1. ## Ideal and homomorphism

1. Let f: R --> S be a ring homomorphism
Let J be an ideal of S. Prove that $\displaystyle f^{-1}(J)$ 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 $\displaystyle f^{-1}(J)$ 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 $\displaystyle J = \{ j = f(i) : i \in I\}$

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

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

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

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 $\displaystyle f^{-1}(J)$ 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 $\displaystyle f^{-1}(J)$ doesn't exist. In this case, the statement would not be true. Am I right?
there's nothing wrong with the question. by definition: $\displaystyle f^{-1}(J)=\{a \in R: \ f(a) \in J \},$ which is defined for any function $\displaystyle f$. to show that $\displaystyle f^{-1}(J)$ is an ideal of R, suppose that $\displaystyle a,b \in f^{-1}(J)$ and

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

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

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

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

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

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

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

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 $\displaystyle J = \{ j = f(i) : i \in I\}$
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 $\displaystyle f(I)$ is an ideal of $\displaystyle f(R)$.

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