Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Thanks.

share|improve this question
3  
See page $38$, example $3$. –  spohreis Apr 26 '12 at 13:59
    
@spohreis whoops thanks :D –  fpqc Apr 26 '12 at 14:01
add comment

1 Answer

up vote 5 down vote accepted

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

share|improve this answer
    
Thanks guess I'm blind :D –  fpqc Apr 26 '12 at 14:08
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.