Fundamental homomorphism theorem (epimorphism)











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Let φ : R → S be a ring epimorphism. Prove that R/kerφ ∼= S.



Is this the fundamental homomorphism theorem? I thought the FHT started with a ring homomorphism and not an epimorphism. Does this change the proof of the theorem ?






ring-theory







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  • If $varphi$ is epi-, then what is $mathrm {Im}, varphi$ in the FHT then?
    – xbh
    Nov 12 at 16:43

















up vote
0
down vote

favorite












Let φ : R → S be a ring epimorphism. Prove that R/kerφ ∼= S.



Is this the fundamental homomorphism theorem? I thought the FHT started with a ring homomorphism and not an epimorphism. Does this change the proof of the theorem ?






ring-theory







share|cite|improve this question






















  • If $varphi$ is epi-, then what is $mathrm {Im}, varphi$ in the FHT then?
    – xbh
    Nov 12 at 16:43















up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let φ : R → S be a ring epimorphism. Prove that R/kerφ ∼= S.



Is this the fundamental homomorphism theorem? I thought the FHT started with a ring homomorphism and not an epimorphism. Does this change the proof of the theorem ?






ring-theory







share|cite|improve this question













Let φ : R → S be a ring epimorphism. Prove that R/kerφ ∼= S.



Is this the fundamental homomorphism theorem? I thought the FHT started with a ring homomorphism and not an epimorphism. Does this change the proof of the theorem ?






ring-theory




ring-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 12 at 16:24









Johnmallu

74




74












  • If $varphi$ is epi-, then what is $mathrm {Im}, varphi$ in the FHT then?
    – xbh
    Nov 12 at 16:43




















  • If $varphi$ is epi-, then what is $mathrm {Im}, varphi$ in the FHT then?
    – xbh
    Nov 12 at 16:43


















If $varphi$ is epi-, then what is $mathrm {Im}, varphi$ in the FHT then?
– xbh
Nov 12 at 16:43






If $varphi$ is epi-, then what is $mathrm {Im}, varphi$ in the FHT then?
– xbh
Nov 12 at 16:43












2 Answers
2






active

oldest

votes

















up vote
0
down vote



accepted










You can state the theorem without the epimorphism assumption replacing $S$ by $mathrm{Im} varphi$. So both formulation are equivalent.






share|cite|improve this answer





















  • Okay thank you !
    – Johnmallu
    Nov 12 at 16:52


















up vote
0
down vote













The general formula for the homomorphism theorem is
$$R/ker(phi)cong mathrm{Im}(phi)$$
In the special case, $phi$ is an epimorphism, then $mathrm{Im}(phi)=S$, but be careful, you cannot deduce the general formula from the special case.






share|cite|improve this answer





















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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    You can state the theorem without the epimorphism assumption replacing $S$ by $mathrm{Im} varphi$. So both formulation are equivalent.






    share|cite|improve this answer





















    • Okay thank you !
      – Johnmallu
      Nov 12 at 16:52















    up vote
    0
    down vote



    accepted










    You can state the theorem without the epimorphism assumption replacing $S$ by $mathrm{Im} varphi$. So both formulation are equivalent.






    share|cite|improve this answer





















    • Okay thank you !
      – Johnmallu
      Nov 12 at 16:52













    up vote
    0
    down vote



    accepted







    up vote
    0
    down vote



    accepted






    You can state the theorem without the epimorphism assumption replacing $S$ by $mathrm{Im} varphi$. So both formulation are equivalent.






    share|cite|improve this answer












    You can state the theorem without the epimorphism assumption replacing $S$ by $mathrm{Im} varphi$. So both formulation are equivalent.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 12 at 16:47









    Can I play with Mathness

    3894




    3894












    • Okay thank you !
      – Johnmallu
      Nov 12 at 16:52


















    • Okay thank you !
      – Johnmallu
      Nov 12 at 16:52
















    Okay thank you !
    – Johnmallu
    Nov 12 at 16:52




    Okay thank you !
    – Johnmallu
    Nov 12 at 16:52










    up vote
    0
    down vote













    The general formula for the homomorphism theorem is
    $$R/ker(phi)cong mathrm{Im}(phi)$$
    In the special case, $phi$ is an epimorphism, then $mathrm{Im}(phi)=S$, but be careful, you cannot deduce the general formula from the special case.






    share|cite|improve this answer

























      up vote
      0
      down vote













      The general formula for the homomorphism theorem is
      $$R/ker(phi)cong mathrm{Im}(phi)$$
      In the special case, $phi$ is an epimorphism, then $mathrm{Im}(phi)=S$, but be careful, you cannot deduce the general formula from the special case.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        The general formula for the homomorphism theorem is
        $$R/ker(phi)cong mathrm{Im}(phi)$$
        In the special case, $phi$ is an epimorphism, then $mathrm{Im}(phi)=S$, but be careful, you cannot deduce the general formula from the special case.






        share|cite|improve this answer












        The general formula for the homomorphism theorem is
        $$R/ker(phi)cong mathrm{Im}(phi)$$
        In the special case, $phi$ is an epimorphism, then $mathrm{Im}(phi)=S$, but be careful, you cannot deduce the general formula from the special case.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 12 at 18:45









        Fakemistake

        1,582714




        1,582714






























             

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