Show that $PS$ is the smallest ideal of $S$ that contains $h(P)$

Question

Let $h: R\to S$ be a ring homomorphism. Let $P\subset R$ be a prime ideal. Denote by $PS$ to the set $$PS=\{s\in S: s=\sum_{i \text{ finite}}h(r_i)s_i, r_i\in R, s_i\in S\}$$

Show that $PS$ is the smallest ideal of $S$ that contains $h(P)$.

I have to prove that $PS=\bigcap_{J\subset S \text{ ideal}, h(P)\subset J}J$ but I do not know how to do this, could someone help me please? Thank you very much.

Answer 1

$\newcommand\Z{\mathbb Z}$It can't be true. As Robert Lewis pointed out in the comments the right hand side is not related to $P$ at all.

Here is an easy counterexample take $R=S=\Z$ and $h$ to be the identity map and $P$ to be the ideal $2\Z$. In this case $PS=\Z$ but the smallest ideal containing $2\Z=h(2\Z)$ is $2\Z$.