Proving the preimage of a ring homomorphism is not empty

Question

I have a question from Abstract Algebra.

Let $R$ and $S$ be commutative rings, and let $\phi:R \to S$ be a ring homomorphism. Show that if $J$ is an ideal of $S$, then $\phi^{-1}(J) := \lbrace r \in R : \phi(r)\in J \rbrace$ is an ideal of $R$.

My Approach

Given any two elements in $\phi^{-1}(J)$, let's say $\phi(r_1), \phi(r_2)$, I can prove that their sum is in $\phi^{-1}(J)$, and given $r_1 \in R, \phi(s) \in \phi^{-1}(J)$, I can prove that their product is in $\phi^{-1}(J)$, thus demonstrating it is an ideal of $R$.

However, my only problem arises in demonstrating that $\phi^{-1}$ is not empty, since that is a necessary condition for the set to be an ideal.

Thank you for your help :-)

Answer 1

Hint: $0 \in J$ for any ideal, since it is a subgroup...

Answer 2

All in one: $\phi^{-1}(J)$ is an ideal because it is the kernel of the composed homomorphism $R\stackrel\phi\longrightarrow S\stackrel{\text{can}}\longrightarrow S/J$