1. ## Ideal and homomorphism

1. Let f: R --> S be a ring homomorphism
Let J be an ideal of S. Prove that $f^{-1}(J)$ is an ideal
of R.
I can see how to show this is true for an isomorphism but not for an homomorphism.
I think there may be an error in the question. If the homomorphism is not bijective (thus is not an isomorphism) than it is possible that $f^{-1}(J)$ doesn't exist. In this case, the statement would not be true. Am I right?

2. Prove that if f is surjective and I is an ideal of R then f(I) is an ideal.
Let $J = \{ j = f(i) : i \in I\}$

We know $0 \in I\$. Therefore, because f is an homomorphism, $f(0)=0$, $0 \in I$ and $0 \in J$

We have $a,b \in I \implies a+b \in I$ and because f is an homomorphism $f(a+b) = f(a) + f(b)$ thus $f(a)+f(b) \in J$

We have $a \in I, r\in R \implies ar \in I$ and because f is an homomorphism $f(ar) = f(a)f(r)$ and that gives $f(a)f(r) \in J$ and conversely s $f(r)f(a) \in J$

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?

2. Originally Posted by vincisonfire
1. Let f: R --> S be a ring homomorphism
Let J be an ideal of S. Prove that $f^{-1}(J)$ is an ideal
of R.
I can see how to show this is true for an isomorphism but not for an homomorphism.
I think there may be an error in the question. If the homomorphism is not bijective (thus is not an isomorphism) than it is possible that $f^{-1}(J)$ doesn't exist. In this case, the statement would not be true. Am I right?
there's nothing wrong with the question. by definition: $f^{-1}(J)=\{a \in R: \ f(a) \in J \},$ which is defined for any function $f$. to show that $f^{-1}(J)$ is an ideal of R, suppose that $a,b \in f^{-1}(J)$ and

$r \in R.$ so $f(a) \in J, \ f(b) \in J.$ thus: $f(a+b)=f(a)+f(b) \in J,$ and: $f(ra)=f(r)f(a) \in J,$ because $J$ is an ideal of S. thus $a+b \in f^{-1}(J)$ and $ra \in f^{-1}(J).$

2. Prove that if f is surjective and I is an ideal of R then f(I) is an ideal.
Let $J = \{ j = f(i) : i \in I\}$

We know $0 \in I\$. Therefore, because f is an homomorphism, $f(0)=0$, $0 \in I$ and $0 \in J$

We have $a,b \in I \implies a+b \in I$ and because f is an homomorphism $f(a+b) = f(a) + f(b)$ thus $f(a)+f(b) \in J$

We have $a \in I, r\in R \implies ar \in I$ and because f is an homomorphism $f(ar) = f(a)f(r)$ and that gives $f(a)f(r) \in J$ and conversely s $f(r)f(a) \in J$

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?
surjectivity of $f$ is needed to prove that for any $s \in S$ and $a \in I$ we have: $sf(a) \in f(I)$: since $f$ is surjective, there exists $r \in R$ such that $s=f(r).$

since $I$ is an ideal of R, we have: $ra \in I.$ thus: $sf(a)=f(r)f(a)=f(ra) \in f(I).$

3. Originally Posted by vincisonfire
2. Prove that if f is surjective and I is an ideal of R then f(I) is an ideal.
Let $J = \{ j = f(i) : i \in I\}$
Try this excercise. Show that the homomorphic image of an ideal is an ideal of the homomorphic image of the ring, i.e. in less fancy language, show that $f(I)$ is an ideal of $f(R)$.

Now, if $f$ is surjective then $f(R) = R$, from here it follows that $f(I)\triangleleft R$