Strange isomorphism: $R/(A + B) cong (R/B)/barA$.

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$R$ is a ring and $A, B$ are ideals of $R.$



I was playing around with some stuff and I was wondering if $R/(A + B) cong (R/B)/barA.$ for $barA$ being the image of $A$ under $R rightarrow R/B.$ I'm pretty sure I have a proof for it. I will post a 'proof' of it as an answer if the isomorphism is correct. I simply don't know if it is correct because I have never seen this before. Or...is it just plain wrong?




abstract-algebra




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  • Are those modules, vector spaces, groups, rings, algebras...?? What is $;overline A;$ in this respect, anyway?! Not all here can read minds...
    – DonAntonio
    Aug 20 at 9:25










  • Perhaps you should specify what kind of objects $R, A, B$ are. The $+$ notation suggests that they are abelian groups?
    – Vincent
    Aug 20 at 9:25






  • 1




    It's correct since $overline A=Acdot R/B=(A+B)/B$.
    – Bernard
    Aug 20 at 9:29















up vote
0
down vote

favorite












$R$ is a ring and $A, B$ are ideals of $R.$



I was playing around with some stuff and I was wondering if $R/(A + B) cong (R/B)/barA.$ for $barA$ being the image of $A$ under $R rightarrow R/B.$ I'm pretty sure I have a proof for it. I will post a 'proof' of it as an answer if the isomorphism is correct. I simply don't know if it is correct because I have never seen this before. Or...is it just plain wrong?




abstract-algebra




share|cite|improve this question






















  • Are those modules, vector spaces, groups, rings, algebras...?? What is $;overline A;$ in this respect, anyway?! Not all here can read minds...
    – DonAntonio
    Aug 20 at 9:25










  • Perhaps you should specify what kind of objects $R, A, B$ are. The $+$ notation suggests that they are abelian groups?
    – Vincent
    Aug 20 at 9:25






  • 1




    It's correct since $overline A=Acdot R/B=(A+B)/B$.
    – Bernard
    Aug 20 at 9:29













up vote
0
down vote

favorite









up vote
0
down vote

favorite











$R$ is a ring and $A, B$ are ideals of $R.$



I was playing around with some stuff and I was wondering if $R/(A + B) cong (R/B)/barA.$ for $barA$ being the image of $A$ under $R rightarrow R/B.$ I'm pretty sure I have a proof for it. I will post a 'proof' of it as an answer if the isomorphism is correct. I simply don't know if it is correct because I have never seen this before. Or...is it just plain wrong?




abstract-algebra




share|cite|improve this question














$R$ is a ring and $A, B$ are ideals of $R.$



I was playing around with some stuff and I was wondering if $R/(A + B) cong (R/B)/barA.$ for $barA$ being the image of $A$ under $R rightarrow R/B.$ I'm pretty sure I have a proof for it. I will post a 'proof' of it as an answer if the isomorphism is correct. I simply don't know if it is correct because I have never seen this before. Or...is it just plain wrong?





abstract-algebra





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edited Aug 20 at 9:28









Bernard

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asked Aug 20 at 9:19









伽罗瓦

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  • Are those modules, vector spaces, groups, rings, algebras...?? What is $;overline A;$ in this respect, anyway?! Not all here can read minds...
    – DonAntonio
    Aug 20 at 9:25










  • Perhaps you should specify what kind of objects $R, A, B$ are. The $+$ notation suggests that they are abelian groups?
    – Vincent
    Aug 20 at 9:25






  • 1




    It's correct since $overline A=Acdot R/B=(A+B)/B$.
    – Bernard
    Aug 20 at 9:29

















  • Are those modules, vector spaces, groups, rings, algebras...?? What is $;overline A;$ in this respect, anyway?! Not all here can read minds...
    – DonAntonio
    Aug 20 at 9:25










  • Perhaps you should specify what kind of objects $R, A, B$ are. The $+$ notation suggests that they are abelian groups?
    – Vincent
    Aug 20 at 9:25






  • 1




    It's correct since $overline A=Acdot R/B=(A+B)/B$.
    – Bernard
    Aug 20 at 9:29
















Are those modules, vector spaces, groups, rings, algebras...?? What is $;overline A;$ in this respect, anyway?! Not all here can read minds...
– DonAntonio
Aug 20 at 9:25




Are those modules, vector spaces, groups, rings, algebras...?? What is $;overline A;$ in this respect, anyway?! Not all here can read minds...
– DonAntonio
Aug 20 at 9:25












Perhaps you should specify what kind of objects $R, A, B$ are. The $+$ notation suggests that they are abelian groups?
– Vincent
Aug 20 at 9:25




Perhaps you should specify what kind of objects $R, A, B$ are. The $+$ notation suggests that they are abelian groups?
– Vincent
Aug 20 at 9:25




1




1




It's correct since $overline A=Acdot R/B=(A+B)/B$.
– Bernard
Aug 20 at 9:29





It's correct since $overline A=Acdot R/B=(A+B)/B$.
– Bernard
Aug 20 at 9:29











2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










This is a consequence of the second isomoprhism theorem (on wikipedia is anyway refered as the third): $B leq A+B leq R$ implies that
$$
R/(A+B) cong (R/B)/((A+B)/B)
$$
Where is not hard to prove that $(A+B)/B$ is the image of $A$ through thr map $pi : R to R / B$






share|cite|improve this answer





























    up vote
    1
    down vote













    We have a map from $R rightarrow (R/B)/barA$ which is a homomorphism as this is the composition of the natural maps $R rightarrow R/B$ and $R/B rightarrow (R/B)/barA.$ It is also surjective as it is the composition of surjective maps. It is not difficult to see that the kernel is $A + B.$ However, @JayTuma's answer seems to be a more direct approach using the second isomorphism theorem.






    share|cite|improve this answer




















    • nice thing here is that you actually proved the second ismorphism theorem :)
      – JayTuma
      Aug 20 at 9:37










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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    This is a consequence of the second isomoprhism theorem (on wikipedia is anyway refered as the third): $B leq A+B leq R$ implies that
    $$
    R/(A+B) cong (R/B)/((A+B)/B)
    $$
    Where is not hard to prove that $(A+B)/B$ is the image of $A$ through thr map $pi : R to R / B$






    share|cite|improve this answer


























      up vote
      1
      down vote



      accepted










      This is a consequence of the second isomoprhism theorem (on wikipedia is anyway refered as the third): $B leq A+B leq R$ implies that
      $$
      R/(A+B) cong (R/B)/((A+B)/B)
      $$
      Where is not hard to prove that $(A+B)/B$ is the image of $A$ through thr map $pi : R to R / B$






      share|cite|improve this answer
























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        This is a consequence of the second isomoprhism theorem (on wikipedia is anyway refered as the third): $B leq A+B leq R$ implies that
        $$
        R/(A+B) cong (R/B)/((A+B)/B)
        $$
        Where is not hard to prove that $(A+B)/B$ is the image of $A$ through thr map $pi : R to R / B$






        share|cite|improve this answer














        This is a consequence of the second isomoprhism theorem (on wikipedia is anyway refered as the third): $B leq A+B leq R$ implies that
        $$
        R/(A+B) cong (R/B)/((A+B)/B)
        $$
        Where is not hard to prove that $(A+B)/B$ is the image of $A$ through thr map $pi : R to R / B$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 20 at 9:39

























        answered Aug 20 at 9:32









        JayTuma

        1,333118




        1,333118




















            up vote
            1
            down vote













            We have a map from $R rightarrow (R/B)/barA$ which is a homomorphism as this is the composition of the natural maps $R rightarrow R/B$ and $R/B rightarrow (R/B)/barA.$ It is also surjective as it is the composition of surjective maps. It is not difficult to see that the kernel is $A + B.$ However, @JayTuma's answer seems to be a more direct approach using the second isomorphism theorem.






            share|cite|improve this answer




















            • nice thing here is that you actually proved the second ismorphism theorem :)
              – JayTuma
              Aug 20 at 9:37














            up vote
            1
            down vote













            We have a map from $R rightarrow (R/B)/barA$ which is a homomorphism as this is the composition of the natural maps $R rightarrow R/B$ and $R/B rightarrow (R/B)/barA.$ It is also surjective as it is the composition of surjective maps. It is not difficult to see that the kernel is $A + B.$ However, @JayTuma's answer seems to be a more direct approach using the second isomorphism theorem.






            share|cite|improve this answer




















            • nice thing here is that you actually proved the second ismorphism theorem :)
              – JayTuma
              Aug 20 at 9:37












            up vote
            1
            down vote










            up vote
            1
            down vote









            We have a map from $R rightarrow (R/B)/barA$ which is a homomorphism as this is the composition of the natural maps $R rightarrow R/B$ and $R/B rightarrow (R/B)/barA.$ It is also surjective as it is the composition of surjective maps. It is not difficult to see that the kernel is $A + B.$ However, @JayTuma's answer seems to be a more direct approach using the second isomorphism theorem.






            share|cite|improve this answer












            We have a map from $R rightarrow (R/B)/barA$ which is a homomorphism as this is the composition of the natural maps $R rightarrow R/B$ and $R/B rightarrow (R/B)/barA.$ It is also surjective as it is the composition of surjective maps. It is not difficult to see that the kernel is $A + B.$ However, @JayTuma's answer seems to be a more direct approach using the second isomorphism theorem.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 20 at 9:34









            伽罗瓦

            934615




            934615











            • nice thing here is that you actually proved the second ismorphism theorem :)
              – JayTuma
              Aug 20 at 9:37
















            • nice thing here is that you actually proved the second ismorphism theorem :)
              – JayTuma
              Aug 20 at 9:37















            nice thing here is that you actually proved the second ismorphism theorem :)
            – JayTuma
            Aug 20 at 9:37




            nice thing here is that you actually proved the second ismorphism theorem :)
            – JayTuma
            Aug 20 at 9:37












             

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