9
$\begingroup$

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.

  • 0
    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

4

First observe that if some $s$ kills $r$, i.e. $sr=0$, then $s^2$ kills $Dr$. Indeed, $s^2Dr=s(sDr)=s(D(sr)-rDs)=-srDs=0.$

Now lets assume $r/w=r'/w'$, and that $s\in W$ kills $rw'-r'w$. I claim that $s^2$ kills $w'^2(rDw-wDr)-w^2(r'Dw'-w'Dr').$ We have some calculation:$w'^2(rDw-wDr)-w^2(r'Dw'-w'Dr')$ $=w'^2rDw-w'^2wDr-w^2r'Dw'+w^2w'Dr'$ $=w'rD(ww')-ww'D(rw')-wr'D(ww')+ww'D(wr')$ $=(rw'-r'w)D(ww')-ww'D(rw'-r'w).$ Since $s^2$ kills both $rw'-r'w$ and $D(rw'-r'w)$, we are done.

1

\rm\displaystyle\ \frac{r}w \equiv 0\:\ \Rightarrow\:\ s\:r = 0\ \Rightarrow\ s^2 \bigg(\!\!\!\frac{r}w\!\!\bigg)' =\ \frac{s\ ((s\:r)'-s'r)}{w}\: -\:\frac{s^2\:r\:w'}{w^2} =\: 0\:\ \Rightarrow\:\ \bigg(\!\!\!\frac{r}w\!\!\bigg)'\!\equiv\: 0

Note $\ $ This proof essentially inlines the lemma that \rm\:(sr/(sw))' = (r/w)'. You may find it more conceptual to proceed indirectly using that lemma (which is how I found the above direct proof).

  • 0
    As far as I can tell, this is not a complete proof. The goal is not to prove that any representative of $0$ gets sent to $0$, but to prove that two representatives of the same fraction get sent to the same value. I do not see how the former implies the latter, as we are lacking linearity of whatever form.2015-10-22
1

[New version. I hadn't noticed the fact that the derivation was module-valued. Sorry. (July 31, 2011, GMT)]

Let $D:R\to M$ be a derivation and $W\subset R$ a multiplicative system. Claim: there is a unique derivation $W^{-1}D:W^{-1}R\to W^{-1}M$ satisfying $(W^{-1}D)\left(\frac{a}{s}\right)=\frac{sD(a)-aD(s)}{s^2}$ for all $a$ in $R$ and $s$ in $W$.

The uniqueness is clear. Let's prove the existence.

Recall that the relation $\sim$ defined on $R\times W$ by $(a,s)\sim(b,t)$ iff $atu=bsu$ for some $u$ in $W$ is an equivalence relation, and that $W^{-1}R$ is defined as the quotient.

Define the relation $\heartsuit$ on $R\times W$ by $(a,s)\heartsuit(b,t)$ iff $b=au$ and $t=su$ for some $u$ in $W$. Then $\sim$ is the equivalence relation generated by $\heartsuit$.

Define $d:R\times W\to W^{-1}M$ by $d(a,s)=\frac{sD(a)-aD(s)}{s^2}\quad.$ One checks that $(a,s)\heartsuit(b,t)$ implies $(W^{-1}D)(a,s)=(W^{-1}D)(b,t)$. Thus $d$ induces a map $W^{-1}D:W^{-1}R\to W^{-1}M$, which is easily seen to be a derivation.

  • 0
    I see. Thanks a lot.2018-04-28