Let $R$ be a commutative ring with unity.
I have read that if $M$ is an injective $R$-module, then $S^{-1}M$ is not necessarily an injective $S^{-1}R$-module. I need an example...
Does last statement true for $P$-injectivity and projectivity?
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What is P-injectivity? – Bernard Aug 08 '15 at 22:18
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$P$-injectivity means if $f: aR\rightarrow M$, then there exist an extension $g:R\rightarrow M$ such that $g=f$ on $aR$, where $aR$ is principal ideal generated by an element $a$ . – Leonardo Aug 09 '15 at 09:04
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For the first part, essentially answered in the comments on Localization of an injective module is also injective, true in the Noetherian case but with references to counterexamples that are non-Noetherian. – hardmath Aug 09 '15 at 14:29
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The localisation of a projective module is projective since a projective module is a direct summand of a free module, and localisation preserves direct summands and freeness.
A counter-example for injective modules was built by E. Dade in 1981, and you can find it in his paper Localization of injective modules, in which he also gives a sufficient condition for this property to be true.
This condition consists of the following $3$ subconditions on the ring $R$ and the multiplicative set $S$:
- $R$ is a coherent ring.
- For any finitely generated ideal $I$ of $R$ and $s\in S$, the chain: $$I\subset (I:s)\subseteq (I:s^2)\subseteq (I:s^3)\subseteq\cdots$$ stabilises.
- Ideals of $R$ are countably generated.
In particular, it is true for noetherian rings.
Bernard
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user26857 counterexample for the statement I wrote is not true always. I thought there is an easy example , but according to Dade's paper, it is not easy .. – Leonardo Aug 09 '15 at 09:07
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If it was found in 1981, while the notion of injectivity dates back to the 40s, it clearly is not a trivial question! For p-injectivity, I know nothing about it. If the ring is a domain, it seems to be the same notion as divisibility? – Bernard Aug 09 '15 at 09:29