Homomorphism in Rings

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I have the following true/false statement from a worked example:



If $F_1, F_2$ are fields and $phi: F_1 rightarrow F_2$ is a homomorphism, then $phi$ is either the zero map or an isomorphism.



The solution says that this is False and gives the identity map $mathbbR rightarrow mathbbC$ as a counterexample.



I am a bit confused by this. Obviously the identity map is not the zero map. When I try to check the conditions for isomoporphism I get confused on the domainrange over which the identity map is defined.

Can someone explain why this example disproves the statement?

Thank you in advance for any answers/comments.




abstract-algebra ring-theory ring-isomorphism




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  • Conventionally, a field homomorphism maps $1$ to $1$, so it can't be the zero map.
    – Bernard
    Aug 20 at 9:27














up vote
2
down vote

favorite












I have the following true/false statement from a worked example:



If $F_1, F_2$ are fields and $phi: F_1 rightarrow F_2$ is a homomorphism, then $phi$ is either the zero map or an isomorphism.



The solution says that this is False and gives the identity map $mathbbR rightarrow mathbbC$ as a counterexample.



I am a bit confused by this. Obviously the identity map is not the zero map. When I try to check the conditions for isomoporphism I get confused on the domainrange over which the identity map is defined.

Can someone explain why this example disproves the statement?

Thank you in advance for any answers/comments.




abstract-algebra ring-theory ring-isomorphism




share|cite|improve this question




















  • Conventionally, a field homomorphism maps $1$ to $1$, so it can't be the zero map.
    – Bernard
    Aug 20 at 9:27












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I have the following true/false statement from a worked example:



If $F_1, F_2$ are fields and $phi: F_1 rightarrow F_2$ is a homomorphism, then $phi$ is either the zero map or an isomorphism.



The solution says that this is False and gives the identity map $mathbbR rightarrow mathbbC$ as a counterexample.



I am a bit confused by this. Obviously the identity map is not the zero map. When I try to check the conditions for isomoporphism I get confused on the domainrange over which the identity map is defined.

Can someone explain why this example disproves the statement?

Thank you in advance for any answers/comments.




abstract-algebra ring-theory ring-isomorphism




share|cite|improve this question












I have the following true/false statement from a worked example:



If $F_1, F_2$ are fields and $phi: F_1 rightarrow F_2$ is a homomorphism, then $phi$ is either the zero map or an isomorphism.



The solution says that this is False and gives the identity map $mathbbR rightarrow mathbbC$ as a counterexample.



I am a bit confused by this. Obviously the identity map is not the zero map. When I try to check the conditions for isomoporphism I get confused on the domainrange over which the identity map is defined.

Can someone explain why this example disproves the statement?

Thank you in advance for any answers/comments.





abstract-algebra ring-theory ring-isomorphism





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asked Aug 20 at 9:18









Nick

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  • Conventionally, a field homomorphism maps $1$ to $1$, so it can't be the zero map.
    – Bernard
    Aug 20 at 9:27
















  • Conventionally, a field homomorphism maps $1$ to $1$, so it can't be the zero map.
    – Bernard
    Aug 20 at 9:27















Conventionally, a field homomorphism maps $1$ to $1$, so it can't be the zero map.
– Bernard
Aug 20 at 9:27




Conventionally, a field homomorphism maps $1$ to $1$, so it can't be the zero map.
– Bernard
Aug 20 at 9:27










2 Answers
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So $mathbbR$ is a subset of $mathbbC$ so the identity map is clearly a homomorphism from $mathbbR$ to $mathbbC.$ As you said, it is clearly not the zero map but the map is not surjective! What maps to $i?$ Hence, it cannot be an isomorphism as it is not bijective.






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    The identity map, by definition is a function $i : mathbbC to mathbbC$. With a little abuse of language we refer to the identity map of $mathbbR$ in $mathbbC$ as the restriction of the above function to $mathbbR$, hence the morphism you are actually considering is
    $$
    i_ : mathbbR to mathbbC
    $$



    which is both non zero and non surjective






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      2 Answers
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      2 Answers
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      up vote
      3
      down vote



      accepted










      So $mathbbR$ is a subset of $mathbbC$ so the identity map is clearly a homomorphism from $mathbbR$ to $mathbbC.$ As you said, it is clearly not the zero map but the map is not surjective! What maps to $i?$ Hence, it cannot be an isomorphism as it is not bijective.






      share|cite|improve this answer
























        up vote
        3
        down vote



        accepted










        So $mathbbR$ is a subset of $mathbbC$ so the identity map is clearly a homomorphism from $mathbbR$ to $mathbbC.$ As you said, it is clearly not the zero map but the map is not surjective! What maps to $i?$ Hence, it cannot be an isomorphism as it is not bijective.






        share|cite|improve this answer






















          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          So $mathbbR$ is a subset of $mathbbC$ so the identity map is clearly a homomorphism from $mathbbR$ to $mathbbC.$ As you said, it is clearly not the zero map but the map is not surjective! What maps to $i?$ Hence, it cannot be an isomorphism as it is not bijective.






          share|cite|improve this answer












          So $mathbbR$ is a subset of $mathbbC$ so the identity map is clearly a homomorphism from $mathbbR$ to $mathbbC.$ As you said, it is clearly not the zero map but the map is not surjective! What maps to $i?$ Hence, it cannot be an isomorphism as it is not bijective.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 20 at 9:22









          伽罗瓦

          934615




          934615




















              up vote
              2
              down vote













              The identity map, by definition is a function $i : mathbbC to mathbbC$. With a little abuse of language we refer to the identity map of $mathbbR$ in $mathbbC$ as the restriction of the above function to $mathbbR$, hence the morphism you are actually considering is
              $$
              i_ : mathbbR to mathbbC
              $$



              which is both non zero and non surjective






              share|cite|improve this answer


























                up vote
                2
                down vote













                The identity map, by definition is a function $i : mathbbC to mathbbC$. With a little abuse of language we refer to the identity map of $mathbbR$ in $mathbbC$ as the restriction of the above function to $mathbbR$, hence the morphism you are actually considering is
                $$
                i_ : mathbbR to mathbbC
                $$



                which is both non zero and non surjective






                share|cite|improve this answer
























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  The identity map, by definition is a function $i : mathbbC to mathbbC$. With a little abuse of language we refer to the identity map of $mathbbR$ in $mathbbC$ as the restriction of the above function to $mathbbR$, hence the morphism you are actually considering is
                  $$
                  i_ : mathbbR to mathbbC
                  $$



                  which is both non zero and non surjective






                  share|cite|improve this answer














                  The identity map, by definition is a function $i : mathbbC to mathbbC$. With a little abuse of language we refer to the identity map of $mathbbR$ in $mathbbC$ as the restriction of the above function to $mathbbR$, hence the morphism you are actually considering is
                  $$
                  i_ : mathbbR to mathbbC
                  $$



                  which is both non zero and non surjective







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 20 at 9:26









                  Bernard

                  111k635103




                  111k635103










                  answered Aug 20 at 9:25









                  JayTuma

                  1,333118




                  1,333118






















                       

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