The following is a statement about extended and contracted ideals under localisation from Atiyah, McDonald
They use that for any ideal I considering the homomorphism $f: A \rightarrow S^{-1}A$ defined by $f(x)= \frac{x}{1}$ we have that $I^e=\Phi (I) S^{-1}R = S^{-1} \Phi (I) = S^{-1} I $. And that $x\in I^{ec}=(S^{-1}I)^c $ iff $\frac{x}{1} = \frac{a}{s}$ for some $s\in S$ and some $a\in I$. That is, there exist $t\in S$ such that $t(sx-a)=0$. Since $0\in I$ and $at\in I$ by construction, this is equivalent to that $tsx\in I$. Hence we have the following equality: $I^{ec}=\bigcup_{s\in S}(I:s)$
Now, this is what confuses me, (proposition 4.8. ii)) let $S$ be a multiplicative closed subset of $R$, and let $Q$ be a $P$-primary ideal of $R$. If $S\bigcap P =\emptyset $ then the contraction of $S^{-1}Q= Q$
In the proof they state that since $s\notin P$ and $as \in Q$ this imply that $a\in Q$. For me this doesn't make sence shouldn't it be $s\notin P$ and $as \in Q$ imply $a\in P$? But then the earlier discussion shows that $(S^{-1}Q)^c=Q^{ec}=\bigcup_{s\in S}(Q:s)=P$ Contradicting the whole proposition...