Need help in a question regarding polynomial oracle reductions Prove the following: 

If there is a polynomial oracle reduction from $S1$ to $S2$:

a. If $S2\in\ P$ so $S1\in\ P$

b. If $S2\notin\ P$ so $S1\notin\ P$

The way I see it - If there is a polynomial oracle reduction from $S1$ to $S2$, you can solve $S1$ with $n^k$ questions to $S2$. If $S2\in\ P$, and $S2$ is solved with $n^k$ actions, it means $S1$ can be solved in $n^{n^k}$ actions - $\in\ P?!$

(I translated the phrase 'polynomial oracle reduction' from hebrew, so ask me if you don't understand what I meant).
Hint: Since $S_1$ runs in polynomial time, say $O(n^a)$, every input to the oracle has polynomial length $O(n^a)$. If $S_2 \in \mathsf{P}$, say can be decided in polynomial times $O(n^b)$, then we can simulate all the oracle calls in $O(O(n^a)^b)$.