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    I have been told that:

    An isomorphism of rings is a surjective and injective ring homomorphism.

    Also:

    A ring homomorphism is injective  \Leftrightarrow ker(\phi) = 0

    where Phi is a homomorphism of rings.

    from this can I take that is  ker(\phi) = 0 the rings are isomorphic??

    Or to put it another way, we know if the rings are homomorphic, and injective, then  ker(\phi) = 0 , and we know that an isomorphism of rings is achieved as stated above, so have I assumed all homomorphisms are surjective?? If so, is this true or not??

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    This isn't right - you need to have surjectivity as well as injectivity to have an isomorphism.

    As an example, consider the map \phi : \mathbb{Z} \rightarrow \mathbb{Z}^2 defined by \phi(n)=(n,0). Clearly \phi is a homomorphism, and the kernel is trivial. But also it should be clear that these two rings aren't isomorphic.
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    (Original post by Mark13)
    This isn't right - you need to have surjectivity as well as injectivity to have an isomorphism.

    As an example, consider the map \phi : \mathbb{Z} \rightarrow \mathbb{Z}^2 defined by \phi(n)=(n,0). Clearly \phi is a homomorphism, and the kernel is trivial. But also it should be clear that these two rings aren't isomorphic.
    I see, so I clearly have assumed all homomorphisms are surjective, which by your example above is shown to be incorrect

    Thanks.
 
 
 
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