Field homomorphism is trivial or injective

Written by Patrick Stevens last updated

Summaries

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  • Summary

    A structure-preserving map between two fields turns out either to be totally trivial (sending every element to ) or it preserves so much structure that its image is an embedded copy of the domain. More succinctly, if is a field homomorphism, then is either the constant map, or it is injective.

  • Technical

    Let and be fields, and let be a field_homomorphism. Then either is the constant map, or it is injective.

  • Let and be fields, and let be a field_homomorphism. Then one of the following is the case:

    • is the constant map: for every , we have .
    • is injective.

    Proof

    Let be non-constant. We need to show that is injective; equivalently, for any pair of elements with , we need to show that .

    Suppose . Then we have ; so because is a field homomorphism and so respects the "subtraction" operation. Hence in fact it is enough to show the following sub-result:

    Suppose is non-constant. If , then .

    Once we have done this, we simply let .

    Proof of sub-result

    Suppose but that is not , so we may find its multiplicative inverse .

    Then ; but is a homomorphism, so , and so .

    But this contradicts that the image of the identity under a group homomorphism is the identity, because we may consider to be a group homomorphism between the multiplicative groups and , whereupon is the identity of , and is the identity of .

    Our assumption on was that , so the contradiction means that if then . This proves the sub-result and hence the main theorem.