How to prove that the image of a prime ideal is also a prime ideal 
If $f:A\rightarrow R$ be a ring homomorphism, where $A$ and $R$ are commutative rings. If $f$ is surjective  and $P$ is a prime ideal in $A$,  how to prove that $f(P)$ is a prime ideal in $R$?

This a an exercise I came across while self-studying joseph rotman's book advanced modern algebra.  I searched this website and find similar questions but they have another condition: $P$ contains the kernel of $f$. However, I want to know how to prove this result without that condition. Give thanks to any useful help！
 A: It's impossible: let $\mathfrak m$ be a maximal ideal of $A$ which doesn't contain $\ker f$.  As $B\simeq A/\ker f$, we have
$$f(\mathfrak m)=\mathfrak m \cdot A/\ker f=(\mathfrak m+\ker f)/\ker f=A/\ker f,$$
which is not a prime ideal in the quotient.
A: Suppose $r,s\in R$ and $rs\in f(P)$. Since $f$ is surjective, you have $r=f(a)$ and $s=f(b)$ for some $a,b\in A$, so the condition is
$$
f(a)f(b)=f(c)
$$
for some $c\in P$. In turn this becomes $ab-c\in\ker f$. If $P\supseteq\ker f$, you can conclude $ab-c\in P$, so also $ab\in P$ and it's easy to finish.
Without the assumption $\ker f\subseteq P$ there's no way to prove the statement. For instance, consider $A=\mathbb{Z}$ and $R=\mathbb{Z}/2\mathbb{Z}$, with $f$ the canonical projection. Then $f(3\mathbb{Z})=R$, which is not a prime ideal.
If you think more about the problem, the assertion would be equivalent to the statement that, if $P$ is a prime ideal of $A$ and $I$ is any ideal of $A$, then $P+I$ is prime: just consider the projection $f\colon A\to A/I$. Saying that $f(P)$ is prime in $A/I$ implies that $f^{-1}(f(P))=P+I$ is prime in $A$.
