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If $R$ is commutative ring, $P_1, P_2, \dots, P_n$ prime ideals of $R$ with the property $P_i \not\subseteq \bigcup _{j \not = i} P_j$, $\forall 1\le i\le n$, and $S:=R\setminus(P_1 \cup \cdots \cup P_n)$, then show: $$S^{-1}R \text{ has exactly } n \text{ maximal ideals}.$$

Definition. $S^{-1}R=${${r \over s} : r \in R , s \in S $}.

  • 2
    Try to use this: the prime ideals of $S^{-1}R$ are of the form $S^{-1}P$, where $P$ is a prime ideal of $R$ with $P\cap S=\emptyset$.2012-11-26
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    Note, that "definition" of $S^{-1}R$ is incomplete. There is no meaning for $\frac{r}{s}$ You have to be more specific (and show some properties of $S$) to prove that $S^{-1}R$ is a well-defined ring.2012-11-26
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    $S^{-1} R$ is well-definitenes it is not important for me, i want to count maximal ideals of $S^{-1}R$2012-11-26
  • 0
    If $S^{-1}R$ is not well-defined how can you talk about maximal ideals?2012-11-26
  • 0
    I am confused as to why $S^{-1}R$ is not well-defined. $S$ is the intersection of multiplicative subsets, hence clearly multiplicative. So $S^{-1}R$ is a well defined ring right?2012-11-26

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