, I was trying to see why $varphi(N)=(N+K)/K$, for the natural A-module homomorphism $varphi:Mto M/K$, and submodules $N,Ksubset M$.



      In doing so, I got a bit confused over why we don't just have $(N+K)/K=N/K$. My thoughts are these:



      $N+K={n+k:nin N, kin K}$, and $(N+K)/K={(n+k)+K:(n+k)in (N+K)}$, $N/K={n+K:nin N}$, as sets.



      And isn't $(n+k)+K=n+K$, for some $nin N, kin K$?



      And then, for the quotients we have equivalence relations, so that



      $N/K nibar{n_1}=bar{n_2}Leftrightarrow n_1 + K=n_2+KLeftrightarrow (n_1-n_2)in K$.



      Well, we also have



      $(N+K)/K nioverline{n_1+k_1}=overline{n_2+k_2}Leftrightarrow n_1 + k_1 + K=n_2+k_2 +KLeftrightarrow (n_1-n_2)+(k_1-k_2)in K Leftrightarrow (n_1-n_2)in K$.



      Thus, intuitively at least, it seems we should have $(N+K)/K=N/K$.



      Now,



      Question1: Where have I gone wrong, misunderstood or forgotten something?



      Question2: Will someone provide a proof sketch of the statement in the title?










      share|cite|improve this question















      In this question modules are A-modules, where A is a unitary commutative ring



      This seems trivial, but beginning with module theory, I was trying to see why $varphi(N)=(N+K)/K$, for the natural A-module homomorphism $varphi:Mto M/K$, and submodules $N,Ksubset M$.



      In doing so, I got a bit confused over why we don't just have $(N+K)/K=N/K$. My thoughts are these:



      $N+K={n+k:nin N, kin K}$, and $(N+K)/K={(n+k)+K:(n+k)in (N+K)}$, $N/K={n+K:nin N}$, as sets.



      And isn't $(n+k)+K=n+K$, for some $nin N, kin K$?



      And then, for the quotients we have equivalence relations, so that



      $N/K nibar{n_1}=bar{n_2}Leftrightarrow n_1 + K=n_2+KLeftrightarrow (n_1-n_2)in K$.



      Well, we also have



      $(N+K)/K nioverline{n_1+k_1}=overline{n_2+k_2}Leftrightarrow n_1 + k_1 + K=n_2+k_2 +KLeftrightarrow (n_1-n_2)+(k_1-k_2)in K Leftrightarrow (n_1-n_2)in K$.



      Thus, intuitively at least, it seems we should have $(N+K)/K=N/K$.



      Now,



      Question1: Where have I gone wrong, misunderstood or forgotten something?



      Question2: Will someone provide a proof sketch of the statement in the title?







      abstract-algebra ring-theory modules






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      edited Nov 22 at 21:34









      user26857

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      asked Nov 22 at 15:46









      Christopher.L

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          The answer to your first question is that $K$ is not necessarily a submodule of $N,$ in which case the expression $N/K$ doesn't make sense! $N + K$ is the smallest submodule of $M$ containing both $N$ and $K,$ so if you want to "take a quotient of $N$ by $K,$" then what you really want is $(N+K)/K.$ If $K$ is already a submodule of $N,$ then $N + K = N$ and no harm is done.



          Proof that $varphi(N)= (N+K)/K$: (you've basically done this already - this is essentially the argument you wanted to make to show that $"N/K" = (N+K)/K$!)



          For $min M,$ $varphi(m) = m+K.$ Thus, $varphi(N) = {n + Kmid nin Nsubseteq M}.$ You've already noted that $(N + K)/K = {(n + k) + Kmid nin N, kin K},$ and that for any $(n + k) + Kin (N+K)/K,$ we have $(n + k) + K = (n + 0) + K = n + K.$ Thus, each element of $(N + K)/K$ has a representative in $N$ (that each element of $N$ represents an element of $(N+K)/K$ is trivially true), and so $varphi(N) = (N+K)/K.$



          Essentially, the argument you've made just shows that the module map
          begin{align*}
          f : varphi(N)&to (N+K)/K\
          n + K&mapsto (n + 0) + K,
          end{align*}

          is an isomorphism. Upon examination, one also sees that $f$ is in fact the identity, considering $varphi(N)$ and $(N+K)/K$ inside $M/K.$






          share|cite|improve this answer



















          • 1




            Silly, of course! I think your first § and the function $f$, which can easily be seen to be the identity map (or easily proven a surjective and injective homom.) sums it up nicely. Basically, $K$ not necessarily submodule of $N$ sums it up enough for me. Sometimes you forget fundamental relations like that, when you're just juggling sets around.
            – Christopher.L
            Nov 23 at 13:21













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          The answer to your first question is that $K$ is not necessarily a submodule of $N,$ in which case the expression $N/K$ doesn't make sense! $N + K$ is the smallest submodule of $M$ containing both $N$ and $K,$ so if you want to "take a quotient of $N$ by $K,$" then what you really want is $(N+K)/K.$ If $K$ is already a submodule of $N,$ then $N + K = N$ and no harm is done.



          Proof that $varphi(N)= (N+K)/K$: (you've basically done this already - this is essentially the argument you wanted to make to show that $"N/K" = (N+K)/K$!)



          For $min M,$ $varphi(m) = m+K.$ Thus, $varphi(N) = {n + Kmid nin Nsubseteq M}.$ You've already noted that $(N + K)/K = {(n + k) + Kmid nin N, kin K},$ and that for any $(n + k) + Kin (N+K)/K,$ we have $(n + k) + K = (n + 0) + K = n + K.$ Thus, each element of $(N + K)/K$ has a representative in $N$ (that each element of $N$ represents an element of $(N+K)/K$ is trivially true), and so $varphi(N) = (N+K)/K.$



          Essentially, the argument you've made just shows that the module map
          begin{align*}
          f : varphi(N)&to (N+K)/K\
          n + K&mapsto (n + 0) + K,
          end{align*}

          is an isomorphism. Upon examination, one also sees that $f$ is in fact the identity, considering $varphi(N)$ and $(N+K)/K$ inside $M/K.$






          share|cite|improve this answer



















          • 1




            Silly, of course! I think your first § and the function $f$, which can easily be seen to be the identity map (or easily proven a surjective and injective homom.) sums it up nicely. Basically, $K$ not necessarily submodule of $N$ sums it up enough for me. Sometimes you forget fundamental relations like that, when you're just juggling sets around.
            – Christopher.L
            Nov 23 at 13:21

















          up vote
          1
          down vote



          accepted










          The answer to your first question is that $K$ is not necessarily a submodule of $N,$ in which case the expression $N/K$ doesn't make sense! $N + K$ is the smallest submodule of $M$ containing both $N$ and $K,$ so if you want to "take a quotient of $N$ by $K,$" then what you really want is $(N+K)/K.$ If $K$ is already a submodule of $N,$ then $N + K = N$ and no harm is done.



          Proof that $varphi(N)= (N+K)/K$: (you've basically done this already - this is essentially the argument you wanted to make to show that $"N/K" = (N+K)/K$!)



          For $min M,$ $varphi(m) = m+K.$ Thus, $varphi(N) = {n + Kmid nin Nsubseteq M}.$ You've already noted that $(N + K)/K = {(n + k) + Kmid nin N, kin K},$ and that for any $(n + k) + Kin (N+K)/K,$ we have $(n + k) + K = (n + 0) + K = n + K.$ Thus, each element of $(N + K)/K$ has a representative in $N$ (that each element of $N$ represents an element of $(N+K)/K$ is trivially true), and so $varphi(N) = (N+K)/K.$



          Essentially, the argument you've made just shows that the module map
          begin{align*}
          f : varphi(N)&to (N+K)/K\
          n + K&mapsto (n + 0) + K,
          end{align*}

          is an isomorphism. Upon examination, one also sees that $f$ is in fact the identity, considering $varphi(N)$ and $(N+K)/K$ inside $M/K.$






          share|cite|improve this answer



















          • 1




            Silly, of course! I think your first § and the function $f$, which can easily be seen to be the identity map (or easily proven a surjective and injective homom.) sums it up nicely. Basically, $K$ not necessarily submodule of $N$ sums it up enough for me. Sometimes you forget fundamental relations like that, when you're just juggling sets around.
            – Christopher.L
            Nov 23 at 13:21















          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          The answer to your first question is that $K$ is not necessarily a submodule of $N,$ in which case the expression $N/K$ doesn't make sense! $N + K$ is the smallest submodule of $M$ containing both $N$ and $K,$ so if you want to "take a quotient of $N$ by $K,$" then what you really want is $(N+K)/K.$ If $K$ is already a submodule of $N,$ then $N + K = N$ and no harm is done.



          Proof that $varphi(N)= (N+K)/K$: (you've basically done this already - this is essentially the argument you wanted to make to show that $"N/K" = (N+K)/K$!)



          For $min M,$ $varphi(m) = m+K.$ Thus, $varphi(N) = {n + Kmid nin Nsubseteq M}.$ You've already noted that $(N + K)/K = {(n + k) + Kmid nin N, kin K},$ and that for any $(n + k) + Kin (N+K)/K,$ we have $(n + k) + K = (n + 0) + K = n + K.$ Thus, each element of $(N + K)/K$ has a representative in $N$ (that each element of $N$ represents an element of $(N+K)/K$ is trivially true), and so $varphi(N) = (N+K)/K.$



          Essentially, the argument you've made just shows that the module map
          begin{align*}
          f : varphi(N)&to (N+K)/K\
          n + K&mapsto (n + 0) + K,
          end{align*}

          is an isomorphism. Upon examination, one also sees that $f$ is in fact the identity, considering $varphi(N)$ and $(N+K)/K$ inside $M/K.$






          share|cite|improve this answer














          The answer to your first question is that $K$ is not necessarily a submodule of $N,$ in which case the expression $N/K$ doesn't make sense! $N + K$ is the smallest submodule of $M$ containing both $N$ and $K,$ so if you want to "take a quotient of $N$ by $K,$" then what you really want is $(N+K)/K.$ If $K$ is already a submodule of $N,$ then $N + K = N$ and no harm is done.



          Proof that $varphi(N)= (N+K)/K$: (you've basically done this already - this is essentially the argument you wanted to make to show that $"N/K" = (N+K)/K$!)



          For $min M,$ $varphi(m) = m+K.$ Thus, $varphi(N) = {n + Kmid nin Nsubseteq M}.$ You've already noted that $(N + K)/K = {(n + k) + Kmid nin N, kin K},$ and that for any $(n + k) + Kin (N+K)/K,$ we have $(n + k) + K = (n + 0) + K = n + K.$ Thus, each element of $(N + K)/K$ has a representative in $N$ (that each element of $N$ represents an element of $(N+K)/K$ is trivially true), and so $varphi(N) = (N+K)/K.$



          Essentially, the argument you've made just shows that the module map
          begin{align*}
          f : varphi(N)&to (N+K)/K\
          n + K&mapsto (n + 0) + K,
          end{align*}

          is an isomorphism. Upon examination, one also sees that $f$ is in fact the identity, considering $varphi(N)$ and $(N+K)/K$ inside $M/K.$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 22 at 20:50

























          answered Nov 22 at 20:27









          Stahl

          16k43353




          16k43353








          • 1




            Silly, of course! I think your first § and the function $f$, which can easily be seen to be the identity map (or easily proven a surjective and injective homom.) sums it up nicely. Basically, $K$ not necessarily submodule of $N$ sums it up enough for me. Sometimes you forget fundamental relations like that, when you're just juggling sets around.
            – Christopher.L
            Nov 23 at 13:21
















          • 1




            Silly, of course! I think your first § and the function $f$, which can easily be seen to be the identity map (or easily proven a surjective and injective homom.) sums it up nicely. Basically, $K$ not necessarily submodule of $N$ sums it up enough for me. Sometimes you forget fundamental relations like that, when you're just juggling sets around.
            – Christopher.L
            Nov 23 at 13:21










          1




          1




          Silly, of course! I think your first § and the function $f$, which can easily be seen to be the identity map (or easily proven a surjective and injective homom.) sums it up nicely. Basically, $K$ not necessarily submodule of $N$ sums it up enough for me. Sometimes you forget fundamental relations like that, when you're just juggling sets around.
          – Christopher.L
          Nov 23 at 13:21






          Silly, of course! I think your first § and the function $f$, which can easily be seen to be the identity map (or easily proven a surjective and injective homom.) sums it up nicely. Basically, $K$ not necessarily submodule of $N$ sums it up enough for me. Sometimes you forget fundamental relations like that, when you're just juggling sets around.
          – Christopher.L
          Nov 23 at 13:21




















           

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