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Let $$D(f)=\operatorname{Spec}A - V(f)=\{p\in \operatorname{Spec} A:f\notin p\}$$ be the principal open subsets of $\operatorname{Spec}A$ with the Zariski topology. I'm trying to prove that $D(f)\backsimeq \operatorname{Spec} A_f$, where $A_f$ is the localization of $A$ at $f$.

If we consider the canonical homomorphism $\varphi:A\to A_f$ given by $\varphi (a) = \frac {a}{1}$, we can define $\varphi':\operatorname{Spec} A_f\to \operatorname{Spec} A$, given by $\varphi'(p)=\varphi^{-1}(p),\forall p\in \operatorname{Spec}A_f$.

I'm having trouble proving that this map is surjective.

Thanks.

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    $\begingroup$ How come you need surjectivity? it's not surjective. You're supposed to show that $Spec A_f=Spec A \setminus V(f)$ and your target should be $Spec A \setminus V(f)$ not the whole space $SpecA.$ $\endgroup$ Commented Sep 13, 2013 at 7:45

2 Answers 2

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Prove more generally that for a multiplicative subset $S \subseteq A$ the maps

$\begin{array}{c} \{\mathfrak{p} \in \mathrm{Spec}(A) : \mathfrak{p} \cap S = \emptyset\} & \cong & \mathrm{Spec}(A_S) \\ \mathfrak{p} & \longrightarrow & \mathfrak{p} A_S \\ \phi^{-1}(\mathfrak{q}) & \longleftarrow & \mathfrak{q} \end{array}$

are well-defined, continuous and inverse to each other.

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  • $\begingroup$ Why $\mathfrak p A_s$ is in $\text Spec(A_S)$? $\endgroup$
    – user42912
    Commented Sep 13, 2013 at 8:57
  • $\begingroup$ Just calculate. Alternatively, $A_S / \mathfrak{p} A_S = (A/\mathfrak{p})_S$ is a localization of an integral domain at a set not containing $0$, hence an integral domain. $\endgroup$ Commented Sep 13, 2013 at 15:10
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It's not surjective. The image will only consist of $D(f)$, i.e., those prime ideals of $A$ that do not contain $f$. To see that the image is indeed $D(f)$, take a prime ideal $p \in D(f)$ and check that $pA_f$ is a prime ideal of $A_f$ whose preimage under $\varphi$ is $p$.

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