$A$ prime in $S$ implies that $\phi^{-1}(A)$ prime in $R$ ; $A$ maximal in $S$ implies that $\phi^{-1}(A)$ maximal in $R$

Question

Suppose $R,S$ are commutative rings with unities. Let $\phi$ be a ring homomorphism mapping $R\to S$ and let $A\subset S$ be an ideal. How can I start the proofs for:

Showing that $A$ prime in $S$ implies that $\phi^{-1}(A)$ prime in $R$

and that

showing that $A$ maximal in $S$ implies that $\phi^{-1}(A)$ maximal in $R$

I feel these questions' proofs will be very similar so that's why they are both listed here under one question.

What I know about these:

By the definition I know that a prime ideal $A$ of $S$ means that if I have $a,b\in S, ab\in A$ then $a\in A$ or $b\in A$.

Also by definition I know that a maximal ideal $A$ in $S$ is a proper ideal such that $A\subset B\subset R$ gives us that $B=A$ or $B=R$ where $B$ is an ideal of $R$.

I also know that since $\phi$ is a homomorphism, the two binary operations are preserved.

If someone could help get me started, it would be appreciated. Thank you.

Answer 1

if you know that $\phi^{-1}A$ is an ideal of $R$, then to show it is prime we argue as follows: $$ ab \in \phi^{-1}A \to \phi(ab) \in A \to \phi(a)\phi(b) \in A \to \phi(a) \in A \lor \phi(b) \in A \\ \to a \in \phi^{-1}A \lor b \in \phi^{-1}A $$