there's nothing wrong with the question. by definition: which is defined for anyfunction. to show that is an ideal of R, suppose that and

so thus: and: because is an ideal of S. thus and

surjectivity of is needed to prove that for any and we have: : since is surjective, there exists such that2. Prove that if f is surjective and I is an ideal of R then f(I) is an ideal.

Let

We know . Therefore, because f is an homomorphism, , and

We have and because f is an homomorphism thus

We have and because f is an homomorphism and that gives and conversely s

Ok I miss some steps because it is tidious to write but you understand what I did. I don't understand where the fact that f is surjective intervene. Can you tell me where and why please?

since is an ideal of R, we have: thus: