# Localisation and prime ideals

If $A$ is a ring and $S=\{1,f,f^2,f^3,...\}$ a multiplicative set of $A$, prove that $\mathrm{Spec}(A_f)=\mathfrak{V}((f))^c$.

Notation: $A_f=S^{-1}A$ and $\mathfrak{V}((f))=\{P \in \mathrm{Spec}(A): P \supset (f)\}$

My attempt:

On one hand, if $P \in \mathrm{Spec}(A)$ and $P \cap S$ is empty then we identify $\mathrm{Spec}(A_f)=\mathrm{Spec}(S^{-1}A)=\{P \in \mathrm{Spec}(A): P\cap S= \emptyset\}$.

On the other hand, $\mathfrak{V}((f))^c=\{P \in \mathrm{Spec}(A): P \nsupseteq (f)\}$.

So the exercise is equivalent to prove: $$\{P \in \mathrm{Spec}(A): P\cap S =\emptyset\}=\{P \in \mathrm{Spec}(A): P \nsupseteq (f)\},$$ but I can't continue so I'd appreciate if somebody could help me.

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If $P\cap S=\emptyset$, then $P\not\supseteq (f)$, and conversely, if $P\cap S\neq \emptyset$, then there exists $n\in \mathbb{N}$ such that $f^n\in P$. Since $P$ is a prime ideal, this implies that $f\in P$ and $P\supseteq (f)$.

Moral: don't forget to use the definition of "prime" in "prime ideal" when establishing properties of the spectrum! It is essential because the spectrum is after all the collection of all prime ideals.

Hope this helps!

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If $P \in \operatorname{Spec}(A)$ and $P \cap S$ is empty, then $P \nsupseteq (f)$ since $f \not\in P$.

If $P \in \operatorname{Spec}(A)$ and $P \cap S$ is nonempty, then $f^i \in P$ for some $i$. Use the fact that $P$ is a prime ideal to conclude $i \geq 1$ and then $f \in P$.

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