every ideal that is not intersect $S$ where $S$ is multiplicative closed is prime ideal? I know that maximal ideals among those are prime ideals. But what about other ideals that is not intersect $S$. Are they also prime ideal?
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Even better: $I\cap{1} = \emptyset$ for any $I\neq R$. – Andrew Dudzik Oct 05 '14 at 17:42
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For starters, you can take any power of those primes. But really, the ($S$-saturated) ideals that do not intersect $S$ correspond ("are") the ideals of $S^{-1}R$. Take $S = \{1, 6, 6^2, \ldots \}$ in $\mathbb{Z}$. The ideals that do not intersect $S$ are $(a)$ where $a$ is divisible by some prime $\ne 2, 3$. The $S$ saturation of that ideal will be $a'$, the part off $2,3$. So you can take $a = 3\cdot 5^2 \cdot 7$, or $a' = 5^2 \cdot 7\ \ $ Of course, the maximal ones that do not intersect $S = \{1, 6, 6^2, \ldots \}$ are $(p)$, where $p$ is a prime $\ne 2,3$.
orangeskid
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