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Let $M,N$ be abelian groups, and let $R$ be a commutative ring with identity and let $S$ be a multiplicatively closed subset of $R$, with $1\in S$.

1) Suppose that $M,N$ are $S^{-1}R$-modules. Is it true that if I have an $S^{-1}R$-isomorphism $f:M\to N$ then it is also an $R$-isomorphism? (here actually I do not need to assume that $M$ and $N$ are $R$-modules, since I think I can give them an $R$-module structure here below).

This seems true to me since I can define the action of $R$ on $M$ and $N$ by $r\cdot m=\frac r 1m,r\cdot n=\frac r 1 n$ for every $m\in M$ and $n\in N$. Hence $f(r\cdot m)=f(\frac r 1 \cdot m)=\frac r 1 f(m)=r\cdot f(m)$. Morphism and bijectivity hold anyway.

2) The converse of (1) is false I think. However, consider the well-known fact: Let $M$ be an $R$-module. Then $S^{-1}M$ is isomorphic to $M\otimes_R S^{-1}R$. But are they isomorphic only as $R$-modules or (more strongly) as $S^{-1}R$-modules and then you deduce also as $R$-modules by (1)?

If you look at the proof, you see that you constructed two $R$-isomorphisms $\phi:S^{-1}M\to M\otimes_RS^{-1}R$ given by $m/s\mapsto m\otimes 1/s$ and $\psi: M\otimes_R S^{-1}R\to S^{-1}M$ given by $m\otimes a/s\mapsto am/s$.

Now on $S^{-1}M$ you have the natural structure as $S^{-1}R$-module, but also on $M\otimes_R S^{-1}R$ you can define an $S^{-1}R$-module structure by defining the action $\frac {r} {s} \cdot (m\otimes_R \frac t u)=m\otimes_R \frac {rt} {su} $.

3) Is this $S^{-1}R$-module structure correct? In this way do the above maps $\phi$ and $\psi$ become $S^{-1}R$-isomorphisms too?

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  • $\begingroup$ For 1), how can you have a $S^{-1}R$-isomorphism? $M$ and $N$ are not $S^{-1}R$-modules. $\endgroup$ – Bernard Mar 24 '17 at 0:02
  • $\begingroup$ In your hypotheses, you should say then ‘Let $M$ be an $S^{-1}R$-module’, &c. $\endgroup$ – Bernard Mar 24 '17 at 0:08
  • $\begingroup$ @user26857 I edited it. Thank you. By the way, do you have a complete answer? $\endgroup$ – Richard Mar 24 '17 at 0:51
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For point 2), you have this result:

If $M,N$ are two $S^{-1}R$-modules, the canonical homomorphism $$\DeclareMathOperator{\Hom}{Hom}\Hom_{S^{-1}R}(M,N)\longrightarrow\Hom_R(M,N)$$ is an isomorphism.

(Bourbaki, Commutative Algebra, ch. II, Rings and modules of fractions, §2 n°8, prop. 19.)

Indeed, if $f$ is an $S^{-1}R$-linear, it is also $R$-linear by restriction of scalars. So the canonical homomorphism is injective (it is the inclusion morphism).

Conversely, any $R$-linear map $f\colon M\longrightarrow N$ is also $S^{-1}R$-linear since for any $s\in S$ and any $m\in M$, $f(m)=f\bigl(s\,\frac ms\bigr)=sf\bigl(\frac ms\bigr)$, whence $\;f\bigl(\frac ms\bigr)=\frac{f(m)}{s}$, and more generally $$f\Bigl(\frac as \,m\Bigr)=f\Bigl(\frac {am} s\Bigr)=\frac{f(am)}s=\frac{af(m)}s=\frac asf(m).$$

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  • $\begingroup$ How does the inverse map work? How can you turn a general $R-morphism$ into an $S^{-1}R-morphism$? $\endgroup$ – Richard Mar 24 '17 at 0:32
  • $\begingroup$ Tensoring with the identity of $S^{-1}R$.It works because $M$ and $N$ are not any $R$-modules, but modules for which multiplication by an element of $S$ is an isomorphism. From this property, you deduce that any $R$-homomorphism is ipso facto an $S^{-1}R$-homomorphism. $\endgroup$ – Bernard Mar 24 '17 at 0:38
  • $\begingroup$ ok but this is too general for me. I can't see it in my problem. I would like to know just if my comments are correct or not. $\endgroup$ – Richard Mar 24 '17 at 0:41
  • $\begingroup$ They're correct, since, as I said, $R$-linear for $S^{-1}R$-modules implies $S^{-1}R$-linear. $\endgroup$ – Bernard Mar 24 '17 at 0:44
  • $\begingroup$ But in my construction I gave an $S^{-1}R$ structure by hand for which I hope $\psi$ and $\phi$ became $S^{-1}R-isomorphisms$ and I can't see that what I have done is the thing that you are claiming, namely tensoring with the identity $\endgroup$ – Richard Mar 24 '17 at 0:47

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