1
$\begingroup$

Let $A$ be a ring and let $X=Spec(A)$ and let $U$ be a basic open set in $X$. (i.e. $U=X_f$ for some $f∈A$). If $U=X_f$, show that the ring $A(U)=A_f$ depends only on $U$ and not on $f$.

My Work:

Must prove that if $U=X_f=X_g$ then $A_f\cong A_g$. If $X_f=X_g$, then we can prove that $r((f))=r((g))$. Hence, $f^n=x_1g$ and $g^m=x_2f$ for some $n,m>0$ and $x_1,x_2\in A$. Here $A_f=S^{-1}A$ where $S=\{f^k\}_{k\geq 0}$ and similarly $A_g$ is defined.

I define $\phi: A_f \rightarrow A_g$ by $\displaystyle \phi(\frac{a}{f^k})=\frac{a}{g^k}$. Then I could show that $\phi$ is bijective. And it is clear that $\displaystyle \phi(\frac{a}{f^k}\frac{b}{f^r})=\phi(\frac{a}{f^k})\phi(\frac{b}{f^r})$.

But in order to show the other property of a homomorphism I must prove that $\displaystyle \phi(\frac{a}{f^k}+\frac{b}{f^r})=\phi(\frac{a}{f^k})+\phi(\frac{b}{f^r})$. But i was fail to prove it. Can anyone please help me to prove it?

$\endgroup$
3
  • $\begingroup$ What do you denote $r((f))=r((g))$? $\endgroup$
    – Bernard
    Commented Mar 8, 2015 at 1:33
  • $\begingroup$ I guess that the OP intends that the radical of the ideal generated by $f$ is equal to the one of the ideal generated by $g$. $\endgroup$
    – Olórin
    Commented Mar 8, 2015 at 1:34
  • $\begingroup$ $r((f))$ is the radical of the ideal generated by $f$. $\endgroup$
    – Extremal
    Commented Mar 8, 2015 at 1:42

1 Answer 1

2
$\begingroup$

There is an elegant way of proving this. You prove that if $M$ is an $A$-module then $S^{-1} A$ (resp. $S^{-1} M$) is canonically isomorphic to $S'^{-1} A$ (resp. $S'^{-1} M$) where $S'$ is the multiplicative subset of $A$ consisting of elements $g\in A$ dividing an element of $S =\{f^n\;|\;n\in\mathbf{N}\}$.

Note that this in fact does not depend on the form of $S$. See EGA I, chapitre 1, 1.3, paragraph (1.3.1) before the lemma (1.3.2) for what you want, as well as EGA I, chapitre 0, paragraphe (1.4.3) for the reason I was stating above. By the way, showing this is quite useful in the "area" I suspect you're actually in, because it is often used for results on modules $\widetilde{M}$'s etc.

In down to earth terms : in you divide an invertible element, you are also invertible.

$\endgroup$
2
  • $\begingroup$ I'm not the OP, but I read this with interest and I never realized that equality of localizations, that is very clever. So I guess the point is the equality of radicals implies every divisor of $g$ is a divisor of $f$ and every divisor of $f$ is a divisor of $g$. So the localizations are equal by your fact. Is that right? $\endgroup$ Commented Mar 8, 2015 at 2:06
  • 1
    $\begingroup$ This is the exact reason why, indeed $\endgroup$
    – Olórin
    Commented Mar 8, 2015 at 2:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .