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I am now going through some problems in Atiyah - Macdonald Chapter 3. In problems 21 and 23 of chapter 3, they use the notation $A_f$ to mean something I don't know. I have not seen this before.

Problem 21 (i): Let $A$ be a ring, $S$ a multiplicatively closed subset of $A$ and $\phi : A \rightarrow S^{-1}A $ the canonical homomorphism. Show that $\phi^{\ast}: \operatorname{Spec} (S^{-1}A) \to \operatorname{Spec} (A)$ is a homeomorphism of $\operatorname{Spec}(S^{-1}A)$ onto its image in $X = \operatorname{Spec}(A)$. Let this image be denoted by $S^{-1}X$. In particular, if $f \in A$, the image of $\operatorname{Spec}(A_f)$ in $X$ is the basic open set $X_f$.

Now $f$ is any element of $A$, so it can't be that $A_f$ means "localisation at $f$" since this is meaningless (unlike $A_\mathfrak{p}$ which means localisation at the prime ideal $\mathfrak{p}$).

In problem 23, this notation $A_f$ is used again in the following way:

23 (i): If $U = X_f$, show that the ring $A(U) = A_f$ depends only on $U$ and not $f$.

$X_f$ in Atiyah - Macdonald means a basic open set.

With these question to provide some context, does anyone know what $A_f$ means? I can't find this notation used anywhere in the text before this.


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See page $38$, example $3$. – spohreis Apr 26 '12 at 13:59
@spohreis whoops thanks :D – user38268 Apr 26 '12 at 14:01
up vote 5 down vote accepted

$A_f$ denotes the localization $S^{-1}A$ where $S = \{ f^n : n \in \mathbb{N} \} = \{1, f, f^2, \dots \}$.

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Thanks guess I'm blind :D – user38268 Apr 26 '12 at 14:08

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