Let $f$ be a surjective ring homomorphism from $R$ to $S$ and $P$ be a prime ideal of $R$ such that $\ker(f)<P$; then $f(P)$ is a prime ideal of $S$.
Any hints? I don't know how to use the property $\ker(f)<P$
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Since $f$ is surjective, the ideals of $S$ have the form $f(I)$, for a unique ideal $I\supseteq\ker(f)$.
Suppose $f(P)\supseteq f(I)f(J)=f(IJ)$. Since both $P$ and $IJ$ contain $\ker(f)$ this is the same as $IJ\subseteq P$. Can you finish?