I have been struggling with this problem for some time, so any help is welcomed.
The problem is as follows,
Let $I, J$ be ideals of a commutative ring $R$ and $S$ a multiplicative subset of $R$. Prove that if $J$ is finitely generated then $(I:J)S^{-1}=(IS^{-1}:JS^{-1})$
The only thing that comes to my mind is prove both containments, so I can take $\frac{a}{s_1}\in(IS^{-1}:JS^{-1})$, then $$\frac{a}{s_1}JS^{-1}\subseteq IS^{-1},$$ and $$\frac{ja}{ts_1}\in I$$ for some $\frac{j}{t}\in J$, but for the cointanment I have $\frac{ja}{ts_1}=\frac{i}{s_2}$. Then, we have $jas_2\cdot w=tsi\cdot w$ for some $w\in S$.
And I'm stuck there. I feel I'm going in the right way but I don't know how to conlude the argument. And that's only one containment.
Thanks in advance for those who can help me.
As you say, take $\frac as\in(IS^{-1}:JS^{-1})$. You need to show that there is $t\in S$ such that $ta\in(I:J)$, since then $\frac as=\frac{ta}{ts}\in(I:J)S^{-1}$. Thus we need $taJ\subseteq I$. Here is where it is key that $J$ is finitely generated. Suppose $b_1,\ldots,b_n$ generate $J$. Instead of trying to multiply the entire ideal $J$ into $I$, try each generator at a time. After all, if $ta$ multiplies $b_i$ into $I$ for all $i$, then it multiplies $J=(b_1,\ldots,b_n)$ into $I$. So can you find a $t_i\in S$ such that $t_iab_i\in I$? Once you have the $t_i$, how can you get a single $t\in S$ that will work for all $i$?