Localisation commutes with taking quotients.

Question

If $A$ is a ring, $S$ a multiplicative set and $I$ an ideal, write $T$ for the image of $S$ in $A / I$. Then $T^{-1}(A/I) \cong S^{-1}A/S^{-1}I$ and in particular, for a prime ideal $P$ we have that $A_P/PA_P$ is isomorphic to Frac$(A/P)$.

My question is regarding the proof, it says: "The quotient ring $A/I$ can be viewed as an $A$-module and then the ring of fractions $T^{-1}(A/I)$ equals the module of fractions $S^{-1}(A/I)$"

Can someone explain to me why the ring of fractions equals the module of fractions? I cant see this, I think I am missing something simple.

Answer 1

You get this even on a set level as $$T^{-1}(A/I) = \{[a]_I/[s]_I;~a ∈ A,~s ∈ S\} = \{1/s·[a]_I;~a ∈ A,~s ∈ S\} = S^{-1}(A/I),$$ if you believe that the $A$-module structure on $T^{-1}(A/I)$ can be extended to a $S^{-1}A$-module structure by setting $1/s·[a]_I = [a]_I/[s]_I$.

But this follows (for example) from the universal property of localizations, viewing the module structure on $T^{-1}(A/I)$ as a ring homomorphism $$A → \mathrm{End}(T^{-1}(A/I)),~ a ↦ ([a]_I/[1]_I~·~),$$ noticing that multiplication by elements of $S$ becomes invertible.

Probably you can also convince yourself of this in some other direct way.

Answer 2

It comes from a more general situation: suppose you have a ring homomorphism f from A to a ring B. Then T=f(S) is a multiplicative set in B, and we can consider the ring of fractions $T^{-1}B$. On another hand, B is an A-module, so we also can consider the module of fractions $S^{-1}B$. What is true is that there is a unique mapping $u \,\colon\,\, T^{-1}B\longrightarrow S^{-1}B$ such that $u\Bigl(\dfrac{b}{f(s)}\Bigr)= \dfrac{b}{s}$.

Moreover $u $ is both an isomorphism for the structures of $S^{-1}A$-modules and of B-modules.

Answer 3

The pair $(S^{-1}A,\lambda_{A,S})$, where $\lambda_{A,S}\colon A\to S^{-1}A$ is defined by $a\mapsto a/1$ is universal with respect to the property that, for every ring homomorphism $g\colon A\to C$ such that $g(s)$ is invertible, there exists a unique ring homomorphism $\hat{g}\colon S^{-1}A\to C$ with $\hat{g}\circ\lambda_{A,S}=g$ and $\hat{g}(a/s)=g(a)g(s)^{-1}$.

Let us show that the pair $(S^{-1}A/S^{-1}I,\mu)$ has the universal property pertaining to $(T^{-1}(A/I),\lambda_{A/I,T})$, where $T=\pi(S)$ (being $\pi\colon A\to A/I$ the canonical map) and $\mu\colon A/I\to S^{-1}A/S^{-1}I$ is defined by $$ \mu(a+I)=\frac{a}{1}+S^{-1}I $$ Note that $\mu$ is well defined, because if $a\in I$, then $a/1\in S^{-1}I$ by definition. It's obviously a ring homomorphism.

Now, suppose $g\colon A/I\to C$ is a ring homomorphism such that $g(s+I)$ is invertible for every $s\in S$. Set $h=g\circ\pi$. Then, by the universal property of $S^{-1}A$, there is a unique ring homomorphism $\hat{h}\colon S^{-1}A\to C$ such that $$ \hat{h}\circ\lambda_{A,S}=h $$ Suppose $a\in I$ and $s\in S$; then $$ \hat{h}\left(\frac{a}{s}\right)=h(a)h(s)^{-1}=h(s)^{-1}g(\pi(a))=0 $$ so that $\hat{h}$ factors through $S^{-1}I$ and it defines a ring homomorphism $\hat{g}\colon S^{-1}A/S^{-1}I\to C$, as requested. The uniqueness of $\hat{g}$ is obvious.