# Contraction of the ideal $S^{-1}Q$

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...

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The real statement, is that if $s \notin P$ and $as \in Q$ would imply $a \in Q$.
But, if $s\notin P$ and $as \in Q$ then this would only imply $a^n\in Q$ since $Q$ is not prime, only primary. This is what I don't understand. Could you explain? –  harajm Nov 5 '12 at 19:15
No. If $a$ is not in $Q$, then $s^n \in Q$ for some $n$ which means that $s \in P$. –  user27126 Nov 5 '12 at 19:16