Question

Prove that a field has no proper homomorphic image?​

Answer 1

It is not true that the homomorphic image of an ideal is an ideal in general. ... map will not be an ideal of F, as there are no proper nonzero ideals in a field. In your proof, you never checked that if a∈A and b∈B, then bf(a)∈f(A)

Step-by-step explanation:

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Answer 2

It is not true that the homomorphic image of an ideal is an ideal in general. Take any ring RR that can be embedded into a field FF (such as ZZ in RR). Then the image of a proper ideal J⊂RJ⊂R under the embedding map will not be an ideal of FF, as there are no proper nonzero ideals in a field.

In your proof, you never checked that if a∈Aa∈A and b∈Bb∈B, then bf(a)∈f(A)bf(a)∈f(A), which is also necessary for an ideal.