3
$\begingroup$

In the paper, Topologically Defined Classes of Commutative Rings, localization of the pullback diagram (with $v,$ surjective) $$ \begin{array} DD & \stackrel{v\ '}{\longrightarrow} & A \\ \downarrow{u' } & & \downarrow{u} \\ B & \stackrel{v}{\longrightarrow} & C \end{array} $$ is given as Proposition 1.9:
... Conversely, if $S_A$ is a multiplicatively closed set of $A$ and if $S_B$ is a multiplicatively closed of $B$ and if $u(S_A)=v(S_B)=S_C$ then $S_A^{-1}A \times_{S_C^{-1}C}S_B^{-1}B \cong (S_A \times_{S_C}S_B)^{-1} D.$

I can not see how a typical element of one side mapped to the other side by this isomorphism.
Can you help please?
Thank you.

$\endgroup$
0

1 Answer 1

1
+50
$\begingroup$

First, you expect that the pullback (as a special kind of limit) should be the target of your map. Thus you expect a map: $$(S_A \times_{S_C}S_B)^{-1} D\longrightarrow S_A^{-1}A \times_{S_C^{-1}C}S_B^{-1}B$$ Denote $S=S_A \times_{S_C}S_B$. In this setting you have $S=v'^{-1}(S_A)\cap u'^{-1}(S_B)$. So that it is naturally a multiplicative subset of $D$. In your context the localization $S^{-1}D$ is the set of pairs or formal quotients $d/s$ modulo identifications. Then $S_A^{-1}A \times_{S_C^{-1}C}S_B^{-1}B\subseteq S_A^{-1}A \times S_B^{-1}B$ is a subset so that you define: $$ d/s\mapsto (v'(d)/v'(s),u'(d)/u'(s))$$ and you verify, that this agrees with the appropriate identifications and that both elements of the pair on the right are mapped to the same element of $S_C^{-1}C$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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