Does a prime ideal map to a prime ideal? I'm stuck in this problem. 

Let $f: R \rightarrow S$ be a surjective homomorphism of rings with kernel $K$ and let $P$ be a prime ideal in $R$ containing $K$. Show that $f(P)$ is a prime ideal in $S$. 

This problem has already been posted here, but I want to prove it without the corresponding between prime ideals. 
My try is here:
Suppose $ab\in f(P)$. Then since f is surjective, there are $p,q\in R$ such that $f(p)=a,f(q)=b$. So, $f(pq)\in f(P)$, and since $K\subset P$, $pq\in P$. 
Now, I want "$pq\in P$ implies $p\in P$ or $q\in P$", but in order to imply this property, not only P is prime, but also R is commutative. However, it is not guaranteed!
How to prove it? Thanks in advance. 
 A: First, show that $f(P)$ is an ideal:


*

*Additive: Suppose that $a\in f(P)$ and $b\in f(P)$, then, since $f$ is surjective, we can find $c,d\in P$ such that $a=f(c)$ and $b=f(d)$.  Moreover, since $c,d\in P$, $c+d\in P$, so $f(c+d)=a+b\in f(P)$.

*Multiplicative.  Suppose that $a\in f(P)$ and $b\in S$.  Since $a\in f(P)$ and $f$ is surjective, there is some $c\in P$ such that $a=f(c)$.  Since $f$ is surjective, there is some $d\in R$ such that $b=f(d)$.  Now, $ab=f(c)f(d)=f(cd)$.  Since $c\in P$, $cd\in P$, so $f(cd)=ab\in f(P)$.
Now, suppose that $A$ and $B$ are ideals of $S$ and $AB\subseteq f(P)$.  Consider $C=f^{-1}(A)$ and $D=f^{-1}(B)$.  We show that $C$ is an ideal (the proof for $D$ is similar):


*

*Additive: Suppose that $a,b\in C$.  Then $f(a),f(b)\in A$.  Since $A$ is an ideal, $f(a)+f(b)\in A$, so $f(a+b)\in A$, so $a+b\in C$.

*Suppose that $a\in C$ and $b\in R$.  Then $f(ab)=f(a)f(b)$.  Since $a\in C$, $f(a)\in A$.  Since $A$ is an ideal, $f(a)f(b)\in A$, so $f(ab)\in A$, so $ab\in C$.
I claim that $CD\subseteq P$.  It is enough to prove that for all $c\in C$ and $d\in D$, $cd\in P$ (normally, one would need to prove this for all sums of this form, but since $P$ is closed under addition, it is enough to prove this without the sum).  


*

*Since $c\in C$ and $d\in D$, there are $a\in A$ and $b\in B$ such that $a=f(c)$ and $b=f(d)$.  Since $AB\subseteq f(P)$, it follows that $ab\in f(P)$.  Therefore, $f(c)f(d)=f(cd)\in f(P)$.  Therefore, there is some $e\in P$ such that $f(cd)=f(e)$; therefore, $f(cd-e)=0$.  Since $P$ contains the kernel, $cd-e\in P$, and since $e\in P$, it follows that $cd\in P$.  Therefore, $CD\subseteq P$.  


Since $P$ is prime, either $C\subseteq P$ or $D\subseteq P$.  Suppose, wlog, that $C\subseteq P$.  I claim that $A\subseteq f(P)$.  


*

*Let $a\in A$, then since $f$ is surjective, there is some $b\in C$ such that $f(b)=a$.  Since $C\subseteq P$, it follows that $b\in P$, so $a=f(b)\in f(P)$.

