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M an A-module, $S\subset A$ a multiplicative subset. Is it possible for $S^{-1}M$ to have an $S^{-1}A$-module structure satisfying $\frac{a}{1}\cdot\frac{m}{1}=\frac{am}{1}$ other than the "usuall" one given by $\frac{a}{s}\cdot\frac{m}{t}=\frac{am}{st}$ ?

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My first inclination on seeing this is to ask, why would you want such a thing? –  Harry Altman Jun 29 '10 at 0:51
    
@Harry Altman It's a long story. –  ashpool Jun 29 '10 at 0:54
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Kwan, are you looking for an example or do you really need to keep all the data (M, A, S) completely general? You're basically asking if you can take a module over a ring and give the underlying additive group of the module another module structure over that ring which is not isomorphic to the first module structure. There is a standard instance of this. Let F be a field and V be a finite-dim. F-vector space. If dim(V) > 1, we can give V lots of nonisomorphic F[x]-module structures: this is the theme of Jordan canonical form (or rational canonical form if F isn't algebraically closed). –  KConrad Jun 29 '10 at 1:08
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Maybe you want to clarify what it is you really want to see. For example, in the "exotic" S^{-1}A-module structure do you want to insist that the A-scaling is the same as it was originally? I suspect you meant to ask for that (which would nullify my previous example), so please write out your question a little more carefully. –  KConrad Jun 29 '10 at 1:11
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Localization solves a very specific universal mapping problem, so if you want to respect that feature of localization then the answer to your question will be no: all possible means of creating a module structure will lead to isomorphic modules. It's like asking if the integers can be given two essentially different fraction fields (it can't). –  KConrad Jun 29 '10 at 1:13
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