Let $R$ be a PID. Let $I$ be an ideal in $S^{-1}R$(the localisation). Then there exists an ideal $J$ in $R$ so that $JS^{-1}R=I$
I am sure I might be missing something but I really can't seem to wrap my head around this.
(I have to do the proof of the fact that if we have $R$ a PID then also $S^{-1}R$ is a PID and I would need this step)
Thanks in advance
Consider the canonical map $i : R \to S^{-1} R$. The obvious thing to do is to consider the ideal $J = i^{-1}(I)$, since taking inverse images is the obvious way of transferring data across ring homomorphisms.
Note that $J S^{-1} R$ is the set $\{i(j) \cdot x \mid j \in J, x \in S^{-1}R\}$.
Now consider some $j \in I$. Write $j = i(k) \cdot s^{-1}$ for some $s \in S$ and $k \in R$. I claim that $k \in J$.
This follows from the fact that $i(k) = j \cdot s = \in I$.
Therefore, $I \subseteq J S^{-1}R$.
Conversely, consider $j \in J$ and $x \in S^{-1}R$. Then $i(j) \in I$ by definition, and thus $i(j) \cdot x \in I$. Therefore, $JS^{-1} R \subseteq I$.
So we see that $I = J S^{-1} R$.
Note that we never used the fact that $R$ is a PID. We only require that $R$ is a ring and that $S$ is a multiplicatively closed set.