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If $A$ is a ring and $S=\{1,f,f^2,f^3,...\}$ a multiplicative set of $A$. Prove that $Spec(A_f)=(\mathfrak{V}((f)))^c$. Notation: $A_f=S^{-1}A$ and $\mathfrak{V}((f))=\{P \in Spec(A): P \supset (f)\}$

My attemp:

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

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

So the exercise is equivalent to prove: $$\{P \in spec(A): P\cap S =\emptyset\}=\{P \in Spec(A): P \nsupseteq (f)\}$$

but I can't continue so I'd appreciate if somebody could help me.

Thanks in advance.

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2 Answers 2

up vote 5 down vote accepted

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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+1 Great answer! I say that because my answer is almost the same as yours and my answer is great :) –  Amitesh Datta Apr 23 at 12:38
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