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I read the book "Algebraic Geometry" by U. Görtz and whenever limits are involved I struggle for an understanding. The application of limits is mostly very basic, though; but I'm new to the concept of limits.

My example (page 60 in the book): Let $A$ be an integral domain. The structure sheaf $O_X$ on $X = \text{Spec}A$ is given by $O_X(D(f)) = A_f$ ($f\in A$) and for any $U\subseteq X$ by

\begin{align} O_X(U) &= \varprojlim_{D(f)\subseteq U} O_X(D(f)) \\ &:= \{ (s_{D(f)})_{D(f)\subseteq U} \in \prod_{D(f)\subseteq U} O_X(D(f)) \mid \text{for all } D(g) \subseteq D(f) \subseteq U: s_{D(f)\big|D(g)} = s_{D(g)}\} \\ &= \bigcap_{D(f)\subseteq U} A_f. \end{align}

I simply don't understand the last equality: In my naive understanding the elements of the last set are "fractions" and the elements of the Inverse Limit are "families of fractions".

Any hint is appreciated.

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Since A is an integral domain, all its localisations embed into its field of fractions. Moreover, let $D(h) \subset D(g)$ and $a, b \in A_g$. Then $a = b$ iff their images in $Frac(A)$ are the same, iff their images in $A_h$ are the same. Hence any element in your "family of fractions" can actually be identified with an element of $Frac(A)$ - and the last line tells you just which fractions you get. –  Tom Bachmann Dec 7 '12 at 17:47
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I think it is a general fact that if you have an inverse system of "sufficiently nice" objects $\left\{A_{i}\right\}_{i\in I}$ in which all the structure maps $\phi_{ij}: A_{j}\to A_{i}$ are injections, then the inverse limit is just the intersection $\bigcap_{i \in I} A_{i}$. –  jmracek Dec 7 '12 at 19:26
    
@TomBachmann -- I feel dumb, but I still don't get it. Let $(s_{D(f)})_{D(f)\subseteq U} \in O_X(U)$. Then each $s_{D(f)}$ of this family can be identified with an element of $\text{Frac}(A)$. But how can the whole family $(s_{D(f)})_{D(f)\subseteq U} \in O_X(U)$ be identified with an element in $\bigcap_{D(f)\subseteq U} A_f$? (The $s_{D(f)}$ of the family don't need to define all the same element in $\text{Frac}(A)$ as the $D(f)$ don't need to be subsets of each other, right?) –  Fred Dec 8 '12 at 8:39
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Yes they do define the same element. This is essentially because your scheme is irreducible, and so all (non-empty) opens meet. In particular, while $D(f)$ and $D(g)$ need not be subsets of each other, $D(fg)$ is a subset of both (and $fg$ is not zero since this is an integral domain). –  Tom Bachmann Dec 9 '12 at 9:33
    
@TomBachmann -- Thank you very much, this is solved. But I can't mark your comment as accepted answer... –  Fred Dec 9 '12 at 11:43
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2 Answers

My personal advice is to study a bit of category theory: it will let you understand all this stuff in a very clearer way. In fact you can easily realize that the first equality is not a definition, but a way to express a limit of an arbitrary presheaf, while the second is an isomorphism, not exactly an equality, given by the universal property defining limits. I started with Hartshorne, but without category theory as a background it's just like wandering in the dark without even a candle with you.

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Is not the best way of defining the structure sheaf. It just uses the fact that $D(f)$ form base for the topology. To see the last isomorphism if you want you can define the sheaf of regular functions $\mathscr{R}(U)=\cap_{x\in X}A_{\frak{p}_x}$. The correct way of understanding this is that for each open it gives you the ring of regular functions on $U$ and that can be seen it as functions that are regular at each point of $U$. Notice that $\mathscr{R}(U)$ is the same as the last intersection and you have a natural map $\mathscr{R}\rightarrow \mathscr{O}_X$ that for $U\subset X$ sends $\phi\in \cap_{D(f)\subset U}A_f$ to $(\phi\mid D(f)\subset U)\in \prod A_f$ that stalk-wise is an isomorphism.

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