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