Ring homomorphism, maximal ideals

Question

Here's a question from my worksheet. I solved subquestion (1) but can use help with the other 2... And also would appreciate any comments on my answer for subquestion (1).

Let $\psi: R\to S$ be a homomorphism of commutative rings. Let $I \subset S$ be an ideal.

We define $J=\psi^{-1}(I) = \{r\in R: \psi(r) \in I\}$.

1) Show that if $I$ is a prime ideal, then $J$ is also a prime ideal.

2) Let us suppose that $I$ is a maximal ideal. In that case, does $J$ have to be a maximal ideal as well?

3) Suppose that $\psi$ is surjective and that $R$ is a principal ideal ring (meaning every ideal in $R$ is a principal ideal). Show that $S$ is also a principal ideal ring.

My solution for 1)

$ab \in J$ implies $\psi(ab) \in I$ which implies $\psi(a) \in I$ or $\psi(b) \in I$ (because $I$ is a prime ideal), that implies $a \in J$ or $b \in J$ due to the definition of $J$. $QED$.

I could use a hand for the other 2 questions.

Edit: I solved question 2 (I think) Please comment on the solution as I am not sure if its a valid solution!

Let's assume $J$ is not a maximal ideal. That means there is an ideal $T$ such that $J \subset T$, but since $\psi(J) = I$ and $\psi(J) \subset \psi(T)$ then $I \subset \psi(T)$ and so $I$ is not maximal ideal. By contraposition, this proves that $J$ must be maximal ideal if $I$ is a maximal ideal.

Answer 1

2. Your proof is valid if $\psi$ is an epimorphism. If not there is no reason for $\psi(T)$ to be an ideal. But you can fix it by taking ideal $I\cap \psi(R)$ instead of of I. Now $\psi :R\rightarrow \psi(R)$.
3. By isomorphism theorem $S\cong R/\ker(\psi)$ and ideals in $S$ are of the form $I/\ker(\psi)$ for some $I\lhd R$ such that $\ker(\psi) \subset I$. Its easy to see that if $I=(a)$ then $I/\ker(\psi)=(a+\ker(\psi))$.