If $I$ is ideal of $R$, $N$ is submodule of $M$, then $Ifrac MN=frac{IM+N}N$












2












$begingroup$


$I$ is ideal of ring $R$. $M,N$ are $R$-modules and $N$ is submodule of $M$, how to prove:



$$Icdotfrac MN=frac{IM+N}N quad? $$



Conclusion:



It's not a "problem" to solve, just a notion. If you meet the same confusion, see answers below :)






abstract-algebra modules







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Note that there is no reason why $N$ is contained in $IM$, but it is contained in $IM+N$ is. Much like the isomorphism $M/(Mcap N) = (M+N)/ N$ makes sense because $Mcap Nsubseteq N$ and $M+Nsupseteq N$.
    $endgroup$
    – Pedro Tamaroff♦
    Dec 25 '18 at 15:07


















2












$begingroup$


$I$ is ideal of ring $R$. $M,N$ are $R$-modules and $N$ is submodule of $M$, how to prove:



$$Icdotfrac MN=frac{IM+N}N quad? $$



Conclusion:



It's not a "problem" to solve, just a notion. If you meet the same confusion, see answers below :)






abstract-algebra modules







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Note that there is no reason why $N$ is contained in $IM$, but it is contained in $IM+N$ is. Much like the isomorphism $M/(Mcap N) = (M+N)/ N$ makes sense because $Mcap Nsubseteq N$ and $M+Nsupseteq N$.
    $endgroup$
    – Pedro Tamaroff♦
    Dec 25 '18 at 15:07
















2












2








2





$begingroup$


$I$ is ideal of ring $R$. $M,N$ are $R$-modules and $N$ is submodule of $M$, how to prove:



$$Icdotfrac MN=frac{IM+N}N quad? $$



Conclusion:



It's not a "problem" to solve, just a notion. If you meet the same confusion, see answers below :)






abstract-algebra modules







share|cite|improve this question











$endgroup$




$I$ is ideal of ring $R$. $M,N$ are $R$-modules and $N$ is submodule of $M$, how to prove:



$$Icdotfrac MN=frac{IM+N}N quad? $$



Conclusion:



It's not a "problem" to solve, just a notion. If you meet the same confusion, see answers below :)






abstract-algebra modules




abstract-algebra modules






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 17:49







Andrews

















asked Dec 25 '18 at 15:04









AndrewsAndrews

1,2812423




1,2812423








  • 2




    $begingroup$
    Note that there is no reason why $N$ is contained in $IM$, but it is contained in $IM+N$ is. Much like the isomorphism $M/(Mcap N) = (M+N)/ N$ makes sense because $Mcap Nsubseteq N$ and $M+Nsupseteq N$.
    $endgroup$
    – Pedro Tamaroff♦
    Dec 25 '18 at 15:07
















  • 2




    $begingroup$
    Note that there is no reason why $N$ is contained in $IM$, but it is contained in $IM+N$ is. Much like the isomorphism $M/(Mcap N) = (M+N)/ N$ makes sense because $Mcap Nsubseteq N$ and $M+Nsupseteq N$.
    $endgroup$
    – Pedro Tamaroff♦
    Dec 25 '18 at 15:07










2




2




$begingroup$
Note that there is no reason why $N$ is contained in $IM$, but it is contained in $IM+N$ is. Much like the isomorphism $M/(Mcap N) = (M+N)/ N$ makes sense because $Mcap Nsubseteq N$ and $M+Nsupseteq N$.
$endgroup$
– Pedro Tamaroff♦
Dec 25 '18 at 15:07






$begingroup$
Note that there is no reason why $N$ is contained in $IM$, but it is contained in $IM+N$ is. Much like the isomorphism $M/(Mcap N) = (M+N)/ N$ makes sense because $Mcap Nsubseteq N$ and $M+Nsupseteq N$.
$endgroup$
– Pedro Tamaroff♦
Dec 25 '18 at 15:07












4 Answers
4






active

oldest

votes


















2












$begingroup$

Let $p_1:Mrightarrow M/N$, $I(M/N)$ is the submodule of $M/N$ generated by $p_1(im), iin I, min M$. $IM+N$ is a submodule of $M$ which contains $N$, we can define the quotient $p_2:IM+Nrightarrow (IM+N)/N$.



Consider $i:IM+Nrightarrow M$ the canonical injection and $p_3=p_1circ i$. $p_3(im+n)=p_1(im+n)=p(im)$. The kernel of $p_3$ contains $N$. This implies that $p_3$ induces a morphism $f: (IM+N)/Nrightarrow I(M/N)$. $f$ is surjective, and $f(p_2(im+n))=0$ is equivalent to $p_1(im)=0$ this is equivalent to saying that $imin N$ and $im+Nin N$, this implies that $p_2(im+n))=0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    In expression $p_1(im)$, you naturally treat $N$ as submodule of $IM$, this nay not be ture if we take $R = mathbb Z$, $I = mmathbb Z$, $M=mathbb Z$, $N=nmathbb Z$, then $IM = mmathbb Z$. However $N=nmathbb Z$ may not contained in $IM = mmathbb Z$.
    $endgroup$
    – Andrews
    Dec 25 '18 at 17:04












  • $begingroup$
    $p_1(im)in M/N$ so $N$ is not involved there.
    $endgroup$
    – Tsemo Aristide
    Dec 25 '18 at 17:11



















1












$begingroup$

Isn't the equality obvious if you observe that
begin{alignat}{2}
I,(M/N)&=Bigl{sum_text{finite}i_k(m_k+N)Bigr}, quad rlap{i_kin I, m_kin M}\
&=Bigl{sum_text{finite}i_km_k+NBigr} &&text{by definition of the product by $i_k$ in $M/N$ }\
&=Bigl{Bigl(sum_text{finite}i_km_kBigr)+NBigr}=(IM+N)/N.
end{alignat}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    let ring $R = mathbb Z$, $I = mmathbb Z$, $M=mathbb Z$, $N=nmathbb Z$, then $IM = mmathbb Z$. However $N=nmathbb Z$ may not contained in $IM = mmathbb Z$. So I think there might be some mistake in your proof.
    $endgroup$
    – Andrews
    Dec 25 '18 at 16:55












  • $begingroup$
    That's precisely because $N=nbf Z$ is not contained in $mbf Z$ that $N$ is added ($m_k+N$ is the congruence class of $m_k$ modulo $N$). All my reasoning and computation is with the congruence classes.
    $endgroup$
    – Bernard
    Dec 25 '18 at 17:11





















1












$begingroup$

$I(M/N)= (IM+N)/N$ is just a definition(don’t overthink it!), because every element in $M/N$ is of the form $m+N$ so when you multiply by an $i in I$ you just get $im + N in IM+N$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Pedro Tamaroff's comment inspires my a lot.



    $N$ may not be contained in $IM$, but it is contained in $IM+N$.



    Eg. $R=mathbb Z$, $I = pmathbb Z$, $M = mathbb Z$, $N = qmathbb Z$, then $IM = p mathbb Z$. However $N = qmathbb Z$ may not contained in $ IM = pmathbb Z$.



    Notion ${IM}/N$ doesn't make much sence while $(IM+N)/N$ does, so we treat $im+N(iin I, min M)$ as element in $(IM+N)/N$. And it's just a matter of notion. .



    Homomorphisms are:



    $phi: Icdot M/Nto (IM+N)/N,quad icdot(m+N)mapsto im+N$.



    $psi:(IM+N)/N to Icdot M/N,quad im+Nmapsto icdot(m+N)$.






    share|cite|improve this answer











    $endgroup$














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      4 Answers
      4






      active

      oldest

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      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Let $p_1:Mrightarrow M/N$, $I(M/N)$ is the submodule of $M/N$ generated by $p_1(im), iin I, min M$. $IM+N$ is a submodule of $M$ which contains $N$, we can define the quotient $p_2:IM+Nrightarrow (IM+N)/N$.



      Consider $i:IM+Nrightarrow M$ the canonical injection and $p_3=p_1circ i$. $p_3(im+n)=p_1(im+n)=p(im)$. The kernel of $p_3$ contains $N$. This implies that $p_3$ induces a morphism $f: (IM+N)/Nrightarrow I(M/N)$. $f$ is surjective, and $f(p_2(im+n))=0$ is equivalent to $p_1(im)=0$ this is equivalent to saying that $imin N$ and $im+Nin N$, this implies that $p_2(im+n))=0$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        In expression $p_1(im)$, you naturally treat $N$ as submodule of $IM$, this nay not be ture if we take $R = mathbb Z$, $I = mmathbb Z$, $M=mathbb Z$, $N=nmathbb Z$, then $IM = mmathbb Z$. However $N=nmathbb Z$ may not contained in $IM = mmathbb Z$.
        $endgroup$
        – Andrews
        Dec 25 '18 at 17:04












      • $begingroup$
        $p_1(im)in M/N$ so $N$ is not involved there.
        $endgroup$
        – Tsemo Aristide
        Dec 25 '18 at 17:11
















      2












      $begingroup$

      Let $p_1:Mrightarrow M/N$, $I(M/N)$ is the submodule of $M/N$ generated by $p_1(im), iin I, min M$. $IM+N$ is a submodule of $M$ which contains $N$, we can define the quotient $p_2:IM+Nrightarrow (IM+N)/N$.



      Consider $i:IM+Nrightarrow M$ the canonical injection and $p_3=p_1circ i$. $p_3(im+n)=p_1(im+n)=p(im)$. The kernel of $p_3$ contains $N$. This implies that $p_3$ induces a morphism $f: (IM+N)/Nrightarrow I(M/N)$. $f$ is surjective, and $f(p_2(im+n))=0$ is equivalent to $p_1(im)=0$ this is equivalent to saying that $imin N$ and $im+Nin N$, this implies that $p_2(im+n))=0$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        In expression $p_1(im)$, you naturally treat $N$ as submodule of $IM$, this nay not be ture if we take $R = mathbb Z$, $I = mmathbb Z$, $M=mathbb Z$, $N=nmathbb Z$, then $IM = mmathbb Z$. However $N=nmathbb Z$ may not contained in $IM = mmathbb Z$.
        $endgroup$
        – Andrews
        Dec 25 '18 at 17:04












      • $begingroup$
        $p_1(im)in M/N$ so $N$ is not involved there.
        $endgroup$
        – Tsemo Aristide
        Dec 25 '18 at 17:11














      2












      2








      2





      $begingroup$

      Let $p_1:Mrightarrow M/N$, $I(M/N)$ is the submodule of $M/N$ generated by $p_1(im), iin I, min M$. $IM+N$ is a submodule of $M$ which contains $N$, we can define the quotient $p_2:IM+Nrightarrow (IM+N)/N$.



      Consider $i:IM+Nrightarrow M$ the canonical injection and $p_3=p_1circ i$. $p_3(im+n)=p_1(im+n)=p(im)$. The kernel of $p_3$ contains $N$. This implies that $p_3$ induces a morphism $f: (IM+N)/Nrightarrow I(M/N)$. $f$ is surjective, and $f(p_2(im+n))=0$ is equivalent to $p_1(im)=0$ this is equivalent to saying that $imin N$ and $im+Nin N$, this implies that $p_2(im+n))=0$.






      share|cite|improve this answer











      $endgroup$



      Let $p_1:Mrightarrow M/N$, $I(M/N)$ is the submodule of $M/N$ generated by $p_1(im), iin I, min M$. $IM+N$ is a submodule of $M$ which contains $N$, we can define the quotient $p_2:IM+Nrightarrow (IM+N)/N$.



      Consider $i:IM+Nrightarrow M$ the canonical injection and $p_3=p_1circ i$. $p_3(im+n)=p_1(im+n)=p(im)$. The kernel of $p_3$ contains $N$. This implies that $p_3$ induces a morphism $f: (IM+N)/Nrightarrow I(M/N)$. $f$ is surjective, and $f(p_2(im+n))=0$ is equivalent to $p_1(im)=0$ this is equivalent to saying that $imin N$ and $im+Nin N$, this implies that $p_2(im+n))=0$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 25 '18 at 17:05

























      answered Dec 25 '18 at 15:16









      Tsemo AristideTsemo Aristide

      60.3k11446




      60.3k11446












      • $begingroup$
        In expression $p_1(im)$, you naturally treat $N$ as submodule of $IM$, this nay not be ture if we take $R = mathbb Z$, $I = mmathbb Z$, $M=mathbb Z$, $N=nmathbb Z$, then $IM = mmathbb Z$. However $N=nmathbb Z$ may not contained in $IM = mmathbb Z$.
        $endgroup$
        – Andrews
        Dec 25 '18 at 17:04












      • $begingroup$
        $p_1(im)in M/N$ so $N$ is not involved there.
        $endgroup$
        – Tsemo Aristide
        Dec 25 '18 at 17:11


















      • $begingroup$
        In expression $p_1(im)$, you naturally treat $N$ as submodule of $IM$, this nay not be ture if we take $R = mathbb Z$, $I = mmathbb Z$, $M=mathbb Z$, $N=nmathbb Z$, then $IM = mmathbb Z$. However $N=nmathbb Z$ may not contained in $IM = mmathbb Z$.
        $endgroup$
        – Andrews
        Dec 25 '18 at 17:04












      • $begingroup$
        $p_1(im)in M/N$ so $N$ is not involved there.
        $endgroup$
        – Tsemo Aristide
        Dec 25 '18 at 17:11
















      $begingroup$
      In expression $p_1(im)$, you naturally treat $N$ as submodule of $IM$, this nay not be ture if we take $R = mathbb Z$, $I = mmathbb Z$, $M=mathbb Z$, $N=nmathbb Z$, then $IM = mmathbb Z$. However $N=nmathbb Z$ may not contained in $IM = mmathbb Z$.
      $endgroup$
      – Andrews
      Dec 25 '18 at 17:04






      $begingroup$
      In expression $p_1(im)$, you naturally treat $N$ as submodule of $IM$, this nay not be ture if we take $R = mathbb Z$, $I = mmathbb Z$, $M=mathbb Z$, $N=nmathbb Z$, then $IM = mmathbb Z$. However $N=nmathbb Z$ may not contained in $IM = mmathbb Z$.
      $endgroup$
      – Andrews
      Dec 25 '18 at 17:04














      $begingroup$
      $p_1(im)in M/N$ so $N$ is not involved there.
      $endgroup$
      – Tsemo Aristide
      Dec 25 '18 at 17:11




      $begingroup$
      $p_1(im)in M/N$ so $N$ is not involved there.
      $endgroup$
      – Tsemo Aristide
      Dec 25 '18 at 17:11











      1












      $begingroup$

      Isn't the equality obvious if you observe that
      begin{alignat}{2}
      I,(M/N)&=Bigl{sum_text{finite}i_k(m_k+N)Bigr}, quad rlap{i_kin I, m_kin M}\
      &=Bigl{sum_text{finite}i_km_k+NBigr} &&text{by definition of the product by $i_k$ in $M/N$ }\
      &=Bigl{Bigl(sum_text{finite}i_km_kBigr)+NBigr}=(IM+N)/N.
      end{alignat}






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        let ring $R = mathbb Z$, $I = mmathbb Z$, $M=mathbb Z$, $N=nmathbb Z$, then $IM = mmathbb Z$. However $N=nmathbb Z$ may not contained in $IM = mmathbb Z$. So I think there might be some mistake in your proof.
        $endgroup$
        – Andrews
        Dec 25 '18 at 16:55












      • $begingroup$
        That's precisely because $N=nbf Z$ is not contained in $mbf Z$ that $N$ is added ($m_k+N$ is the congruence class of $m_k$ modulo $N$). All my reasoning and computation is with the congruence classes.
        $endgroup$
        – Bernard
        Dec 25 '18 at 17:11


















      1












      $begingroup$

      Isn't the equality obvious if you observe that
      begin{alignat}{2}
      I,(M/N)&=Bigl{sum_text{finite}i_k(m_k+N)Bigr}, quad rlap{i_kin I, m_kin M}\
      &=Bigl{sum_text{finite}i_km_k+NBigr} &&text{by definition of the product by $i_k$ in $M/N$ }\
      &=Bigl{Bigl(sum_text{finite}i_km_kBigr)+NBigr}=(IM+N)/N.
      end{alignat}






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        let ring $R = mathbb Z$, $I = mmathbb Z$, $M=mathbb Z$, $N=nmathbb Z$, then $IM = mmathbb Z$. However $N=nmathbb Z$ may not contained in $IM = mmathbb Z$. So I think there might be some mistake in your proof.
        $endgroup$
        – Andrews
        Dec 25 '18 at 16:55












      • $begingroup$
        That's precisely because $N=nbf Z$ is not contained in $mbf Z$ that $N$ is added ($m_k+N$ is the congruence class of $m_k$ modulo $N$). All my reasoning and computation is with the congruence classes.
        $endgroup$
        – Bernard
        Dec 25 '18 at 17:11
















      1












      1








      1





      $begingroup$

      Isn't the equality obvious if you observe that
      begin{alignat}{2}
      I,(M/N)&=Bigl{sum_text{finite}i_k(m_k+N)Bigr}, quad rlap{i_kin I, m_kin M}\
      &=Bigl{sum_text{finite}i_km_k+NBigr} &&text{by definition of the product by $i_k$ in $M/N$ }\
      &=Bigl{Bigl(sum_text{finite}i_km_kBigr)+NBigr}=(IM+N)/N.
      end{alignat}






      share|cite|improve this answer









      $endgroup$



      Isn't the equality obvious if you observe that
      begin{alignat}{2}
      I,(M/N)&=Bigl{sum_text{finite}i_k(m_k+N)Bigr}, quad rlap{i_kin I, m_kin M}\
      &=Bigl{sum_text{finite}i_km_k+NBigr} &&text{by definition of the product by $i_k$ in $M/N$ }\
      &=Bigl{Bigl(sum_text{finite}i_km_kBigr)+NBigr}=(IM+N)/N.
      end{alignat}







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 25 '18 at 15:32









      BernardBernard

      124k741117




      124k741117












      • $begingroup$
        let ring $R = mathbb Z$, $I = mmathbb Z$, $M=mathbb Z$, $N=nmathbb Z$, then $IM = mmathbb Z$. However $N=nmathbb Z$ may not contained in $IM = mmathbb Z$. So I think there might be some mistake in your proof.
        $endgroup$
        – Andrews
        Dec 25 '18 at 16:55












      • $begingroup$
        That's precisely because $N=nbf Z$ is not contained in $mbf Z$ that $N$ is added ($m_k+N$ is the congruence class of $m_k$ modulo $N$). All my reasoning and computation is with the congruence classes.
        $endgroup$
        – Bernard
        Dec 25 '18 at 17:11




















      • $begingroup$
        let ring $R = mathbb Z$, $I = mmathbb Z$, $M=mathbb Z$, $N=nmathbb Z$, then $IM = mmathbb Z$. However $N=nmathbb Z$ may not contained in $IM = mmathbb Z$. So I think there might be some mistake in your proof.
        $endgroup$
        – Andrews
        Dec 25 '18 at 16:55












      • $begingroup$
        That's precisely because $N=nbf Z$ is not contained in $mbf Z$ that $N$ is added ($m_k+N$ is the congruence class of $m_k$ modulo $N$). All my reasoning and computation is with the congruence classes.
        $endgroup$
        – Bernard
        Dec 25 '18 at 17:11


















      $begingroup$
      let ring $R = mathbb Z$, $I = mmathbb Z$, $M=mathbb Z$, $N=nmathbb Z$, then $IM = mmathbb Z$. However $N=nmathbb Z$ may not contained in $IM = mmathbb Z$. So I think there might be some mistake in your proof.
      $endgroup$
      – Andrews
      Dec 25 '18 at 16:55






      $begingroup$
      let ring $R = mathbb Z$, $I = mmathbb Z$, $M=mathbb Z$, $N=nmathbb Z$, then $IM = mmathbb Z$. However $N=nmathbb Z$ may not contained in $IM = mmathbb Z$. So I think there might be some mistake in your proof.
      $endgroup$
      – Andrews
      Dec 25 '18 at 16:55














      $begingroup$
      That's precisely because $N=nbf Z$ is not contained in $mbf Z$ that $N$ is added ($m_k+N$ is the congruence class of $m_k$ modulo $N$). All my reasoning and computation is with the congruence classes.
      $endgroup$
      – Bernard
      Dec 25 '18 at 17:11






      $begingroup$
      That's precisely because $N=nbf Z$ is not contained in $mbf Z$ that $N$ is added ($m_k+N$ is the congruence class of $m_k$ modulo $N$). All my reasoning and computation is with the congruence classes.
      $endgroup$
      – Bernard
      Dec 25 '18 at 17:11













      1












      $begingroup$

      $I(M/N)= (IM+N)/N$ is just a definition(don’t overthink it!), because every element in $M/N$ is of the form $m+N$ so when you multiply by an $i in I$ you just get $im + N in IM+N$






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        $I(M/N)= (IM+N)/N$ is just a definition(don’t overthink it!), because every element in $M/N$ is of the form $m+N$ so when you multiply by an $i in I$ you just get $im + N in IM+N$






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          $I(M/N)= (IM+N)/N$ is just a definition(don’t overthink it!), because every element in $M/N$ is of the form $m+N$ so when you multiply by an $i in I$ you just get $im + N in IM+N$






          share|cite|improve this answer











          $endgroup$



          $I(M/N)= (IM+N)/N$ is just a definition(don’t overthink it!), because every element in $M/N$ is of the form $m+N$ so when you multiply by an $i in I$ you just get $im + N in IM+N$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 25 '18 at 17:27









          Andrews

          1,2812423




          1,2812423










          answered Dec 25 '18 at 17:12









          Sorin TircSorin Tirc

          1,865213




          1,865213























              0












              $begingroup$

              Pedro Tamaroff's comment inspires my a lot.



              $N$ may not be contained in $IM$, but it is contained in $IM+N$.



              Eg. $R=mathbb Z$, $I = pmathbb Z$, $M = mathbb Z$, $N = qmathbb Z$, then $IM = p mathbb Z$. However $N = qmathbb Z$ may not contained in $ IM = pmathbb Z$.



              Notion ${IM}/N$ doesn't make much sence while $(IM+N)/N$ does, so we treat $im+N(iin I, min M)$ as element in $(IM+N)/N$. And it's just a matter of notion. .



              Homomorphisms are:



              $phi: Icdot M/Nto (IM+N)/N,quad icdot(m+N)mapsto im+N$.



              $psi:(IM+N)/N to Icdot M/N,quad im+Nmapsto icdot(m+N)$.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Pedro Tamaroff's comment inspires my a lot.



                $N$ may not be contained in $IM$, but it is contained in $IM+N$.



                Eg. $R=mathbb Z$, $I = pmathbb Z$, $M = mathbb Z$, $N = qmathbb Z$, then $IM = p mathbb Z$. However $N = qmathbb Z$ may not contained in $ IM = pmathbb Z$.



                Notion ${IM}/N$ doesn't make much sence while $(IM+N)/N$ does, so we treat $im+N(iin I, min M)$ as element in $(IM+N)/N$. And it's just a matter of notion. .



                Homomorphisms are:



                $phi: Icdot M/Nto (IM+N)/N,quad icdot(m+N)mapsto im+N$.



                $psi:(IM+N)/N to Icdot M/N,quad im+Nmapsto icdot(m+N)$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Pedro Tamaroff's comment inspires my a lot.



                  $N$ may not be contained in $IM$, but it is contained in $IM+N$.



                  Eg. $R=mathbb Z$, $I = pmathbb Z$, $M = mathbb Z$, $N = qmathbb Z$, then $IM = p mathbb Z$. However $N = qmathbb Z$ may not contained in $ IM = pmathbb Z$.



                  Notion ${IM}/N$ doesn't make much sence while $(IM+N)/N$ does, so we treat $im+N(iin I, min M)$ as element in $(IM+N)/N$. And it's just a matter of notion. .



                  Homomorphisms are:



                  $phi: Icdot M/Nto (IM+N)/N,quad icdot(m+N)mapsto im+N$.



                  $psi:(IM+N)/N to Icdot M/N,quad im+Nmapsto icdot(m+N)$.






                  share|cite|improve this answer











                  $endgroup$



                  Pedro Tamaroff's comment inspires my a lot.



                  $N$ may not be contained in $IM$, but it is contained in $IM+N$.



                  Eg. $R=mathbb Z$, $I = pmathbb Z$, $M = mathbb Z$, $N = qmathbb Z$, then $IM = p mathbb Z$. However $N = qmathbb Z$ may not contained in $ IM = pmathbb Z$.



                  Notion ${IM}/N$ doesn't make much sence while $(IM+N)/N$ does, so we treat $im+N(iin I, min M)$ as element in $(IM+N)/N$. And it's just a matter of notion. .



                  Homomorphisms are:



                  $phi: Icdot M/Nto (IM+N)/N,quad icdot(m+N)mapsto im+N$.



                  $psi:(IM+N)/N to Icdot M/N,quad im+Nmapsto icdot(m+N)$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 25 '18 at 17:45

























                  answered Dec 25 '18 at 17:25









                  AndrewsAndrews

                  1,2812423




                  1,2812423






























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