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I have spent several hours on this, apparently straightforward issue. This is with reference to page 17 in the following notes

http://www.math.lsa.umich.edu/~hochster/615W10/615.pdf

Suppose, $R$ is a commutative ring, $W$ a multiplicatively closed subset in $R$, $M$ an $R$-module. If $D:R\to M$ is a derivation, then $W^{-1}D: W^{-1}R\to W^{-1}M$ is a derivation where $W^{-1}D$ acts on $\frac{r}{w}$ by the quotient rule, i.e. maps $\frac{r}{w}$ to $\frac{wD(r)-rD(w)}{w^2}$.

I have tried several manipulations, but I am unable to show that this map is well defined. I would appreciate if anyone can help me see what I am missing here.

Thanks.

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    It seems to me that it suffices to check $(W^{-1}D)\frac{r}{w}=(W^{-1}D)\frac{xr}{xw}$ for $w,x$ in $W$, $r$ in $R$, and that this is straightforward.2011-07-29
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    More precisely, let $W^{-1}D$ act on $W\times R$ by the quotient rule, and check $(W^{-1}D)(xr,xw)\sim(W^{-1}D)(r,w)$.2011-07-29
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    @Pierre Yes, that's essentially what's behind my answer - which I had hope to elaborate but hadn't had the time.2011-07-29
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    Same question as http://math.stackexchange.com/questions/1052153/extension-of-r-linear-derivation-to-localization , where Hanno gave a beautiful answer.2015-11-08

3 Answers 3