If $f(x) \in f(p)$ then $x \in p$, for $p$ a prime ideal and $f$ a ring homomorphism

Question

Let $f: A\to B$ be a ring homomorphism and $p$ be a prime ideal in $A$. Here $A,B$ are commutative rings. Prove or disprove that if $x\in A, f(x) \in f(p)$ then $x\in p$.

Def. [Introduction to commutative algebra. ATIYAH & MACDONALD] $f$ is a ring hommomorphism if: $ i) f(x+y)=f(x)+f(y) \\ ii) f(xy)=f(x)f(y) \\ iii) f(1)=1.$

This question just came up in my mind but I don't know whether it is true or not. Can you prove or give me a counter example?

Answer 1

This is false if $P\subsetneq \ker{f}$.

For example, take the usual map $f:\mathbb{Z}\to\mathbb{Z}/2$, and consider $x=2$, $P=(0)$.