Algebra - Homomorphism












0












$begingroup$


When checking if two rings are isomorphic, we check if mapping is homomorphism and then we check if it is bijective (injective and surjective).
In some tasks when checking if isomorphisms, we checked if it is homomorphism, surjective and instead of injective we questioned the kernel (Ker) if it is ideal.



Do we need to question all of this or there are some cases?
I think that if we have ring factor, we check homomorphism, surjective and Ker,
but if we have just ring on the left side, we check if mapping is homomorphism, surjective and injective.



Can anyone explain? Thank you so much!






ring-homomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    Kernel is always an ideal. For injectivity, the thing you check for the kernel is whether or not it is trivial.
    $endgroup$
    – Anurag A
    Dec 6 '18 at 22:37










  • $begingroup$
    So when checking if isomorphism, we look if mapping is homomorphism, surjective, injective and we check kernel. All of this?
    $endgroup$
    – Haus
    Dec 6 '18 at 22:39






  • 1




    $begingroup$
    for isomorphism you need 3 things: homomorphic property, surjectivity and injectivity. Usually kernel helps with injectivity.
    $endgroup$
    – Anurag A
    Dec 6 '18 at 22:40








  • 1




    $begingroup$
    A homomorphism is injective if and only if its kernel is trivial (proof: one direction is obvious. For the other, if $f(a) = f(b)$, then $f(a - b) = f(0) = 0$, so $a - b$ lies in the kernel).
    $endgroup$
    – user3482749
    Dec 6 '18 at 22:40










  • $begingroup$
    I got it now. Thank you! :)
    $endgroup$
    – Haus
    Dec 6 '18 at 22:40
















0












$begingroup$


When checking if two rings are isomorphic, we check if mapping is homomorphism and then we check if it is bijective (injective and surjective).
In some tasks when checking if isomorphisms, we checked if it is homomorphism, surjective and instead of injective we questioned the kernel (Ker) if it is ideal.



Do we need to question all of this or there are some cases?
I think that if we have ring factor, we check homomorphism, surjective and Ker,
but if we have just ring on the left side, we check if mapping is homomorphism, surjective and injective.



Can anyone explain? Thank you so much!






ring-homomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    Kernel is always an ideal. For injectivity, the thing you check for the kernel is whether or not it is trivial.
    $endgroup$
    – Anurag A
    Dec 6 '18 at 22:37










  • $begingroup$
    So when checking if isomorphism, we look if mapping is homomorphism, surjective, injective and we check kernel. All of this?
    $endgroup$
    – Haus
    Dec 6 '18 at 22:39






  • 1




    $begingroup$
    for isomorphism you need 3 things: homomorphic property, surjectivity and injectivity. Usually kernel helps with injectivity.
    $endgroup$
    – Anurag A
    Dec 6 '18 at 22:40








  • 1




    $begingroup$
    A homomorphism is injective if and only if its kernel is trivial (proof: one direction is obvious. For the other, if $f(a) = f(b)$, then $f(a - b) = f(0) = 0$, so $a - b$ lies in the kernel).
    $endgroup$
    – user3482749
    Dec 6 '18 at 22:40










  • $begingroup$
    I got it now. Thank you! :)
    $endgroup$
    – Haus
    Dec 6 '18 at 22:40














0












0








0





$begingroup$


When checking if two rings are isomorphic, we check if mapping is homomorphism and then we check if it is bijective (injective and surjective).
In some tasks when checking if isomorphisms, we checked if it is homomorphism, surjective and instead of injective we questioned the kernel (Ker) if it is ideal.



Do we need to question all of this or there are some cases?
I think that if we have ring factor, we check homomorphism, surjective and Ker,
but if we have just ring on the left side, we check if mapping is homomorphism, surjective and injective.



Can anyone explain? Thank you so much!






ring-homomorphism







share|cite|improve this question











$endgroup$




When checking if two rings are isomorphic, we check if mapping is homomorphism and then we check if it is bijective (injective and surjective).
In some tasks when checking if isomorphisms, we checked if it is homomorphism, surjective and instead of injective we questioned the kernel (Ker) if it is ideal.



Do we need to question all of this or there are some cases?
I think that if we have ring factor, we check homomorphism, surjective and Ker,
but if we have just ring on the left side, we check if mapping is homomorphism, surjective and injective.



Can anyone explain? Thank you so much!






ring-homomorphism




ring-homomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 22:49









Bernard

120k740116




120k740116










asked Dec 6 '18 at 22:35









HausHaus

307




307












  • $begingroup$
    Kernel is always an ideal. For injectivity, the thing you check for the kernel is whether or not it is trivial.
    $endgroup$
    – Anurag A
    Dec 6 '18 at 22:37










  • $begingroup$
    So when checking if isomorphism, we look if mapping is homomorphism, surjective, injective and we check kernel. All of this?
    $endgroup$
    – Haus
    Dec 6 '18 at 22:39






  • 1




    $begingroup$
    for isomorphism you need 3 things: homomorphic property, surjectivity and injectivity. Usually kernel helps with injectivity.
    $endgroup$
    – Anurag A
    Dec 6 '18 at 22:40








  • 1




    $begingroup$
    A homomorphism is injective if and only if its kernel is trivial (proof: one direction is obvious. For the other, if $f(a) = f(b)$, then $f(a - b) = f(0) = 0$, so $a - b$ lies in the kernel).
    $endgroup$
    – user3482749
    Dec 6 '18 at 22:40










  • $begingroup$
    I got it now. Thank you! :)
    $endgroup$
    – Haus
    Dec 6 '18 at 22:40


















  • $begingroup$
    Kernel is always an ideal. For injectivity, the thing you check for the kernel is whether or not it is trivial.
    $endgroup$
    – Anurag A
    Dec 6 '18 at 22:37










  • $begingroup$
    So when checking if isomorphism, we look if mapping is homomorphism, surjective, injective and we check kernel. All of this?
    $endgroup$
    – Haus
    Dec 6 '18 at 22:39






  • 1




    $begingroup$
    for isomorphism you need 3 things: homomorphic property, surjectivity and injectivity. Usually kernel helps with injectivity.
    $endgroup$
    – Anurag A
    Dec 6 '18 at 22:40








  • 1




    $begingroup$
    A homomorphism is injective if and only if its kernel is trivial (proof: one direction is obvious. For the other, if $f(a) = f(b)$, then $f(a - b) = f(0) = 0$, so $a - b$ lies in the kernel).
    $endgroup$
    – user3482749
    Dec 6 '18 at 22:40










  • $begingroup$
    I got it now. Thank you! :)
    $endgroup$
    – Haus
    Dec 6 '18 at 22:40
















$begingroup$
Kernel is always an ideal. For injectivity, the thing you check for the kernel is whether or not it is trivial.
$endgroup$
– Anurag A
Dec 6 '18 at 22:37




$begingroup$
Kernel is always an ideal. For injectivity, the thing you check for the kernel is whether or not it is trivial.
$endgroup$
– Anurag A
Dec 6 '18 at 22:37












$begingroup$
So when checking if isomorphism, we look if mapping is homomorphism, surjective, injective and we check kernel. All of this?
$endgroup$
– Haus
Dec 6 '18 at 22:39




$begingroup$
So when checking if isomorphism, we look if mapping is homomorphism, surjective, injective and we check kernel. All of this?
$endgroup$
– Haus
Dec 6 '18 at 22:39




1




1




$begingroup$
for isomorphism you need 3 things: homomorphic property, surjectivity and injectivity. Usually kernel helps with injectivity.
$endgroup$
– Anurag A
Dec 6 '18 at 22:40






$begingroup$
for isomorphism you need 3 things: homomorphic property, surjectivity and injectivity. Usually kernel helps with injectivity.
$endgroup$
– Anurag A
Dec 6 '18 at 22:40






1




1




$begingroup$
A homomorphism is injective if and only if its kernel is trivial (proof: one direction is obvious. For the other, if $f(a) = f(b)$, then $f(a - b) = f(0) = 0$, so $a - b$ lies in the kernel).
$endgroup$
– user3482749
Dec 6 '18 at 22:40




$begingroup$
A homomorphism is injective if and only if its kernel is trivial (proof: one direction is obvious. For the other, if $f(a) = f(b)$, then $f(a - b) = f(0) = 0$, so $a - b$ lies in the kernel).
$endgroup$
– user3482749
Dec 6 '18 at 22:40












$begingroup$
I got it now. Thank you! :)
$endgroup$
– Haus
Dec 6 '18 at 22:40




$begingroup$
I got it now. Thank you! :)
$endgroup$
– Haus
Dec 6 '18 at 22:40










2 Answers
2






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$begingroup$

A mapping is an isomorphism if and only if it is a homomorphism and a bijection. In particular, a homomorphism is injective if and only if its kernel is trivial, so these are exactly equivalent.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    To check a map $f$ is isomorphism you need to do



    (1) $f$ is a homomorphism.



    (2) $f$ is surjective.



    (3) $f$ is injective.



    Equivalently, you can check



    (1) $f$ is a homomorphism.



    (2) $f$ is surjective.



    (3') $ker f=0$.



    That is because (3) and (3') are equivalent once you prove (1).






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      1












      $begingroup$

      A mapping is an isomorphism if and only if it is a homomorphism and a bijection. In particular, a homomorphism is injective if and only if its kernel is trivial, so these are exactly equivalent.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        A mapping is an isomorphism if and only if it is a homomorphism and a bijection. In particular, a homomorphism is injective if and only if its kernel is trivial, so these are exactly equivalent.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          A mapping is an isomorphism if and only if it is a homomorphism and a bijection. In particular, a homomorphism is injective if and only if its kernel is trivial, so these are exactly equivalent.






          share|cite|improve this answer









          $endgroup$



          A mapping is an isomorphism if and only if it is a homomorphism and a bijection. In particular, a homomorphism is injective if and only if its kernel is trivial, so these are exactly equivalent.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 22:39









          user3482749user3482749

          4,266919




          4,266919























              1












              $begingroup$

              To check a map $f$ is isomorphism you need to do



              (1) $f$ is a homomorphism.



              (2) $f$ is surjective.



              (3) $f$ is injective.



              Equivalently, you can check



              (1) $f$ is a homomorphism.



              (2) $f$ is surjective.



              (3') $ker f=0$.



              That is because (3) and (3') are equivalent once you prove (1).






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                To check a map $f$ is isomorphism you need to do



                (1) $f$ is a homomorphism.



                (2) $f$ is surjective.



                (3) $f$ is injective.



                Equivalently, you can check



                (1) $f$ is a homomorphism.



                (2) $f$ is surjective.



                (3') $ker f=0$.



                That is because (3) and (3') are equivalent once you prove (1).






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  To check a map $f$ is isomorphism you need to do



                  (1) $f$ is a homomorphism.



                  (2) $f$ is surjective.



                  (3) $f$ is injective.



                  Equivalently, you can check



                  (1) $f$ is a homomorphism.



                  (2) $f$ is surjective.



                  (3') $ker f=0$.



                  That is because (3) and (3') are equivalent once you prove (1).






                  share|cite|improve this answer









                  $endgroup$



                  To check a map $f$ is isomorphism you need to do



                  (1) $f$ is a homomorphism.



                  (2) $f$ is surjective.



                  (3) $f$ is injective.



                  Equivalently, you can check



                  (1) $f$ is a homomorphism.



                  (2) $f$ is surjective.



                  (3') $ker f=0$.



                  That is because (3) and (3') are equivalent once you prove (1).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 22:42









                  Eclipse SunEclipse Sun

                  7,5021437




                  7,5021437






























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