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I am doing exercise 3.23 in Atiyah Macdonald and in the first part of the problem they ask to show that the ring $A_f = S^{-1}A$ where $S = \{1,f,f^2 \ldots \}$ depends only on the choice of the basic open set $X_f$ and not on $f$. For reference,

$$X_f \stackrel{\text{def}}{\equiv} \{P \in \operatorname{Spec}(A) : f \notin P\}.$$

I interpret this as saying that if there is another $g \in A$ such that $X_g = X_f$, then $A_f \cong A_g$. Since $X_f = X_g$ we deduce that $\operatorname{rad}\Big((f)\Big) = \operatorname{rad}\Big((g) \Big).$ From these equalities of radicals, this means that there are equations

\begin{eqnarray*} f^n &=& ag \hspace{5mm} \\ g^m &=& bf \end{eqnarray*}

for some $n,m \in \Bbb{N}$ such that $n,m >0$ and $a,b \in A$. Now we may suppose that $f$ and $g$ are not nilpotent for then $A_f \cong A_g \cong 0$. To show that $A_f \cong A_g$ I am trying to use Corollary 3.3 of Atiyah - Macdonald:

${\color{blue}{\text{Corollary 3.3: If $\tilde{g}: A \rightarrow B$ is a ring homomorphism such that}}}$

${\color{blue}{\text{(i) $s \in S \implies \tilde{g}(s)$ is a unit }}}$

${\color{blue}{\text{(ii) $\tilde{g}(a) = 0\implies as = 0$ for some $s \in S$}}}$

${\color{blue}{\text{(iii) Every element in $B$ is of the form $\tilde{g}(a)\tilde{g}(s)^{-1}$ for some $a\in A$ and $s \in S$ }}}$

${\color{blue}{\text{then there is a unique isomorphism $h:S^{-1}A \rightarrow B$ such that $\tilde{g}$ factorises }}}$ ${\color{blue}{\text{through the localisation $S^{-1}A$.}}}$

In our case we can set $B = P^{-1}A$ where $P = \{1,g,g^2, \ldots\}$.

$\textbf{Difficulty:}$ I have been spending some time now trying to define such a map $\tilde{g} : A \longrightarrow P^{-1}A$. Now it seems to me from those two equations above that there are only choices (up-to multiplication by some constant) orf such a map $\tilde{g}$. The problem is every time I define my $\tilde{g}$, property ${\color{blue}{(iii)}}$ is not satisfied. For example, I have tried to define for $x \in A$

$$ \tilde{g}(x) = \frac{bx}{a} \hspace{4mm} \text{or} \hspace{4mm} \tilde{g}(x) = \frac{x^n}{ab} \hspace{4mm} \text{or} \hspace{4mm} \tilde{g}(x) = \frac{bx}{ag^{m-1}}.$$

Suppose I try to write some $x/g^k$ in $P^{-1}A$ as $g(x')g(s)^{-1}$ for some $a' \in A$ and $s \in S$. Then one candidate we have is $x' =x$ and $s = f$. But then we only get

$$g(x)g(f)^{-1} = \frac{bx}{g^m}.$$

If I try say $g(xf)g(f)^{-1}$ to remove the $g^m$ in the denominator the problem is that I just end up with $x$. The problems are similar for the second and third $\tilde{g}$ I defined above. How can I get around this? Please do not give it all away as I would like to enjoy some of the fun!

Thanks.

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Truly, your dedication to that book is moving. :) –  Galois Group May 9 '12 at 13:31
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@FortuonPaendrag what do you mean? –  user38268 May 9 '12 at 13:34
    
Your interpretation is correct. I think you are overcomplicating things. All you have to do is show that an $A$-algebra in which $f$ is invertible also has an inverse for $g$ and vice-versa. Then you will have shown that $A_f$ has the universal property of $A_g$ and vice-versa, so they must be isomorphic. –  Zhen Lin May 9 '12 at 13:37
    
@ZhenLin Do you mean to say I should define a map directly from $A_f$ to $A_g$? I do not understand what you mean by "an $A$ - algebra in which $f$ is invertible". –  user38268 May 9 '12 at 13:39
    
@BenjaminLim: Look up the definition of $A$-algebra, and it should be clear what I mean. I absolutely do not mean for you define a map directly – that it what the universal property does for you. –  Zhen Lin May 9 '12 at 14:06

2 Answers 2

up vote 2 down vote accepted

I think you have to define $\tilde g\colon A \to A_g$ to be the canonical map $\tilde g(x) = x/1$. What is the inverse of $\tilde{g}(f)$? In order to check (iii), you just have to show that $1/g$ is of the desired form. Look at $\tilde{g}(a)\tilde g(f^n)^{-1}$ and use both of your equations. [This seems like a reasonable guess in view of $f^n = ag$, and indeed it does work.]

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I hope I haven't given too much away -- I think there's still a bit to untangle here for you. And now I'm late for the train! –  Dylan Moreland May 9 '12 at 13:55
    
Sorry I meant that indeed it does, for example the element $x/g^k \in A_g$ is the image of $\tilde{g}(a^k x)\tilde{g}((f^n)^k)^{-1}$. –  user38268 May 9 '12 at 23:18
    
I just realised something. If $\tilde{g}$ is our canonical map then $\tilde{g}(f) = \frac{f}{1}$. Now for this to be invertible we need that $f \in \{1,g,g^2,\ldots \}$. Now we have that $g^m = bf$ so that if I can write $f = \frac{g^m}{b}$ I should be done. The problem is that I don't know if $b \in \{1,g,g^2,\ldots \}$ so how can I do this? –  user38268 May 10 '12 at 4:18
    
@BenjaminLim Hopefully I'm not misinterpreting you: it is not necessarily the case that the only invertible elements of $A_g$ are the $g^i$. I'll try to think of a better reason for why $f$ is invertible, beyond "I can write down an inverse, $b/g^m$." –  Dylan Moreland May 10 '12 at 5:39
    
It's ok I got it now with Arturo's comment above. Up to now I was assuming this tacitly all along! –  user38268 May 10 '12 at 5:59

You can use universal properties.

You know that $f^n = ag$ for some $a$; that means that $g$ is invertible in $A_f$, and therefore there is a unique homomorphism $\varphi\colon A_g\to A_f$ that commutes with the canonical maps $A\to A_g$ and $A\to A_f$; symmetrically, since $g^m=bf$ for some $b$, $f$ is invertible in $A_g$ so there is a unique homomorphism $\psi\colon A_f\to A_g$ that commutes with the canonical maps.

Now consider $\psi\circ\varphi\colon A_f\to A_f$. This is the unique map that commutes with the canonical map $A\to A_f$, hence $\psi\circ\varphi$ is the identity. Likewise, $\varphi\circ\psi\colon A_g\to A_g$ must be the identity, by the uniqueness clause of the universal property of localizations. Therefore, $\varphi=\psi^{-1}$, so $A_f\cong A_g$, as desired.

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Thanks for your answer. How do you know that $f$ is invertible in $A_g$ from the equation $g^m = bf$? Don't I need that $b \in \{1,g,\ldots\}$? –  user38268 May 10 '12 at 4:21
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@Benjamin: $g$ is invertible in $A_g$, hence $g^m$ is invertible in $A_g$ (being a product of units). Since $f$ divides an invertible element, it is itself invertible: if $x$ divides $u$ and $u$ is invertible, then there exists $v$ with $uv=1$, there exists $y$ with $xy=u$, so $x(yv) = (xy)v = uv = 1$. –  Arturo Magidin May 10 '12 at 4:23
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@Benjamin: No, there are other elements besides those in $\{1,g,\ldots\}$ that may be invertible. Take the case of $\mathbb{Z}$, with $g=22$. Then $121$ is invertible, even though $121\neq 22^k$ for any $k$, since $\frac{1}{121} = \frac{4}{(121)(4)} = \frac{4}{22^2}$. –  Arturo Magidin May 10 '12 at 4:27
    
Thanks man I learned something from you! That example was really good. –  user38268 May 10 '12 at 4:34

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