How to localize $J$ at $a_{i}$?

Question

Assume $I$ is a finitely presented $R$ module such that for $\langle a_{i}\rangle=(1),a_{i}\in R$, we have $I[\frac{1}{a_{i}}]=R[\frac{1}{a_{i}}]$. Define $J=Hom_{R}(I,R)$, what is a good way to show $$\displaystyle J[\frac{1}{a_{i}}]=Hom_{R[\frac{1}{a_{i}}]}(I[\frac{1}{a_{i}}],R[\frac{1}{a_{i}}])$$

I guess probably I do not even need to use the conditions given and this in fact holds for more general cases. One side is easy - any homomorphism from $I$ to $R$ automatically give a trivial homomorphism over the localized rings $I[\frac{1}{a_{i}}]$ and $R[\frac{1}{a_{i}}])$. However given a $R[\frac{1}{a_{i}}]$ homomorphism between the localized rings, how do I show I can some how "chop off" the extra localized part to get a $R$-homomorphism back.

Jacob Lurie commented that:

We can choose a finite presentation $R^{m}\rightarrow R^{n}\rightarrow I\rightarrow 0$, which leads to a sequence $$0\rightarrow J\rightarrow Hom(R^{n},R)\rightarrow Hom(R^{m},R)$$ It follows that the formation of $J$ commutes with localization.

This proof avoids the above issue and looks much better. So I just want to ask if the same contention holds when the extra condition is removed.

Answer 1

The concise proof of the result does not work if $I$ is not a finitely presented $R$ module. We start with the exact sequence of $R$-modules $$R^m \to R^n \to I \to 0.$$ Apply the functor Hom($-$,$R$) followed by the functor $S^{-1}$. We get an exact sequence of $S^{-1}R$ modules $$0 \to S^{-1}{\rm{Hom}}(I,R) \to S^{-1}{\rm{Hom}}(R^n, R) \to S^{-1}{\rm{Hom}}(R^m,R)$$

Now let's start from the beginning again and apply the functor $S^{-1}$ first, then the functor Hom($-$,$S^{-1}R$). We get $$0 \to {\rm{Hom}}(S^{-1}I, S^{-1}R) \to {\rm{Hom}}(S^{-1}(R^n), S^{-1}R) \to {\rm{Hom}}(S^{-1}(R^m), S^{-1}R)$$

Comparing these 2 exact sequences, our goal is to show that the second terms in these sequences are equal, and to do that, it is enough to show that the third and fourth terms of each are isomorphic via an isomorphism that commutes with the exact sequences. In other words, we must prove that $$S^{-1}{\rm{Hom}}(R^n, R) \cong {\rm{Hom}}(S^{-1}(R^n),S^{-1}R)$$ via an isomorphism that commutes with the exact sequences.

Given an element $s^{-1}f$ on the left side, we define an element $g$ on the right side by $$g(r_1/t, \ldots, r_n/t) = 1/(st) \cdot f(r_1,\ldots,r_n).$$

In the other direction, given a $g$ on the right side, restrict* it to the subring $R^n$. The image of the restriction in $S^{-1}R$ is finitely generated and hence there is a common denominator $t$ that we can multiply everything in the image by to get an element of $R$. Then $t^{-1} \cdot (tg)$ is an element on the left side. Here we used the finiteness of $n$.

*If $R$ is not a domain, $R$ may not be a subring of $S^{-1}R$. Instead, apply the argument to $T^n$ where $T$ is the subring $\{r/1 : r \in R\}$ of $S^{-1}R$. Then $tg$ is a map in Hom($T^n$, $R$) which you can compose with the natural map $R^n \to T^n$ to get a map in Hom($R^n$, $R$).