Let $R$ be a ring with $1\neq 0$. Let $\varphi: R \rightarrow S$ be a homomorphism of Commutative Rings. If $P$ is a prime ideal of $S$, then prove that $\varphi^{-1}(P)=R$ or $\varphi^{-1}(P)$ is a prime ideal of $R$.
My Question:
Why it has been assumed that $\varphi (R)\cap P \neq\emptyset$ for any homomorphism $ \varphi?$
Moreover, Can you please give me hints on how to proceed to this question?
Also, It has not been specified whether $S$ contains $1\neq 0.$
You want a hint: say that $ab\in \varphi^{-1}(P)$, then you need to deduce that either $a\in \varphi^{-1}(P)$ or $b\in\varphi^{-1}(P)$. Well, what can you say about $\varphi(a),\varphi(b)$ and $\varphi(ab)$? Then $\varphi^{-1}(P)$ is either prime, or is all of $R$.