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?

1$\begingroup$ Yes. This is true more or less by definition. $\endgroup$ – Qiaochu Yuan Jan 21 '10 at 15:22

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 Lichtenstein Jan 21 '10 at 15:59

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 closeworthy 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 Geraschenko Jan 22 '10 at 2:59

$\begingroup$ I deleted my statement asking for a vote for closure and have now voted to close. $\endgroup$ – Harry Gindi Mar 9 '10 at 1:43

$\begingroup$ I agree with Anton. But oh well, if the question is bothering you so much, just go ahead :) $\endgroup$ – Wanderer Mar 9 '10 at 1:58
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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.