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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.

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What does $\langle a_i \rangle = (1)$ mean ? There is a sequence of elements $a_1, a_2, \ldots$ that generate the unit ideal? I don't see how that's relevant to the rest of the question, which seems to only use one $a_i$ at a time. – Ted Jan 23 '13 at 4:18
No, I state it only for completeness' sake. – Bombyx mori Jan 23 '13 at 4:31
It feels like, in order to explicitly construct the isomorphism as you're proposing, you basically need to repeat the proof that localization commutes with exact sequences (which is what the concise argument uses). By the way, what is the "extra condition" that you refer to at the end ? – Ted Jan 23 '13 at 4:38
@Ted: Yes that's what I think. The extra condition is $I$ is a finitely presented $R$ module. – Bombyx mori Jan 23 '13 at 22:02
Why do you think you can remove that condition? Even the concise proof uses that condition. If you work through the proof, eventually you need the condition $(S^{-1}R)^n = S^{-1}(R^n)$ (here $S$ is a multiplicative subset of $R$), which is only true in general if $n$ is finite (if there are infinitely many components, you can't put everything "under a common denominator".) – Ted Jan 24 '13 at 4:42
up vote 1 down vote accepted

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($R^n$, $R$).

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Thank you! I did not expect such detailed help, so really appreciate for what you wrote. – Bombyx mori Jan 24 '13 at 6:23
I read this carefully and really feel I learned something. Thank you. – Bombyx mori Jan 24 '13 at 6:41
You're welcome! Glad to hear it was helpful. – Ted Jan 24 '13 at 6:48
I think one thing I learned is writing out such details can be helpful if one cannot do it in 30 seconds in mind. I thought I finished the "concise proof" somehow yesterday in my brain, but when I read your proof I realized I did not. – Bombyx mori Jan 24 '13 at 7:03
Minor correction - going from left to right in the isomorphism does not use the finiteness of $n$ (it's $S^{-1}(R^n)$, not $(S^{-1} R)^n$). – Ted Jan 24 '13 at 7:14

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