If $Q$ is a radical ideal of $R$ with $Q∩S=∅$, show $QR_S$ is a radical ideal of $R_S$.
Here is what I have done, since $Q$ is radical, let $q\in Q$ , then $q^n\in I$ for some $I$ ideal. let $a\in QR_S$, then $a=q\cdot r/s$ for some $q\in Q$, $r\in R$ and $s\in S$. then $a^n=q^n\cdot r^n/s^n\in IR_S$. since $IR_S$ is a ideal of $R_S$. we are done.
Am I right? and what I have confuse is why we need the condition of $Q∩S=∅$? I didn't use it at all. please help, thank you.
On
You can do this with or without resorting to elements. We outline how to do this with the structure of ideals.
We have $Q$ is radical iff $Q=\bigcap\{P\ |\ Q\subseteq P,\ P\text{ is a prime ideal in }R\}$. An ideal in $R_S$ is prime iff is of the form $PR_S$ for prime ideal $P$ in $R$ such that $P\cap S=\emptyset$. Since $Q\cap S=\emptyset$, the extended ideal $QR_S$ is proper and thus contained in a maximal ideal. Localization preserves the lattice structure of the ideals which are disjoint from $S$, hence $QR_S$ is the intersection of the primes in $R_S$ containing it, i.e., it is a radical ideal.
I prefer this approach since it is conceptual rather than computational intensive.
$Q$ is a radical ideal if $\forall x\in R$ and $n\in \mathbb{N}$ such that $x^n\in Q$ it follows that $x\in Q$. The condition for $Q\cap S=\emptyset$, assuming $S$ is a multiplicatively closed system has to be required for all ideals $P$ for which you construct $PR_S$ (as not to be equal to the whole ring) which can be proved to be an ideal of $R_S$. Otherwise, assume $\exists x\in Q\cap S\Rightarrow x\cdot 1/x\in QR_S\Rightarrow 1\in QR_S\Rightarrow QR_S=R_S$. Now for the radical part, take $q\in R_S$, $q=a/s$, $a\in R$, $s\in S$ such that $\exists n\in\mathbb N$ such that $q^n\in QR_S\Rightarrow a^n/s^n=b/t, b\in Q, t\in S\Rightarrow a^nt=bs^n\in Q\Rightarrow b^ns^n\in Q\Rightarrow a^nt\in Q\Rightarrow $
$a^nt^n\in Q\Rightarrow$ since $Q$ radical, $at\in Q\Rightarrow at/st=a/s\in QR_S$.