If we have a morphism between two affine Schemes $f: X \rightarrow Y$ with $X = $ Spec $A$, and $Y = $ Spec $B$, is it true that $f^{-1}(D(g)) = D(f'(g))$? (where $f'$ is the associated map on the structure sheaves) If so, is there a simple proof? Otherwise, is there any other way to characterize the preimages of distinguished open sets?
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1$\begingroup$ Yes. This is true more or less by definition. $\endgroup$– Qiaochu YuanJan 21, 2010 at 15:22
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5$\begingroup$ The idea of the proof is that D(g) is the locus where g, thought of as a function on Y, doesn't vanish. f'(g) is the function gf on X, so its nonvanishing locus is exactly the preimage of D(g). $\endgroup$– Sam LichtensteinJan 21, 2010 at 15:59
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3$\begingroup$ Though I agree that this isn't a very good question (the answer is, "just work through the definitions"), it doesn't feel close-worthy to me. It's something I can imagine a mathematician outside of algebraic geometry being confused about, and I don't get the feeling that the asker is trying to get somebody else to do his work for him. Though it's elementary (and a bit lazy), it feels like it's fundamentally okay. Then again, maybe I'm just in a good mood. $\endgroup$– Anton GeraschenkoJan 22, 2010 at 2:59
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$\begingroup$ I deleted my statement asking for a vote for closure and have now voted to close. $\endgroup$– Harry GindiMar 9, 2010 at 1:43
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$\begingroup$ I agree with Anton. But oh well, if the question is bothering you so much, just go ahead :) $\endgroup$– WandererMar 9, 2010 at 1:58
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2 Answers
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Yes. You may want to look at Hartshorne's 'Algebraic Geometry' section II.2. For example Proposition II.2.3 discusses this matter.
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This can be also be done using abstract nonsense.
Regard $\operatorname{Spec}$ as a functor $\mathbf{CRing}^{\operatorname{op}}\to\mathbf{Sch}$.
As a left adjoint, $\operatorname{Spec}$ converts pushouts in $\mathbf{CRing}$ to pullbacks.
The pushout of $A\leftarrow B\rightarrow B_g$ is $A\otimes_B B_g\simeq A_{\overline{g}}$ where $\overline{g}$ the image of $g$ in $A$.
The pullback of $\operatorname{Spec}A\rightarrow \operatorname{Spec}B \leftarrow \operatorname{Spec}B_g$ is $f^{-1}\operatorname{Spec}B_g$.
Therefore, $\operatorname{Spec}{A_\overline{g}}$ is isomorphic to $f^{-1}\operatorname{Spec}B_g$.
To show that the underlying sets are the same, the following will suffice :
- The maps $\operatorname{Spec}{A_\overline{g}}\to\operatorname{Spec}A$, $f^{-1}\operatorname{Spec}B_g\to\operatorname{Spec}A$, are inclusions.
- $\operatorname{Spec}A_{\overline{g}}$ is contained in $f^{-1}\operatorname{Spec}B_g$.
