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 if $I\subset R$ is an ideal then $I\subset h^{-1}(IS)$. Give an example in the that inclusion is proper
I am confused with this problem, I think that $IS=\{s\in S: s=\sum_{i \text{ finite}}h(a_i)s_i, a_i\in I, s_i\in S\}$ but I do not know if I am reasoning correctly, in the case where $IS=\{s\in S: s=\sum_{i \text{ finite}}h(a_i)s_i, a_i\in I, s_i\in S\}$,
Take $a\in I$, then $a=h^{-1}(h(a))$ where $h(a)\in IS$ ?
What's wrong with my attempt to prove the other contention?
we take $s\in IS$ then $s=\sum_{i \text{ finite}}h(a_i)s_i$, so $h^{-1}(s)=h^{-1}(\sum_{i \text{ finite}}h(a_i)s_i)=\sum_{i \text{ finite}}a_ih^{-1}(s_i)\in I$ and so P then P and P is an ideal.
What is the example that serves to make the inclusion proper? Thank you very much.
The obvious problem with your proof is that $h$ need not hit some of the $s_i$ (so $h^{-1}(s_i)$ is empty) but $\sum h(a_i)s_i$ could be in the image of $h$. A easy way to get this is to have $s_j=-s_i\notin h(R)$.
Instead, the definition says $IS$ is the ideal of $S$ generated by $h(I)$ (i.e., the smallest ideal that contains the set $h(I)$). Then $I\subset h^{-1}(IS)$ is just a set-theory consequence of how inverse image of function behaves.
A trivial example where the inclusion is proper is where $I$ is not a maximal ideal but $S$ is a field.