4
$\begingroup$

I'm trying to solve this item of the question 23:

23. Let $A$ be a ring, let $X = \operatorname{Spec}(A)$ and let $U$ be a basic open set in $X$ (i.e., $U = X_f$ for some $f \in A$: Chapter 1, Exercise 17).
i) If $U = X_f$, show that the ring $A(U) = A_f$ depends only on $U$ and not on $f$.

In another words, we have to prove the sheaf $A$ is well defined. Well, if $U=X_f=X_g$ we should prove that $A_f\cong A_g$. I'm using the obvious mapping $F:A_f\to A_g$, $F(a/f^n)=a/g^n$, but I couldn't prove this mapping is well-defined and an homomorphism with respect to addition neither (the multiplicative part is trivial).

I need help.

Thanks a lot.

$\endgroup$
1
  • $\begingroup$ Use the universal property of localizations! Alternatively, the search function (the same question has been asked at least two times on math.SE). $\endgroup$ Commented Oct 19, 2013 at 0:46

1 Answer 1

4
$\begingroup$

$X_f = X_g$ is equivalent to $r((f)) = r((g))$ (ch.1, exc.17iv, which is where $X_f$ is defined in Atiyah-Macdonald). Therefore $f^n = h g$ and $g^m = k f$ for some $n, m \in {\mathbb N}$ and some $h, k \in A$.

Now, for simplicity, assume for the moment that $f$ and $g$ are not zerodivisors. In that case, $A_f$ and $A_g$ can be considered subrings of the total ring of fractions.

Now $A_f$ contains $1/g = h f^{-n}$ and $A_g$ contains $1/f = k g^{-m}$, so $A_f = A_g$. (Note: they're equal if you consider them as subrings of the total ring of fractions, not just isomorphic).

In general, the reasoning is the same, you just need a more careful formulation.

Write $i \colon A \to A_f$ and $j \colon A \to A_g$ for the localisation maps. Because $j(f)$ is invertible (with inverse $j(k) j(g)^{-m}$), by the universal property of localisation, there is a (unique) homomorphism $\phi \colon A_f \to A_g$ such that $j = \phi \circ i$. Similarly, there is a (unique) homomorphism $\psi \colon A_g \to A_f$ with $i = \phi \circ j$.

enter image description here

Now both $1_{A_f}$ and $\psi \circ \phi$ make the outer diagram above commute (i.e., $i = (\psi \circ \phi) \circ i$ and $i = 1_{A_f} \circ i$). By the universal property of localisation, there is a unique such map; therefore $\psi \circ \phi = 1_{A_f}$. Likewise $\phi \circ \psi = 1_{A_g}$.

So $\phi$ and $\psi$ are each others inverse, showing that $A_f \cong A_g$.

$\endgroup$
5
  • $\begingroup$ Doesn't my map work? $\endgroup$
    – user42912
    Commented Oct 18, 2013 at 22:07
  • $\begingroup$ @user42912 No, that's in general not a homomorphism. Just look at the case where $f = g^2$; then your map sends $1/f$ to $1/g$, but it should go to $1/g^2$ (i.e., to $1/f$ itself). $\endgroup$ Commented Oct 18, 2013 at 22:09
  • 1
    $\begingroup$ @Magdiragdag: The localizations are isomorphic via a unique isomorphism of $A$-algebras. But they are not equal. It doesn't really matter if they are equal or not, because this is only a matter of set theory. At least, with the usual construction of the set $A_S$ as a quotient set of $A \times S$, the localizations are not equal. Perhaps they are equal via some other construction ... $\endgroup$ Commented Oct 19, 2013 at 11:30
  • $\begingroup$ Thinking of a localization as a quotient is always such a bother - imagine you'd do that for ${\mathbb Q}$ all the time. I am nearly always thinking of these localizations as subrings of the total ring of fractions - but maybe my argumentation misses some borderline cases. $\endgroup$ Commented Oct 19, 2013 at 12:21
  • 1
    $\begingroup$ The ring $A_f$ is not a subring of the total ring of fractions $Tot(A)$ of $A$ if $f$ is a zero divisor. Actually, there does not even exist a natural morphism $A_f\to Tot(A)$ . $\endgroup$ Commented May 7, 2014 at 11:12

You must log in to answer this question.

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