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I'm working on an assignment problem and I'm stuck.

Statement of Problem:

Let $R$ be a Noetherian ring, and $S\subset R$ a multiplicatively closed subset. Show that if $M$ and $N$ are $R$-modules, with one of them finitely generated, then show that

$S^{-1}Hom_{R}(M,N)$ is isomorphic to $Hom_{S^{-1}R}(S^{-1}M, S^{-1}N)$

The "natural isomorphism" that I tried to define (that is, take an element $\frac{f}{s}$ in $S^{-1}Hom_{R}(M,N)$ and map it to the function $\phi$ defined by $\phi(\frac{m}{s}) = \frac{f(m)}{s}$) is not injective ($\phi$ "forgets" about the $s$ i started with in the denominator).

I'm actually really stumped on how to show this. For starters, the hypothesis of one of $M$ and $N$ being finitely generated throws me off as I'm not sure how that will be involved. Additionally, I'm not even sure if I should be trying to show it directly even. On a previous assignment, I proved an isomorphism existed by constructing the mapping and showing it was an isomorphism. After two pages, the solution was two lines using an exact sequence. :P

Any advice?

2 Answers 2

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You need to define the map $\phi(m/s') = f(m)/ss'$ for it to be well defined (This observation was due to Hurkyl).

Okay here is one part:

Suppose $M$ is finitely generated and let $m_1,...,m_n$ be a finite set of generators of $M$. I will work with the map you defined.

Let $f/s \in S^{-1}Hom_R(M,N)$ such that $\phi(m/s')=f(m)/ss' = 0$ for all $m/s' \in S^{-1}M$. In particular $\phi(m_i/1) = 0 = f(m_i)/s$, for all $i \in \{1,...,n\}$. Thus, for all $i, \exists s_i \in S$ such that $s_if(m_i) = 0$ in $S^{-1}N$. Take $s''=s_1...s_n$. Then $s''f = 0$ in $Hom_R(M,N)$. Thus, $f/s = 0$ in $S^{-1}Hom_R(M,N)$. Thus your map is injective in this case.

Hope this gives you a hint as to how to proceed.

Added Later

Here is injectivity when $N$ is Noetherian. Let $N' = im(f)$. Then $N'$ is finitely generated with finite generators say $n_1,...,n_m$. Now, for all $m/s' \in S^{-1}M, \phi(m/s') = f(m)/ss' = 0$ means that for all $i$, $n_i/s = 0$. Hence, repeating the argument above, one can find an $s'' \in S$ such that $s''im(f) = 0$. But, this means that $s''f = 0$ in $Hom_R(M,N)$. So, $f/s$ is zero in $S^{-1}Hom_R(M,N)$, which proves injectivity again.

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    Ahhh! Nice! We actually used that $s''$ trick before on a previous assignment. Thank you very much I'll try the other case now.2012-02-28
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    Actually you also need to show surjectivity in this case. I have not thought about surjectivity, but I think here you have to somehow use the Noetherian hypothesis on R. Just remember that localization of a Noetherian ring is Noetherian. Also, any finitely generated module over a Noetherian ring is a Noetherian module.2012-02-28
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    I hate to push this further since you've helped me so much already. But for the other possibility, when $N$ is finitely generated, then I still have, for each $m\in M$, an $s\in S$ such that $sf(m) = 0$. From this, I need to get a fixed $s'\in S$ such that $[s'f](m) = 0$ for all $m\in M$. I can write $f(m) = r_1n_1 + ... r_ln_l$ for some $r_{i}\in R$ (if I take $n_{i}$'s to be the generators of $N$). But now the $r_{i}$'s but this doesn't help anymore since I can't use either the $r_i$'s or the $n_i$'s to make my $s'$.2012-02-28
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    Update: Just had an idea. I think if I assume for a contradiction that for every $s\in S$ there is an $m$ in $M$ for which $sf(m)\neq 0$, I can construct an ascending chain of submodules of $M$ with no maximal element! I haven't worked out all the details yet, so I better just calm down for now....2012-02-28
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    Hmm.. I forgot that only $N$ is Noetherian in this case....2012-02-28
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    Okay, if $N$ is Noetherian, then work with the image of $F: M \rightarrow N$. The image is finitely generated as an R-module (because it is a submodule of a Noetherian module). Again, I have not completely thought it out, but I think it should help.2012-02-28
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    I added to the proof. I think the argument works again.2012-02-28
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    I run into difficulties constructing any sort of ascending chain because for all I know, the image of $F$ is already all of $N$. Where can I go from there? I also tried considering the family of $\{Image(f/s) : s\in S\}$, and chose a maximal element $Image(f/t)$. I thought this $t$ may be the "s" I needed but it doesn't seem to help me.2012-02-28
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/2632/discussion-between-borninthe80s-and-rankeya)2012-02-28
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Your putative natural isomorphism isn't well-defined. If I let $\theta$ denote your homomorphism, then observe these two examples:

$$ \theta(f/s)(m/s) = f(m)/s $$ $$ \theta(g/1)(n/1) = g(n)/1 $$

Now, if you let $f = sg$ and $m = sn$, the first equation simplifies to

$$ \theta(g/1)(n/1) = f(n) = s g(n)$$

What you're missing, I think, is the idea to have division work inversely to multiplication. $\theta(sf/1) = s \theta(f/1)$, so conversely, you want $\theta(f/s) = (1/s) \theta(f/1)$

It would be most natural to let your natural isomorphism have the property that

$$ \theta(f/1)(m/1) = f(m)/1$$

that is, $\theta$ just acts as the identity operator whenever it makes sense to think of it that way. Then, you just fill everything else in by multiplicativity:

$$ \theta(f/s)(m/s') = f(m)/(ss')$$

Hopefully you can take it from here. (Don't forget to verify that $\theta(f/s)$ is a well-defined homomorphism, and also that $\theta$ is a well-defined homomorphism!)

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    Thanks for observing this. I completely overlooked this.2012-02-28